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§ Features
This book is strictly according to CBSE 2024
board exam syllabus.
Due to new syllabus, only those previous year
questions has been included in this Question
Bank that are according to 2024 board exam
syllabus so that students don’t get confused
by old question papers.
Your board exam will be in March 2024 and
our book will be available from October 1st
week or 2nd week within the price range of
Rs. 58 to 61. Till then if you can wait then wait
else you can buy other publishers book but
other publishers question bank will not
contain CBSE 2023 questions [This is
challenge].
Why 2023 questions is important for you
CBSE has reduced the syllabus in comparison
to previous years and increased their level of
questions so that you also score well in CUET
UG which is the new entrance exam that has
been started in 2022 for admission in any top
colleges or top universities.
After syllabus reduction and change in
question pattern, 2023 previous years
questions is the only source that represents
the new question pattern.
As 2022 board examination was conducted in
two phases:
Term I (Half Syllabus) (MCQ based)
Term II (Half Syllabus) (Subjective)
So 2022 had different question pattern.
That’s the reason why you need to practice
CBSE 2023 questions.
Keep in mind that from 2023 CBSE is
preparing the question paper in such pattern
so that students can also score good in CUET
UG Entrance exams.
We have included all the previous
year’s questions from all the question
papers till 2023 to 2019 including
compartment and 2024 official sample
question paper that are strictly
according to CBSE 2024 syllabus in one
book so that students don’t get
exhausted and confused with so many
question papers.
In 2021 main board examination was
not conducted due to covid and
students were passed on the basis of
their academic scores. But 2021
compart examination was conducted
for those students who failed or were
absent due to COVID cases. We have
even included those questions too in
our book.
*
Unit I
Electrostatics
Published by :
www.cbse.page
www.cuet.pw
’ Chapter 1
™ Electric Charges and
Fields
Weightage for CBSE Board 2024
Unit I and II
Combined
Weightage is
16 Marks
À Chapter 1
À Chapter 2
À Chapter 3
It’s not fixed which chapters will
have more weightage, but overall weightage
will be 16 Marks. It may vary too in your
question paper.
` Syllabus
Electric charges, Conservation of charge,
Coulomb’s law-force between two- point charges,
forces between multiple charges; superposition
principle and continuous charge distribution.
Electric field, electric field due to a point charge,
electric field lines, electric dipole, electric field
due to a dipole, torque on a dipole in uniform
electric field.
Electric flux, statement of Gauss’s theorem and
its applications to find field due to infinitely long
straight wire, uniformly charged infinite plane
sheet and uniformly charged thin spherical shell
(field inside and outside).
§ 2024 SQP
š Question 1
An electric dipole placed in an electric field of
5
intensity 2 × 10 N/C at an angle of 30°
experiences a torque equal to 4Nm. The charge
on the dipole of dipole length 2 cm is
(a) 7 𝜇 C
(b) 8 mC
(c) 2 mC
(d) 5 mC
A Solution
𝜏
𝑞=
2𝑎 𝐸 sin 𝜃
4
=
2 × 10−2 × 2 × 105 sin 30◦
−3
= 2 × 10 C = 2 mC
¥ 1 Marks
[𝜏 = 𝑞𝐸 × 2𝑎 sin 𝜃]
This type of same question was asked in 2023
question paper 55/5/1
§ 2023 55(B) Compart
š Question 2
The electric flux through a Gaussian spherical
surface enclosing a point charge 𝑞 is 𝜙. If the
charge is replaced by an electric dipole,
magnitude of its dipole moment being 2𝑞𝑎, the
flux through the surface will be :
(a) 2𝜙
(b) 𝜙
(c)
A Answer
𝜙
2
(d) Zero
¥ Set 1, 1 Marks
Zero
š Question 3
9
10 electrons are transferred to a pith ball with
charge 0 · 16nC. Its charge now is :
−10
(a) Zero
(b) −3 · 2 × 10 C
−9
−10
(c) −1.6 × 10 C
(d) 3 · 2 × 10 C
A Answer
Zero
¥ Set 1, 1 Marks
š Question 4
Assertion (A) : The electrostatic field E is a
conservative field.
Reason (R) : Line integral of electric field E
around a closed path is non zero.
(a) Both Assertion (A) and Reason (R) are true
and Reason (R) is the correct explanation of
the Assertion (A).
(b) Both Assertion (A) and Reason (R) are true,
but Reason (R) is not the correct explanation
of the Assertion (A).
(c) Assertion (A) is true, but Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is also
false.
A Answer
¥ Set 1, 1 Marks
Assertion (A) is true, but Reason (R) is false.
š Question 5 [OR Section]
(i) Why do we use a small test charge to measure
an electric field ?
(ii) A small stationary positively charged particle
is free to move in an electric field. In which
direction will it begin to move ?
(iii) Two point charges 𝑄1 ( 40 𝜇𝐶) and 𝑄2 (−16 𝜇𝐶)
are placed along 𝑥−axis at 0 cm and 24 cm
from the origin respectively. Calculate the net
force on a third charge 𝑄3 (−2.5 𝜇𝐶) placed at
𝑥 = 36cm from the origin.
A Answer
¥ 5 Marks
(i) So that it does not disturb the distribution of
charges whose electric field is to be
measured. Otherwise field will be different
from the actual field.
(ii) It will move in the direction of electric field.
𝑘𝑄2𝑄3 𝑘𝑄1𝑄3
(iii) F = F2 − F1 =
−
2
2
𝑟23
𝑟13
9
𝐹=
9 × 10 × 16 × 2.5 × 10
2
−
2
10 )
−12
−
( 12 ×
9
−12
9 × 10 × 40 × 2.5 × 10
( 36 ×
9
2
−
2
10 )
−12
𝐹 = ( 9 × 10 × 10
𝐹 = 25 − 6.94
𝐹 = 18.06 N
)
16 × 2.5
( 12 ×
−
2
10 )
−
2
40 × 2.5
( 36 ×
2
−
2
10 )
§ 2023 55/C Compart
š Question 6
When a negative charge (−𝑄) is brought near
one face of a metal cube, the
(a) cube becomes positively charged
(b) cube becomes negatively charged
(c) face near the charge becomes positively
charged and the opposite face becomes
negatively charged
(d) face near the charge becomes negatively
charged and the opposite face becomes
positively charged
A Answer
¥ Set 2, 1 Marks
Face near the charge becomes positively charged
and the opposite face becomes negatively
charged
š Question 7
An electric dipole with dipole moment
𝑃® = 𝑃0𝑖ˆ − 𝑃0 𝑗ˆ is placed in an electric field
𝐸® = 𝐸1𝑖ˆ + 𝐸2 𝑗ˆ, where 𝑃0, 𝐸1 and 𝐸2 are constants.
The torque 𝜏® acting on the dipole is :
(a) 𝑃0 (𝐸2 − 𝐸1) 𝑘ˆ
(b) 𝑃0 (𝐸2 + 𝐸1) 𝑘ˆ
(c) −𝑃0 (𝐸2 + 𝐸1) 𝑘ˆ
(d) 𝑃0 (𝐸1 − 𝐸2) 𝑘ˆ
A Answer
¥ Set 3, 1 Marks
𝑃0 (𝐸2 + 𝐸1) 𝑘ˆ is the torque acting on the dipole
š Question 8
The inward and the outward electric flux
through a Gaussian surface are 2𝜙 and 𝜙
respectively.
(a) What is the net charge enclosed by the
surface ?
(b) If the net outward flux through the surface
were zero, can it be concluded that there
were no charges inside the surface ? Justify
your answer.
A Answer
¥ 2 Marks
(a) We know that, electric flux or electric field
lines entering in a closed surface ( 2𝜙) is −ve
and electric flux or electric field lines leaving
a closed surface (𝜙) is +ve.
Hence, net electric flux enclosed by the
surface, Φ = −2𝜙 + 𝜙
Now, according to Gauss theorem,
𝑞
Φ=
Therefore, 𝑞 = Φ𝜀0
𝜀0
𝑞 = (−2𝜙 + 𝜙) 𝜀0 = −𝜙 𝜀0
(b) No. Net flux piercing out through a body
depends on the net charge contained in the
body. If net flux is zero, then it can be said
that net charge inside the body is zero. The
body may have equal amount of positive and
negative charges.
š Question 9
A sphere 𝑆1 of radius 𝑟1 enclosing a charge 𝑄1 is
surrounded by another concentric sphere 𝑆2 of
radius 𝑟2 (𝑟2 > 𝑟1) . If there is a charge −𝑄2 in the
space between 𝑆1 and 𝑆2, find the ratio of electric
flux through 𝑆1 and 𝑆2.
A Answer
¥ Set 2, 2 Marks
𝑄
𝑄1 − 𝑄2
Φ1 =
Φ2 =
𝜀𝑜
𝜀𝑜
Φ1
𝑄1
=
Φ 2 𝑄1 − 𝑄2
š Question 10
A uniform
electric
field
is
represented
as
𝑁
3
®
𝐸 = 3 × 10
𝑖ˆ. Find the electric flux of this
𝐶
field through a square of side 10 cm when the :
(a) plane of the square is parallel to 𝑦 − 𝑧 plane,
and
(b) the normal to plane of the square makes an
◦
angle of 60 with the 𝑥−axis.
A Answer
¥ Set 3, 2 Marks
(a) Φ = 𝐸® · 𝐴𝑖ˆ = 𝐸𝐴
−4
−1 2
3
= 3 × 10 × 100 × 10 = 30 NC m
(b) Φ = 𝐸𝐴 cos 60
−1 2
= 15NC m
◦
§ 2023 55/1/1 All Sets
š Question 11
A point charge situated at a distance 𝑟 from a
short electric dipole on its axis, experiences a
force 𝐹®. If the distance of the charge is 2𝑟 , the
force on the charge will be
(a)
®
F
16
A Answer
(b)
®
F
8
(c)
®
F
(d)
4
®
F
2
¥ Set 1, 1 Marks
Let charge q is placed at the distance r from the
dipole. Electric field due to the dipole at an axial
point is
1 2p
E=
3
4𝜋𝜀0 r
Force on a charge q placed at distance 𝑟 is
1 2pq
F = qE =
3
4𝜋𝜀0 r
If we doubled the distance
1
2pq
′
F =
3
4𝜋𝜀0 ( 2r)
′
F =
1 2pq 1
4𝜋𝜀0 r3
8
∴
" #
®
F
8
š Question 12
(i) State Coulomb’s law in electrostatics and write
it in vector form, for two charges.
(ii) Gauss’s law is based on the inverse-square
dependence on distance contained in the
Coulomb’s law. Explain.
(iii) Two charges 𝐴 (charge 𝑞) and 𝐵 (charge 2𝑞)
are located at points ( 0, 0) and (a, a)
respectively. Let 𝑖ˆ and 𝑗ˆ be the unit vectors
along 𝑥−axis and 𝑦−axis respectively. Find the
force exerted by 𝐴 on 𝐵, in terms of 𝑖ˆ and 𝑗ˆ.
A Answer
¥ 5 Marks
(i) Force between two point charges varies
inversely with the square of distance between
the charges and is directly proportional to the
product of magnitude of the two charges and
acts along the line joining the two charges.
−→
𝐹12 =
1
𝑞1𝑞2 −
→
𝑟
12
3
4𝜋 ∈𝑜 𝑟12
−
→
Where 𝑟12 is a vector from charge 𝑞2 to charge
𝑞1 .
(ii) In derivation of Gauss’s law, flux is calculated
using Coulomb’s law and surface area. Here
1
coulomb’s law involves 2 factor and surface
2
𝑟
area involves 𝑟 factor. When product is
taken, the two factors cancel out and flux
becomes independent of 𝑟 .
−−→
(iii) 𝑟® = 𝐴𝐵 = 𝑎b
𝑖 + 𝑎b
𝑗
√
−−→ √ 2
2
r = | 𝐴𝐵| = 𝑎 + 𝑎 = 2𝑎
1 𝑞1𝑞2
ˆ
𝐹® =
𝑟
2
4𝜋 ∈ 𝑜 𝑟
1
b
𝑞
×
2𝑞 (𝑎b
𝑖
+
𝑎
𝑗)
𝐹® =
× √
× √
4𝜋𝜀𝑜 ( 2𝑎) 2
2𝑎
2
b
b
1
2𝑞
(
𝑖
+
𝑗)
𝐹® =
× 2× √
4𝜋𝜀𝑜 2𝑎
2
𝐹® =
2
1
4𝜋𝜀𝑜
𝐹® = √
𝑞
𝑞
×√
2𝑎
2
4 2𝜋𝜀𝑜 𝑎2
1
2
× (b
𝑖+b
𝑗)
(b
𝑖+b
𝑗)
2
𝑞
® =
| 𝐹|
2
4𝜋 ∈ 𝑜 𝑎
š OR
(i) Derive an expression for the electric field at a
point on the equatorial plane of an electric
dipole consisting of charges 𝑞 and −𝑞
separated by a distance 2a.
(ii) The distance of a far off point on the
equatorial plane of an electric dipole is
halved. How will the electric field be affected
for the dipole?
(iii) Two identical electric dipoles are placed
along
the
diagonals
of
a
square
ABCD
of
side
√
2 m as shown in the figure. Obtain the
magnitude and direction of the net electric
field at the centre ( O) of the square.
A Answer
¥ 5 Marks
(i) The magnitudes of the electric fields due to
the two charges +𝑞 and −𝑞 are given by
𝑞
1
E+𝑞 =
× 2
4𝜋𝜀𝑜 𝑟 + 𝑎2
1
𝑞
× 2
E−𝑞 =
2
4𝜋𝜀𝑜 𝑟 + 𝑎
The components normal to dipole axis cancel
away. The components along the dipole axis add
up.
Total electric field is opposite to dipole moment.
𝐸® = − 𝐸+𝑞 + 𝐸−𝑞 cos 𝜃 b
𝑝
=
=
−2𝑞𝑎
2
2
4𝜋 ∈0 (𝑟 + 𝑎 )
b
𝑝
3/2
− 𝑝®
2 3/2
2
4𝜋 ∈0 (𝑟 + 𝑎 )
(ii) At far off point 𝑟 ≫ 𝑎
®
−
𝑝
𝐸® =
3
4𝜋 ∈𝑜 𝑟
When distance is halved.
®
−
𝑝
−
8 𝑝®
𝐸® =
𝑟 3 =
3
4𝜋 ∈ 𝑜 𝑟
4𝜋 ∈𝑜
2
𝐸® becomes 8 times
(iii)
𝑝1 = 𝑞 × 2𝐶𝑚 (along OA)
𝑝2 = 𝑞 × 2𝐶𝑚 (along OD)
√︃
√
2
𝑝net = 𝑝1 + 𝑝22 = 2 2𝑞𝐶𝑚
Electric field at centre O
𝐸=
𝑘𝑝𝑛𝑒𝑡
2
2 3/2
(𝑟 + 𝑎 )
at point O, r = 0, a = 1 m
√
√
𝑘 × 2 2𝑞
𝐸=
=
2
2
𝑘𝑞
3
1
√
2 2𝑞
=
Along DC
4𝜋𝜀𝑜
Alternatively
𝑘𝑞
𝐸= 2
𝑟
AC = BD = 2 m
r = OA = OB = OC = OD = 1 m
Electric field at O due to charges at B and D
E1 = EB + ED =
𝑘𝑞
2
1
+
= 2𝑘𝑞 along OB
𝑘𝑞
2
1
Electric field at O due to charges at A and C
E2 = EA + EC
𝑘𝑞
𝑘𝑞
+
=
2
𝑘𝑞
along
OC
12 12
√︃
2
2
Enet = 𝐸1 + 𝐸2
√
√
2 2𝑞
= 2 2kq =
Along DC
4𝜋𝜀𝑜
E2 =
Alternatively
Considering AB as dipole, electric field at O
E1 =
2𝑘𝑞 × 𝑎
2𝑘𝑞𝑎
=
!
3
/
2
3/2
2 2
1
1
1
1
+
+ √
√
2 2
2
2
= 2𝑘𝑞𝑎
Similarly considering DC as another dipole,
electric field at O
2𝑘𝑞𝑎
2𝑘𝑞 × 𝑎
E2 =
=
!
3
/
2
3/2
2 2
1
1
1
1
+
+ √
√
2 2
2
2
= 2𝑘𝑞𝑎
Enet = E1 + E2 = 4kqa
=
1
4𝜋𝜀𝑜
√
1
×4×√ ×𝑞
= 2 2kq =
√
2
2 2𝑞
4𝜋𝜀𝑜
Along DC
§ 2023 55/2/1 All Sets
š Question 13
The magnitude of the electric field due to a point
charge object at a distance of 4.0 m is 9 N/C.
From the same charged object the electric field
of magnitude, 16 N/C will be at a distance of
(a) 1 m
(b) 2 m
(c) 3 m
(d) 6 m
A Answer
¥ Set 1, 1 Marks
As per given in your NCERT Book,
(Topic: Coulomb’s Law)
1
9
𝑘=
= 9 × 10
4𝜋𝜀𝑜
9
9 × 10 × 𝑞
𝑘𝑞
𝐸1 = 2
∴ ⇒9=
2
4
𝑟1
( 9 × 16)
−9
𝑞=
=
16
×
10
( 9 × 109)
For electric field 16 N/C
𝑘𝑞
𝐸2 = 2
𝑟2
9
16 =
−9
9 × 10 × 16 × 10
2
2
𝑟
2
2
𝑟 =9
⇒ 𝑟2 = 3 m
š Question 14
An isolated point charge particle produces an
® at a point 3 m away from it. The
electric field E
® /4)
distance of the point at which the field is ( E
will be
(a) 2 m
(b) 3 m
(c) 4 m
A Answer
(d) 6 m
¥ Set 2, 1 Mark
The electric field E produced by a point charge at
a distance 𝑟 from it is given by the equation:
𝑘𝑄
𝐸= 2
𝑟
Where 𝑘 is Coulomb’s constant and 𝑄 is the
charge on the particle.
Let the distance of the point where the electric
field is
𝐸
be 𝑥 metres away from the particle.
4
So, the electric field at this point can be
calculated as:
𝐸
𝑘𝑄
= 2
4
𝑥
Also, we know that the electric field at a point 3
metres away from the particle is E. So we can
write:
𝐸=
𝑘𝑄
2
3
Equating these two equations, we get:
𝑘𝑄
𝑘𝑄
=
2
2
𝑥
3 ×4
By solving, we get:
2
2
⇒ 𝑥 = 3 × 4 = 36
⇒ 𝑥 = 6m
š Question 15
Assertion (A) : Work done in moving a charge
around a closed path, in an electric field is
always zero.
Reason (R) : Electrostatic force is a conservative
force.
(a) Both Assertion (A) and Reason (R) are true
and Reason (R) is the correct explanation of
the Assertion (A).
(b) Both Assertion (A) and Reason (R) are true,
but Reason (R) is not the correct explanation
of the Assertion (A).
(c) Assertion (A) is true, but Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is also
false.
A Answer
¥ Set 1, 1 Marks
Both Assertion (A) and Reason (R) are true and
Reason (R) is the correct explanation of
Assertion(A).
š Question 16
(i) Use Gauss’ law to obtain an expression for the
electric field due to an infinitely long thin
straight wire with uniform linear charge
density 𝜆 .
(ii) An infinitely long positively charged straight
wire has a linear charge density 𝜆 . An
electron is revolving in a circle with a
constant speed 𝑣 such that the wire passes
through the centre, and is perpendicular to
the plane, of the circle. Find the kinetic
energy of the electron in terms of magnitudes
of its charge and linear charge density 𝜆 on
the wire.
(iii) Draw a graph of kinetic energy as a function
of linear charge density 𝜆 .
A Answer
¥ 5 Marks
(i) Flux through the Gaussian Surface is equal to
flux through the curved cylindrical part
= E × 2𝜋𝑟𝑙
The surface charge equal to 𝜆𝑙
Δ𝑄
where, 𝜆 =
= linear charge density
Δ𝑙
According to Gauss’s law
𝑞 𝜆𝑙
𝐸 × 2𝜋𝑟𝑙 = =
𝜀0 𝜀0
𝜆
𝐸=
2𝜋𝜀0𝑟
𝜆
(ii) As 𝐸 =
2𝜋𝜀0𝑟
The revolving electron experiences an
electrostatic force and provides necessarily
centripetal force.
2
𝑚𝑣
⇒
= 𝑒𝐸
𝑟
1
2
∴ Kinetic energy 𝐾 = 𝑚𝑣
2
By Substituting:
1
1
1
𝜆𝑟
𝑒𝜆
2
= 𝑚𝑣 = 𝑒𝐸𝑟 = 𝑒
=
2
2
2 2𝜋𝜀0𝑟 4𝜋𝜀0
This Question was repeated in 2019 (B), 2013
𝑒𝜆
(iii) Kinetic energy 𝐾 =
4𝜋𝜀0
∴𝐾∝𝜆
This Question was repeated in 2013
§ 2023 55/3/1 All Sets
š Question 17
−16
An electron experiences a force 1.6 × 10
→
−
→
−
N î
in an electric field E . The electric field E is
N
N
3
3
(a) 1.0 × 10
î
(b) − 1.0 × 10
î
C
C
N
−3
(c) 1 · 0 × 10
î
C
N
−3
(d) − 1.0 × 10
î
C
A Answer
𝐹® = (−𝑞) 𝐸®
𝐸® =
¥ Set 1, 1 Marks
−16
1.6 × 10
N î
−
19
10
−1 . 6 ×
C
N
3
= − 1.0 × 10
î
C
š Question 18
An infinitely long uniformly charged wire
−1
4
produces an electric field of 18 × 10 NC at a
distance of 1.0 cm. The linear charge density on
the wire is
(a) 1.12 × 10
−14
−1
Cm
(b) 3.08 × 10
−15
−1
Cm
−1
−9
−1
−7
(c) 1.0 × 10 Cm
(d) 1.0 × 10 Cm
A Answer
¥ Set 1, 1 Marks
𝜆
𝐸=
2𝜋𝜀𝑜𝑟
Therefore 𝜆 = 𝐸 × 2𝜋𝜀𝑜𝑟
𝑑 = 1cm = 0.1m
−12
𝜀0 = 8.854 × 10
2
−1
C N
4
−2
m
𝜆 = 18 × 10 × 2 × 3.14 × 8.854 × 10
−7
−12
× 0.1
−1
= 1.0 × 10 Cm
š Question 19
(i) Define electric flux and write its SI unit.
(ii) Use Gauss’ law to obtain the expression for
the electric field due to a uniformly charged
infinite plane sheet.
(iii) A cube of side L is kept in space, as shown in
N
→
−
the figure. An electric field E = ( A𝑥 + B) î
C
exists in the region. Find the net charge
enclosed by the cube.
A Answer
¥ 5 Marks
(i) Electric flux is the number of electric field
lines passing through an area normally.
𝜙 = 𝐸® · 𝐴®
2
S.I. unit of electric flux Nm /C or V-m.
(ii) From
∫ Gauss’s law:
−
→
𝑞
𝜙=
𝐸® · 𝑑 𝐴 =
𝜀𝑜
𝜎𝐴
𝜎
2𝐸𝐴 =
𝐸=
𝜀𝑜
2𝜀 𝑜
(iii) 𝜙𝐿 = 𝐸 Δ𝑆 cos 180 = −𝐸 Δ𝑆 = −𝐵𝐿
◦
2
𝜙𝑅 = 𝐸 Δ𝑆 cos 0 = 𝐸 Δ𝑆
2
3
2
= (𝐴𝐿 + 𝐵)𝐿 = 𝐴𝐿 + 𝐵𝐿
◦
3
Net flux = 𝜙𝐿 + 𝜙𝑅 = 𝐴𝐿 + 𝐵𝐿
𝑞
3
Net flux = 𝐴𝐿 =
𝜀𝑜
2
− 𝐵𝐿
2
3
Net charge enclosed by the cube is 𝑞 = 𝐴𝐿 𝜀0
§ 2023 55/4/1 All Sets
š Question 20
Two charges 𝑞1 and 𝑞2 are placed at the centres
of two spherical conducting shells of radius 𝑟1
and 𝑟2 respectively. The shells are arranged such
that their centres are 𝑑 [> (𝑟1 + 𝑟2)] distance
apart. The force on 𝑞2 due to 𝑞1 is :
1 q1q2
1
q1q2
(a)
(b)
2
2
4𝜋𝜀0 d
4𝜋𝜀0 ( d − r1)
𝑞1 𝑞2
1
(c) Zero
(d)
2
4𝜋𝜀0 [𝑑 − (𝑟1 + 𝑟2)]
A Answer
¥ Set 1, 1 Marks
(c) Zero is the correct option
š Question 21
A point charge 𝑞 is kept at a distance 𝑟 from an
infinitely long straight wire with charge density
𝜆 . The magnitude of the electrostatic force
experienced by charge 𝑞 is :
(a) Zero
𝑞𝜆
(b)
2𝜋𝜀0𝑟
𝑞𝜆
(c)
4𝜋𝜀0𝑟
𝑞𝜆
(d)
𝜀0𝑟
A Answer
¥ Set 2, 1 Marks
𝑞𝜆
(b)
is the correct option.
2𝜋𝜀0𝑟
š Question 22
A particle of mass 𝑚 and charge −𝑞 is moving
with a uniform speed 𝑣 in a circle of radius 𝑟 ,
with another charge 𝑞 at the centre of the circle.
The value of r is
1
q
1
q 2
(a)
(b)
4𝜋𝜀0 m v
4𝜋𝜀0 m v
m q 2
m q
(d)
(c)
4𝜋𝜀0 v
4𝜋𝜀0 v
A Answer
(b)
1
¥ Set 3, 1 Marks
q 2
4𝜋𝜀0 m v
is the correct option
š Question 23
Two identical dipoles are arranged in x-y plane
as shown in the figure. Find the magnitude and
the direction of net electric field at the origin O.
A Answer
¥ 2 Marks
→
−
Dipole moment due to dipole BA is 𝑝1 and dipole
→
−
moment due to dipole DC is 𝑝2.
→
−
Electric field 𝐸®1 due to 𝑝1 is along OB.
→
−
→
−
Electric field 𝐸2 due to 𝑝2 is along OD.
Magnitude of resultant Electric field
√
−−→
𝐸𝑛𝑒𝑡 = 2 2𝐸
𝑞
1
→
−
→
−
Since 𝐸1 = 𝐸2 = 𝐸 =
· 2
4𝜋𝜀0 𝑎
√
1
2 2𝑞
𝐸®𝑛𝑒𝑡 =
· 2
4𝜋𝜀0
𝑎
Direction of 𝐸®𝑛𝑒𝑡 is 225° to 𝑥− axis.
Alternatively:
Considering CB as dipole, electric field at O
√
𝐸1 = "
2𝑘𝑞 × (𝑎/ 2)
2
𝑎
𝑎
+ √
√
2
2
√
2𝑘𝑞 × 𝑎
𝐸1 = 3/2
1
1
3
+
𝑎
2 2
√
2𝑘𝑞
𝐸1 =
2
𝑎
#
2 3/2
Similarly considering AD as another dipole,
electric field at O
√
𝐸2 = "
2𝑘𝑞 × (𝑎/ 2)
2
𝑎
𝑎
+ √
√
2
2
√
2𝑘𝑞 × 𝑎
𝐸2 = 3/2
1
1
3
𝑎
+
2 2
√
2𝑘𝑞
𝐸2 =
2
𝑎
√
𝐸net = 𝐸1 + 𝐸2 =
#
2 3/2
2𝑘𝑞
2
𝑎
√
+
2𝑘𝑞
2
√
=
2 2𝑘𝑞
2
𝑎
𝑎
Direction of 𝐸®𝑛𝑒𝑡 is 225° to 𝑥− axis.
š Question 24
Use Gauss’s law to obtain the expression for
electric field due to uniformly charged infinite
plane thin sheet.
A Answer
¥ Set 2, 2 Marks
Net flux through the Gaussian surface𝜙 = 2𝐸𝐴
𝑞
By Gauss Law 𝜙 =
𝜀0
𝑞
= 2𝐸𝐴
where, 𝑞 = 𝜎𝐴
𝜀0
𝜎𝐴
= 2𝐸𝐴
𝜀0
𝜎
𝜎
= 2𝐸 ⇒ 𝐸 =
𝜀0
2𝜀0
š Question 25
Use Gauss’s law to obtain the expression for
electric field due to a thin infinitely long straight
uniformly charged wire.
A Answer
¥ Set 3, 2 Marks
Flux through the Gaussian surface
= flux through the curved cylindrical part of the
surface
= 𝐸 × 2𝜋𝑟𝑙
The surface includes charges equal to 𝜆𝑙 ,
𝜆𝑙
Gauss law then gives 𝐸 × 2𝜋𝑟𝑙 =
𝜀0
𝜆
𝐸=
2𝜋𝜀0𝑟
§ 2023 55/5/1 All Sets
š Question 26
An electric dipole of length 2 cm is placed at an
◦
5
angle of 30 with an electric field 2 × 10 N/C. If
−3
the dipole experiences a torque of 8 × 10 Nm,
the magnitude of either charge of the dipole, is
(a) 4 𝜇 C
(b) 7 𝜇 C
(c) 8mC
A Answer
¥ Set 1, 1 Marks
𝜏 = 𝑝𝐸 sin 𝜃
𝑝=
8 × 10
(d) 2mC
𝑝=
−3
2 × 105 ×
1
𝜏
𝐸 sin 𝜃
= 8 × 10
2
−2
𝑑 = 2 cm = 2 × 10
m
Therefore,
−8
𝑝 8 × 10
𝑞= =
=
4
𝜇
C
−
2
𝑑 2 × 10
−8
𝑝=𝑞×𝑑
š Question 27
A charge 𝑄 is placed at the centre of a cube. The
electric flux through one of its face is
𝑄
𝑄
(a)
(b)
𝜀0
6𝜀0
A Answer
𝑄
(c)
8𝜀0
𝑄
(d)
3𝜀0
¥ Set 2, 1 Marks
As the charge at the center of the cube, the flux
through each surface is same.
𝑄
Using Gauss’s law, 6𝜙 =
𝜀0
𝑄
⇒𝜙=
6𝜀 0
š Question 28
→
−
An electric dipole of dipole moment ( p ) is kept
→
−
in a uniform electric field E Show graphically
the variation of torque acting on the dipole (𝜏)
with its orientation (𝜃) in the field. Find the
orientation in which torque is
(i) zero
(ii) maximum.
A Answer
¥ Set 2, 2 Marks
(i) Torque is zero for orientation corresponding
◦
◦
to 𝜃 = 0 and 𝜃 = 180
(ii) Torque is maximum for orientation
𝜋
3𝜋
corresponding to 𝜃 = and 𝜃 =
2
2
§ 2023 55(B)
š Question 29
The charge on a body is 8 × 10
the body has :
−12
(a) lost 8 × 10
electrons
10
(b) gained 4 × 10 electrons
8
(c) gained 2 × 10 electrons
7
(d) lost 5 × 10 electrons
A Answer
−12
C. It means that
¥ Set 1, 1 Marks
7
(d) lost 5 × 10 electrons is the correct option.
š Question 30
(i) What is an electric dipole ? Derive an
expression for the torque acting on an electric
dipole in a uniform electric field.
(ii) An electric dipole with dipole moment
−9
6 × 10 C−m is aligned at an angle of 30° with
the direction of a uniform electric field of
4
−1
magnitude 4 × 10 NC . Calculate magnitude
of the torque acting on the electric dipole.
A Answer
¥ 5 Marks
(i) Electric Dipole:An arrangement of two equal and opposite
charges separated by a small distance. Consider
an electric dipole of charges ±𝑞 and length 2𝑙
kept in a uniform electric field E. The dipole
makes an angle 𝜃 with the uniform electric field.
Forces on two charges due to uniform electric
field, 𝑞𝐸 and −𝑞𝐸 are in opposite directions. The
distance between the line of action of two force
is 2𝑙 sin 𝜃. Torque due to the couple
𝜏 = force × perpendicular distance between the
two force
𝜏 = 𝑞𝐸( 2𝑙 sin 𝜃)
𝜏 = = 𝑞( 2𝑙) · 𝐸 sin 𝜃 = 𝑝𝐸 sin 𝜃
(ii) Magnitude of the torque on the electric dipole
𝜏 = 𝑝𝐸 sin 𝜃
−9
4
◦
𝜏 = 6 × 10 × 4 × 10 × sin 30
−4
𝜏 = 1.2 × 10 Nm
š OR
(a) State Gauss’s Law in electrostatics. Using it,
derive an expression for the electric field due
to a uniformly charged thin spherical shell of
radius R at a point
(i) outside, and
(ii) inside the shell.
(b) A point charge of 4 𝜇 C is at the centre of a
cubic Gaussian surface, 1 m on edge. Find the
electric flux through one of the faces of the
Gaussian surface.
A Answer
¥ 5 Marks
(a) Gauss’s law in electrostatics states that net
electric flux linked with a closed surface
equals 1/𝜀𝑜 times the charge enclosed by the
surface.
Consider a conducting spherical shell of radius
𝑅, charge given to it is 𝑞.
(i) For point P outside the shell construct a
concentric sphere(Gaussian surface) of
radius𝑟(𝑟 > 𝑅)
Electric flux linked with the Gaussian sphere
2
𝜙𝐸 = 𝐸 · 4𝜋𝑟
Charge enclosed by the Gaussian spherical
surface = 𝑞
According to Gauss’s Law
𝑞
𝜙𝐸 =
𝜀𝑜
𝑞
2
𝐸 · 4𝜋𝑟 =
𝜀𝑜
𝑞
𝐸=
2
4𝜋𝜀𝑜𝑟
(ii) For a point inside the shell Charge enclosed by
the Gaussian spherical surface of radius
′ ′
𝑟 (𝑟 < 𝑅) = 0
𝑞
𝑖𝑛
′2
=0
∴ 𝐸 · 4𝜋𝑟 =
𝜀𝑜
𝐸=0
(b) Net electric flux linked with the cube
′
𝜙E(net)
𝑞
=
𝜀𝑜
Electric flux linked with one face
𝜙
𝑞
E(net)
′
𝜙𝐸 =
=
𝜀𝑜
6𝜀 𝑜
′
𝜙𝐸
=
4 × 10
−6
6 × 8.854 × 10−12
4
2 −1
= 7.5 × 10 Nm C
Question is asked to find the electric flux
through one of the faces of the Gaussian
surface not the net electric flux through the
surface.
That’s why flux has been divided by 6
š Question 31
After centuries of efforts, careful studies,
experiments and analysis by different scientists,
it was concluded that there are two kinds of
entities called the electric charge. The property
which differentiates the two kinds is called the
polarity of charge. The two kinds of charges
were named as positive (+) and negative (−) by
American scientist Benjamin Franklin.
A small sphere S1 with charge 8𝑞 is 1.6 m away
from another identical sphere S2 with charge
+2𝑞. The two spheres are brought in contact
with each other and then separated by a
distance 1.6 m. Initially the force between the
−4
two spheres was 8.1 × 10 N. Based on the above
facts, answer the following questions :
(i) Which sphere will transfer the electrons to the
other sphere after they were brought in
contact ?
A Answer
¥ Set 1, 1 Marks
Electrons will be transferred from sphere S1 to
sphere S2 when brought in contact.
(ii) How does the net electric field at the midpoint
on the line joining the two spheres change
after contact ?
A Answer
¥ Set 1, 1 Marks
Net electric field changes from non zero to zero
at the midpoint of line joining the centres of the
two spheres.
(iii) What was initial charge on spheres S1 and S2 ?
A Answer
¥ 2 Marks
1 𝑞1 𝑞2
=
𝐹
4𝜋𝜀0 𝑟 2
( 8𝑞)( 2𝑞)
−4
9
= 8.1 × 10 N
9 × 10 ×
1.6 × 1.6
−4
8.1 × 1.6 × 1.6 × 10
2
16𝑞 =
9
9 × 10
−4
8
.
1
×
1
.
6
×
1
.
6
×
10
2
𝑞 =
9
9 × 10 × 16
2
−16
−8
𝑞 = 9 × 16 × 10
∴ 𝑞 = 12 × 10 C
−8
Charge on S1 = −8𝑞 = −96 × 10 C
−8
Charge on S2 = 2𝑞 = 24 × 10 C
OR
(iii) What is the charge on spheres S1 and S2 after
contact ?
A Answer
¥ 2 Marks
Charges on S1 and S2 after were brought in
contact −3𝑞 and −3𝑞
§ 2022 Term I 55/2
š Question 32
A negatively charged object X is repelled by
another charged object Y. However an object Z is
attracted to object Y. Which of the following is
the most possibility for the object Z?
(a) positively charged only
(b) negatively charged only
(c) neutral or positively charged
(d) neutral or negatively charged
A Answer
¥ Set 1, 1 Marks
(c) neutral or positively charged
Explanation
▲ X is negatively charged. Since it is repelled by
Y, Y is also negatively charged.
▲ Now, if Z is neutral or positively charged then
it will be attracted by Y.
š Question 33
In an experiment three microscopic latex
spheres are spread into a chamber and became
charged with charges +3e,+5e and –3e
respectively. All the three spheres came in
contact simultaneously for a moment and got
separated. Which one of the following are
possible values for the final charge on the
spheres?
(a) +5e, –4e, +5e
(b) +6e, +6e, –7e
(c) –4e, +3.5e, +5.5e
(d) +5e, –8e, +7e
A Answer
¥ Set 1, 1 Marks
(b) +6e, +6e, –7e
Explanation
▲ When three spheres are brought in contact
the net charge will be –3e + 3e + 5e = 5e
▲ When separated this 5e charge will be
distributed.
▲ In option (A), Total charge is 6e. So, this option
is not correct.
▲ In option (B), total charge is 5e. So, this option
is correct.
▲ In option (C), Total charge is 5e. But this
violates quantization of charge principle.
Hence this option is not correct.
▲ In option (D), total charge is -10e, Hence this
option is not correct.
š Question 34
18
An object has charge of 1 C and gains 5.0 × 10
electrons. The net charge on the object becomes:
(a) −0.80 C
(b) +0.80 C
(c) +1.80 C
A Answer
(d) +0.20 C
¥ Set 1, 1 Marks
d) +0.20 C
Explanation
Charge of 1 electron =−1.6 × 10
−19
𝐶
18
So, Charge of 5.0 × 10 electrons
18
−19
= −5.0 × 10 × 1.6 × 10 𝐶 = −0.8𝐶
Already existing charge = 1C
So, net charge at present = 1 − 0.8 = 0.2𝐶
š Question 35
The magnitude of electric field due to a point
charge 2𝑞, at distance 𝑟 is E. Then the magnitude
of electric field due to a uniformly charged thin
spherical shell of radius R with total charge 𝑞 at
a distance
(a)
𝐸
𝑟
2
(𝑟 >> 𝑅) will be
(b) 0
4
A Answer
(c) 2E
(d) 4E
¥ Set 1, 1 Marks
(c) 2E
Explanation
Electric field due to the point charge
𝐾 × 2𝑞
=𝐸=
𝑟
2
Electric field due to the spherical shell
𝐾 ×𝑞
=𝐸 =
=
2
𝐸
(𝑟/2) 2
′
š Question 36
A square sheet of side 𝑎 is lying parallel to XY
plane at 𝑧 = 𝑎. The electric field in the region is
→
−
2ˆ
E = 𝑐𝑧 𝑘 . The electric flux through the sheet is
1
1
4
3
4
(a) 𝑎 𝑐
(b) 𝑎 𝑐
(c) 𝑎 𝑐
(d) 0
3
3
A Answer
¥ Set 1, 1 Marks
4
(a) 𝑎 𝑐
Explanation
→
− →
−
Flux = E · A
2ˆ
2ˆ
2 2
= 𝑐𝑧 𝑘 · 𝑎 𝑘 = 𝑐𝑧 𝑎
𝑧=𝑎
2
2
𝑧 =𝑎
š Question 37
Three charges 𝑞, −𝑞 and 𝑞0 are placed as shown
in figure. The magnitude
of
the
net
force
on
the
1
charge 𝑞0 at point 𝑂 is 𝑘 =
( 4𝜋𝜀0)
(a) 0
(b)
2𝑘𝑞𝑞0
𝑎2
√
(c)
2𝑘𝑞𝑞0
𝑎2
A Answer
√
2𝑘𝑞𝑞0
(c)
2
𝑎
1 𝑘𝑞𝑞0
(d) √
2
𝑎
2
¥ Set 1, 1 Marks
Explanation
Positions of 𝑞0, −𝑞 and 𝑞 are shown. Both −𝑞 and
𝑞 charges are equidistant from 𝑞0. So, the
magnitude of both the forces on 𝑞0 will be equal.
◦
The angle between the forces will be 90 as
shown in the diagram. Hence the resultant force
√
2
F
2
F
+2F×F×
√
√
𝑘𝑞𝑞0 √
1
𝑞𝑞0
= 2F= 2× 2 = 2×
× 2
𝑎
4𝜋𝜀0
𝑎
=
+
◦
cos 90
š Question 38
Four objects W, X, Y and Z, each with charge +q
are held fixed at four points of a square of side d
as shown in the figure. Objects X and Z are on
the midpoints of the sides of the square. The
electrostatic force exerted by object W on object
X is F. Then the magnitude of the force exerted
by object W on Z is
(a)
𝐹
(b)
7
A Answer
(b)
𝐹
(c)
5
𝐹
(d)
3
5
Explanation
Force on X by W is
2
2
4𝑘𝑞
𝑘𝑞
=
𝐹=
2
2
(𝑑/2)
𝑑
√︂
2
√︁
5
𝑑
2
2
𝑊𝑍 = 𝑑 + (𝑑/2) =
Force on Z by W is
2
𝑘𝑞
4
𝑘𝑞
𝐹
′
𝐹 = 2=
=
2
5𝑑
5
5𝑑
4
2
¥ Set 1, 1 Marks
𝐹
2
𝐹
4
š Question 39
Assertion (A) : A negative charge in an electric
field moves along the direction of the electric
field.
Reason (R) : On a negative charge a force acts in
the direction of the electric field.
(a) Both Assertion (A) and Reason (R) are true
and Reason (R) is the correct explanation of
the Assertion (A).
(b) Both Assertion (A) and Reason (R) are true,
but Reason (R) is not the correct explanation
of the Assertion (A).
(c) Assertion (A) is true, but Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is also
false.
A Answer
¥ Set 1, 1 Marks
Assertion (A) is false and Reason (R) is also false.
Explanation
Assertion is wrong since electron moves in
opposite direction of the electric field.
Reason is also false since on negative charge
force acts in the opposite direction of the electric
field.
§ 2020 55/C Compart
š Question 40
(a) A uniformly charged
large
plane
sheet
has
1
−15
2
charge density 𝜎 =
× 10 C/m .
18𝜋
Find the electric field at point A which is
50 cm from the sheet.
Consider a straight line with three points P, Q
and R, placed 50 cm from the charged sheet
on the right side as shown in the figure. At
which of these points, does the magnitude of
the electric field due to the sheet remain the
same as that at point 𝐴 and why?
(b) Two small identical conducting spheres
carrying charge 10 𝜇 C and −20 𝜇 C when
separated by a distance of r, experience a
force F each. If they are brought in contact
𝑟
and then separated to a distance of , what is
2
the new force between them in terms of F ?
A Answer
¥ 3 Marks
(a) Electric field due to a uniformly charged sheet
𝜎
−6
E=
= 1 × 10 N/C outward
2𝜀 𝑜
At point Q For finite plane sheet, electric field
is uniform in the middle. At the edges it will
be curved.
(b) q1 = q2 =
′
′
𝐹 =
′
𝐹
2
10 + (−20)
2
= −5 𝜇 C
§ 2020 55/5/1 All Sets
š Question 41
The electric flux through a closed Gaussian
surface depends upon
(a) Net charge enclosed and permittivity of the
medium
(b) Net charge enclosed, permittivity of the
medium and the size of the Gaussian surface
(c) Net charge enclosed only
(d) Permittivity of the medium only
A Answer
¥ Set 1, 1 Marks
Net Charge enclosed and permittivity of the
medium
š Question 42
A point charge is placed at the centre of a hollow
conducting sphere of internal radius r and outer
radius 2r. The ratio of the surface charge density
of the inner surface to that of the outer surface
will be
A Answer
¥ Set 1, 1 Marks
𝑄
Surface charge density 𝜎 =
𝐴
𝑞
Surface charge density on 𝜎1 =
4𝜋𝑟 2
𝑞
Surface charge density on 𝜎2 =
2
4𝜋( 2𝑅)
𝑞
2
𝜎1
4
4𝜋𝑟
=
=
𝑞
𝜎2
1
2
4𝜋( 2𝑅)
š Question 43
Derive the expression for the torque acting on
an electric dipole, when it is held in a uniform
electric field. Identify the orientation of the
dipole in the electric field, in which it attains a
stable equilibrium.
A Answer
¥ 2 Marks
Force on 𝑞 is 𝑞E and a force on −𝑞 is −𝑞E. Hence
torque
𝜏 = 𝑞𝐸 × 2𝑎 sin 𝜃
𝜏 = 𝑃𝐸 sin 𝜃
𝜏® = 𝑃® × 𝐸®
For stable equilibrium 𝜃 = 0°
š Question 44
(a) Write two important characteristics of
equipotential surfaces.
(b) A thin circular ring of radius 𝑟 is charged
uniformly so that its linear charge density
becomes 𝜆 . Derive an expression for the
electric field at a point P at a distance 𝑥 from
it along the axis of the ring. Hence, prove that
at large distances (𝑥 >> 𝑟) , the ring behaves
as a point charge.
A Answer
¥ 5 Marks
(a) For equipotential surfaces
▲ Potential has the same value at all points on
the surface.
▲ Electric field is normal to the equipotential
surface at all points
▲ Work done in moving any charge between any
two points on the equipotential surface is zero
(b) Electric field due to any elemental (point)
charge 𝑑𝑞, at point P.
= dE =
1
dq
2
2
4𝜋𝜀0 (𝑥 + 𝑟 )
This is directed along AP Its component along
the axis OP of the ring is
𝑥
= dE cos 𝜃 = dE √
𝑥 2 + 𝑟2
The component, perpendicular to the axis gets
cancelled by the elemental electric field due to
another elemental charge symmetrically located
on the other side of the axis. Hence total electric
field∫
𝑑 E cos 𝜃
∫
1
𝑑𝑞
𝑥
=
√
2
2
(𝑥 + 𝑟 ) 𝑥 2 + 𝑟 2
4𝜋𝜀0
∫
1
𝑥
=
𝜆
d
𝑙
4𝜋𝜀0 (𝑥 2 + 𝑟 2) 3/2
𝐸=
=
1
𝑥𝜆
4𝜋𝜀0 (𝑥 2 + 𝑟 2)
×
2
𝜋𝑟
3/2
𝑄
𝑥
=
4𝜋𝜀0 (𝑥 2 + 𝑟 2) 3/2
Where 𝑄 = 𝜆 × 2𝜋𝑟
= total charge on the ring
This field is directed along the axis.
When 𝑥 much larger than 𝑟 , we have
𝑥
𝑄
1 𝑄
𝐸=
=
2
3
/
2
2
4𝜋𝜀0 (𝑥 )
4𝜋𝜀0 𝑥
This corresponds to the expression for the
electric field due to a point charge. Thus at large
distances the ring behaves like a point charge.
š OR
(a) State Gauss’s law on electrostatics and derive
an expression for the electric field due to a
long straight thin uniformly charged wire
(linear charge density 𝜆 ) at a point lying at a
distance r from the wire.
−1
(b) The magnitude of electric field (in NC ) in a
region varies with the distance r (in m) as
E = 10 r + 5. By how much does the electric
potential increase in moving from point at
r = 1 m to a point at r = 10 m.
A Answer
¥ 5 Marks
(a) Electric flux through of a closed surface is 1/𝜀◦
times the charge enclosed by the surface.
𝑞
Therefore; 𝜙𝐸 =
𝜀◦
Let the charge be uniformly distributed on a
wire
∫
∫
∫
∫
𝜙=
𝑑𝜙 =
𝐸® · 𝑑 𝑠®1 +
𝐸® · 𝑑 𝑠®2 +
𝐸® · 𝑑 𝑠®3
∫
∫
∫
◦
◦
◦
=
𝐸𝑑𝑠1 cos 0 + 𝐸𝑑𝑠2 cos 90 + 𝐸𝑑𝑠3 cos 90
∫
=𝐸
𝑑𝑠1 = 𝐸 · 2𝜋𝑟𝑙
𝑞
= 𝐸 · 2𝜋𝑟𝑙
by Gauss’s law
𝜀0
𝑞
1 𝜆
=
E=
2𝜋𝜀0𝑟𝑙 2𝜋𝜀0 𝑟
(b) E = 10r + 5
dV = −E · dr
∫
dV = −
∫
1
10
® d®r = −
E
∫
1
10
( 10r + 5) dr
V=−
"∫
V = 10
10
10𝑟 dr +
1
2 10
𝑟
2
1
+ 5 (𝑟)
∫
1
#
10
5dr
10
1
V = −5 [ 100 − 1] + 5 [ 10 − 1]
V = −5 × 99 + 5 × 9 = −540 V
§ 2019 55/5/1 All Sets
š Question 45
Derive an expression for the torque acting on an
electric dipole of dipole moment 𝑝® placed in a
uniform electric field 𝐸®. Write the direction
along which the torque acts.
A Answer
¥ 2 Marks
Force on either charge F = qE
Magnitude of torque = Either of force × ⊥
distance between them.
𝜏 = qE2a sin 𝜃
𝜏 = pE sin 𝜃
𝜏® = 𝑝® × 𝐸®
Direction is normal to the paper coming out of it.
š OR
Derive an expression for the electric field at a
point on the axis of an electric dipole of dipole
moment 𝑝®. Also write its expression when the
distance 𝑟 >> the length 𝑎 of the dipole.
A Answer
𝑞
®
𝐸− =
along
(−)
𝑝
2
4𝜋𝜀0 (𝑟 + 𝑎)
𝑞
®
𝐸+ =
along
(+)
𝑝
2
4𝜋𝜀0 (𝑟 + 𝑎)
∴ Total field at 𝑃, 𝐸 = 𝐸− − 𝐸+
𝑞
1
1
=
−
4𝜋𝜀0 (𝑟 − 𝑎) 2 (𝑟 + 𝑎) 2
𝑞
2 𝑝𝑟
=
2
2
2
4𝜋𝜀0 (𝑟 − 𝑎 )
For 𝑟 >> 𝑎
1 2𝑝
𝐸=
4𝜋𝜀0 𝑟 3
¥ 2 Marks
š Question 46
Why is the direction of the electric field due to a
charged conducting sphere at any point
perpendicular to its surface ?
A Answer
¥ Set 2, 1 Marks
If electric field is not perpendicular but has a
component tangential to the surface of the
conductor, it will exert force on charge and make
them more. It means electrostatic condition is
violated.
š Question 47
Why can the interior of a conductor have no
excess charge in the static situation ?
A Answer
¥ Set 3, 1 Marks
Take any arbitrary closed surface S inside the
conductor. Since electric field inside the
conductor is zero, electric flux through S is also
zero. Hence by Gauss’s law there is no net charge
enclosed by S
OR
Electric field inside the conductor is zero
§ 2019 55 (B)
š Question 48
Why are electric field lines perpendicular at a
point on an equipotential surface of a conductor?
A Answer
¥ Set 1, 1 Marks
The work done in moving a charge from one
point to another on an equipotential surface is
zero. If the field is not normal to an
equipotential surface, it would have a non zero
component along the surface. This would imply
that work would have to be done to move a
charge on the surface which is contradictory to
the definition of equipotential surface.
Alternatively
Work done to move a charge 𝑑𝑞 on a surface can
be expressed as
−
→
𝑑𝑊 = 𝑑𝑞( 𝐸® · 𝑑𝑟)
But 𝑑𝑊 = 0 on an equipotential surface
∴
−
→
𝐸® ⊥ 𝑑𝑟
Equipotential surfaces for a charge −𝑞 is shown
alongside.
š Question 49
Using Gauss law, derive an expression for the
electric field at a point near an infinitely long
straight uniformly charged wire.
A Answer
¥ 3 Marks
Flux through the Gaussian Surface is equal to
flux through the curved cylindrical part
= E × 2𝜋𝑟𝑙
The surface includes charge equal to 𝜆𝑙
Δ𝑄
= linear charge density
where, 𝜆 =
Δ𝑙
Gauss’s Law then gives
𝜆𝑙
𝐸 × 2𝜋𝑟𝑙 =
𝜀◦
𝜆
𝐸=
2𝜋𝜀◦𝑟
š OR
(i) An electric dipole of dipole moment 𝑝® is held
in a uniform electric field 𝐸®. Show that the
torque acting on the dipole is given by 𝑝® × 𝐸®.
(ii) How much work is required in turning the
electric dipole from the position of most
stable equilibrium to that of most unstable ?
A Answer
¥ 3 Marks
(i) Magnitude of torque = 𝑞𝐸 × 2𝑎 sin 𝜃
= 2𝑞𝑎𝐸 sin 𝜃
= 𝑝𝐸 sin 𝜃
Its direction is normal to the plane of the paper,
coming out of it.
The magnitue of 𝑃® × 𝐸® Is also PE sin 𝜃 and its
direction is normal to the paper, Thus 𝜏® = 𝑃® × 𝐸®
(ii) Work done in turning the dipole from most
stable euilibirium to most unstable
equilibrium position
W = 𝑈 𝑓 −𝑈𝑖
U = −pE cos 𝜃
◦
𝑈𝑖 = −pE cos 0 = −pE
𝑈 𝑓 − = −pE cos 180 = pE
W = pE − (−p𝐸) = 2pE
◦
*
Unit I
Electrostatics
Published by :
www.cbse.page
www.cuet.pw
’ Chapter 2
™ Electrostatic
Potential and
Capacitance
Weightage for CBSE Board 2024
Unit I and II Combined À Chapter 1
Weightage is
À Chapter 2
16 Marks
À Chapter 3
It’s not fixed which chapters will
have more weightage, but overall weightage
will be 16 Marks. It may vary too in your
question paper.
` Syllabus
Electric potential, potential difference, electric
potential due to a point charge, a dipole and
system of charges; equipotential surfaces,
electrical potential energy of a system of
two-point charges and of electric dipole in an
electrostatic field.
Conductors and insulators, free charges and
bound charges inside a conductor. Dielectrics
and electric polarization, capacitors and
capacitance, combination of capacitors in series
and in parallel, capacitance of a parallel plate
capacitor with and without dielectric medium
between the plates, energy stored in a capacitor
(no derivation, formulae only).
§ 2024 SQP Official
š Question 1
Which of the following is not the property of an
equipotential surface?
(a) They do not cross each other.
(b) The work done in carrying a charge from one
point to another on an equipotential surface
is zero.
(c) For a uniform electric field, they are
concentric spheres.
(d) They can be imaginary spheres.
A Answer
¥ 3 Marks
Option (c) is not the property of an equipotential
surface.
š Question 2
Assertion (A) : An electron has a higher
potential energy when it is at a location
associated with a negative value of potential and
has a lower potential energy when at a location
associated with a positive potential.
Reason (R) : Electrons move from a region of
higher potential to a region of lower potential.
(a) Both Assertion (A) and Reason (R) are true
and Reason (R) is the correct explanation of
the Assertion (A).
(b) Both Assertion (A) and Reason (R) are true,
but Reason (R) is not the correct explanation
of the Assertion (A).
(c) Assertion (A) is true, but Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is also
false.
A Answer
¥ Set 1, 1 Marks
Assertion (A) is true, but Reason (R) is false.
š Question 3
Charges (+𝑞) and (−𝑞) are placed at the points A
and B respectively which are a distance 2L apart.
C is the midpoint between A and B. What is the
work done in moving a charge +𝑄 along the
semicircle CRD.
A Answer
¥ 3 Marks
VC = 0
h
i
𝑞 𝑞
−𝑞
1
=
−
VD =
4𝜋𝜀0 3L L
6𝜋𝜀0L
−𝑄𝑞
W = Q [ VD − VC ] =
6𝜋𝜀0L
š Question 4
(i) Derive an expression for the capacitance of a
parallel plate capacitor with air present
between the two plates.
(ii) Obtain the equivalent capacitance of the
network shown in figure. For a 300 V supply,
determine the charge on each capacitor.
A Answer
¥ 5 Marks
(i) Let the two plates be kept parallel to each
other separated by a distance 𝑑 and
cross-sectional area of each plate is 𝐴.
Electric field by a single thin plate
𝜎
𝐸=
[𝜎 = surface charge density]
2𝜖 𝑜
Outer region I (region above the plate 1),
𝜎
𝜎
𝐸=
−
=0
2𝜖 𝑜 2𝜖 𝑜
Outer region II (region below the plate 2)
𝜎
𝜎
𝐸=
−
=0
2𝜖 𝑜 2𝜖 𝑜
Therefore the total electric field due to inner
region between the plates 1 and 2, the electric
fields due to the two charged plates add up
𝜎
𝜎
𝜎
𝑄
+
= =
𝐸=
2𝜖 𝑜 2𝜖 𝑜 𝜖 𝑜 𝜖 𝑜 𝐴
Potential difference between the plates
𝑉 = 𝐸𝑑 =
1 𝑄𝑑
𝜀0 𝐴
The capacitance 𝐶 of the parallel plate capacitor
is then
𝑄 𝜀0 𝐴
𝐶= =
𝑉
𝑑
which depends only on the geometry of the
system
(ii)
The equivalent capacitance =
Charge on C4 =
200
× 10
−12
200
3
pF
−8
× 300 = 2 × 10 C
3
Potential difference across
−12
200 × 10
× 300
𝐶4 =
=
200
V
−
12
3 × 100 × 10
Potential difference across
𝐶1 = 300 − 200 = 100 V
−12
−8
Charge on 𝐶1 = 100 × 10
× 100 = 1 × 10 C
Potential difference across 𝐶2 and 𝐶3 series
combination = 100 V
Potential difference across 𝐶2 and 𝐶3 each = 50 V
Charge on 𝐶2 and 𝐶3 each
−12
−8
= 200 × 10 × 50 = 1 × 10 C
š [OR Section]
(i) A dielectric slab of thickness 𝑡 is kept between
the plates of a parallel plate capacitor with
plate separation 𝑑 (𝑡 < 𝑑) . Derive the
expression for the capacitance of the
capacitor.
(ii) A capacitor of capacity C1 is charged to the
potential of V0. On disconnecting with the
battery, it is connected with an uncharged
capacitor of capacity C2 as shown in the
adjoining figure. Find the ratio of energies
before and after the connection of switch S.
A Answer
¥ 5 Marks
(i)
The capacitance 𝐶 of the parallel plate capacitor
is
𝑄 𝜀0 𝐴
𝐶= =
𝑉
𝑑
where 𝐴 is the area of parallel plates.
Suppose that the capacitor is connected to a
battery, an electric field E0 is produced.
Now if we insert the dielectric slab of thickness
𝑡 = less than 𝑑 , the electric field reduces to E.
Now the gap between the plates is divided in two
parts, for distance 𝑡 there is electric field E and
for the remaining distance (𝑑 − 𝑡) , the electric
field is E0.
If V be the potential difference between the
plates of the capacitor, then
V = E𝑡 + E0 (𝑑 − 𝑡)
E0
[ Given: 𝑡 < 𝑑]
= 𝐾 −→ ratio of dielectric constant
E
E0
1
V = 𝑡 + E0 (𝑑 − 𝑡) = E0𝑡
− 1 + E0 𝑑
𝐾
𝐾
𝜎
𝑞
As E0 =
=
𝜀0 𝜀0 𝐴
𝑞
1
𝑞
V=
·𝑡
−1 +
·𝑑
𝜀0 𝐴
𝐾
𝜀0 𝐴
1
𝑞
V=
𝑡
−1 +𝑑
𝜀0 𝐴
𝐾
So finally, the capacitance when the slab of
thickness 𝑡 is inserted between the parallel
plates of the capacitor is
𝜀0A
𝑞
𝜀0 A 𝐾
=
𝑡 − 𝐾𝑡 + 𝐾𝑑
= 𝑉
1
−1 +𝑑
𝑡
𝐾
𝜀0 A 𝐾
𝜀0A 𝐾
⇒𝐶=
or
𝑡 − 𝐾 (𝑡 + 𝑑)
𝑡( 1 − 𝐾) + 𝐾𝑑
𝐶=
Same type of question was asked in CBSE
2013 Board Exam
A slab of material of dielectric constant 𝐾 has
the same area as that of the plates of a parallel
plate capacitor but has the thickness 𝑑/2, where
𝑑 is the separation between the plates. Find out
the expression for its capacitance when the slab
is inserted between the plates of the capacitor.
[2013]
A Solution
The capacitance 𝐶 of the parallel plate capacitor
is
𝑄 𝜀0 𝐴
𝐶= =
𝑉
𝑑
where 𝐴 is the area of parallel plates.
Suppose that the capacitor is connected to a
battery, an electric field E0 is produced.
Now if we insert the dielectric slab of thickness
𝑡 = 𝑑/2, the electric field reduces to E.
Now the gap between the plates is divided in two
parts, for distance 𝑡 there is electric field E and
for the remaining distance (𝑑 − 𝑡) , the electric
field is E0.
If V be the potential difference between the
plates of the capacitor, then
V = E𝑡 + E0 (𝑑 − 𝑡)
V=
E𝑑
E0 𝑑
𝑑
𝑑
+
= ( E + E0 )
Given: 𝑡 =
2
2
2
2
E0
= 𝐾 −→ ratio of dielectric constant
E
𝑑 E0
𝑑 E0
V=
+ E0 =
( 1 + 𝐾)
2 𝐾
2𝐾
𝜎
𝑞
𝑑
𝑞
=
and V =
·
( 1 + 𝐾)
E0 =
𝜀0 𝜀0 A
2 𝐾 𝜀0 A
𝑞
2 𝐾𝜀0A
𝐶= =
𝑉 𝑑 ( 1 + 𝐾)
Same type of questions can come in your exam
by just twisting the question as thickness of the
slab to 𝑑/4, 3/4𝑑 , etc.
(ii) Let V1 be the common potential of both
capacitor after the complete flow of charge.
From conservation of charge we have
C1V0 = ( C1 + C2) V1
C1
V0
V1 =
( C1 + C2)
Energy before connecting the switch
1
2
U1 = C1V0
2
Energy after connection switch
1
2
U2 = ( C1 + C2) V1
2
2 2
C1V0
1
U2 = ( C1 + C2)
2
2
( C1 + C2)
2 2
C1V0
1
U2 = ×
2
2 ( C1 + C2)
Now ratio is given by,
U2
U1
=
2 2
C 1 V0
2 ( C1 + C2)
×
2
2
2
C1V0
Therefore, the ratio of energies before and after
the connection of switch S
U2
C1
U1
=
( C1 + C2)
§ 2023 55(B) Compart
š Question 5
(i) How will the capacitance of a parallel plate
capacitor change if :
(1) the plates area is doubled ?
(2) the separation between the plates is doubled ?
(ii) The effective capacitance of three capacitors
of the same capacitance connected in series is
1 F. Find the :
(1) effective capacitance if they are connected in
parallel.
(2) ratio of energy stored in the parallel
combination of the capacitors to that in the
series combination, if the combinations are
connected to the same source one by one.
A Answer
¥ 5 Marks
(i.1) 𝐶 ∝ 𝐴 Capacitance gets doubled
(i.2) 𝐶 ∝
1
𝑑
Capacitance reduces to half
(ii) Given that the effective capacitance of three
capacitors of the same capacitance
connected in series is 1 F.
𝐶
= 1𝜇𝐹
∴ 𝐶 = 3𝜇𝐹
3
(1) Effective capacitance in parallel
𝐶 𝑃 = 3𝐶 = 3 × 3𝜇𝐹
𝐶 𝑃 = 9𝜇𝐹
1
(2) 𝑈 = 𝐶𝑉
2
2
𝑈𝑃 𝐶 𝑝 9
=
= =3:1
𝑃𝑆 𝐶 𝑆 3
š Question 6
Case Study :
A charged body inside an electric field. A
−13
charged latex sphere of mass 1.85 × 10
kg is
held stationary in between two horizontal plates
which are separated by a distance 0·62 cm. The
potential difference between the plates is
3
1.24 × 10 V with the upper plate being positive.
Based on the above facts, answer the following
questions :
(i) What is the nature of charge on the latex
sphere ?
(ii) What is the direction of electric field between
the plates ?
(iii) What is the magnitude of electric field
between the plates ?
OR
What is the magnitude of charge on the latex
2
sphere ? (Take 𝑔 = 10 ms )
A Answer
(i) Negative Charge
(ii) Downwards
¥ 4 Marks
𝑉
(iii) 𝐸 =
𝑑
3
𝐸=
1.24 × 10 V
0.62 ×
−
2
10
m
5
−1
= 2 × 10 Vm
OR
𝑞𝐸 = 𝑚𝑔
−13
𝑚𝑔 1.85 × 10 × 10
𝑞=
=
5
𝐸
2 × 10
−18
= 9.25 × 10 C
§ 2023 55/C Compart
š Question 7
A uniform electric field E of 500 N/C is directed
along +𝑥 axis. O, B and A are three points in the
field having 𝑥 and 𝑦 coordinates (in cm) ( 0, 0),
( 4, 0) and ( 0, 3) respectively. Calculate the
potential difference between the points
(i) O and A
(ii) O and B.
A Answer
¥ 2 Marks
(i) VOA = E (𝑥2 − 𝑥1) = 500 × 0 = 0 V
(ii) VOB = −E (𝑥2 − 𝑥1) = −500 × 4 × 10
−2
= −20 V
š OR
Three point charges 1 𝜇 C, −1 𝜇 C and 2 𝜇 C are kept
at the vertices A, B and C respectively of an
equilateral triangle of side 1 m.A1, B1 and C1 are
the midpoints of the sides AB, BC and CA
respectively. Calculate the net amount of work
done in displacing the charge from 𝐴 to 𝐴1, from
𝐵 to 𝐵1 and from 𝐶 to 𝐶1.
A Answer
¥ 2 Marks
Initial electrostatic potential energy of the
system
𝑘
−12
Ui = [ 1 × (−1) + (−1) × 2 + ( 1) × ( 2)] × 10
𝑟
9
9 × 10
−12
=
[−1 − 2 + 2] × 10
1
= −9 × 10
−3
J
Now A1 B1 = B1C1 = A1C1 =
1
m
2
Final electrostatic potential energy of the system
−9 × 10
𝑈𝑓 =
1/2
−9
= −18 × 10
−3
J
Amount of work done W = 𝑈 𝑓 − 𝑈𝑖
−3
W = −18 × 10
+ 9 × 10
−3
= −9 × 10
−3
J
š Question 8
A parallel plate capacitor is an arrangement of
two identical metal plates kept parallel, a small
distance apart. The capacitance of a capacitor
depends on the size and separation of the two
plates and also on the dielectric constant of the
medium between the plates. Like resistors,
capacitors can also be arranged in series or
parallel or a combination of both. By virtue of
electric field between the plates, charged
capacitors store energy.
(a) The capacitance of a parallel plate capacitor
increases from 10 𝜇 F to 80 𝜇 F on introducing a
dielectric medium between the plates. Find
the dielectric constant of the medium.
A Answer
¥ Set 1, 1 Marks
𝐶 80 𝜇𝐹
𝐾=
=
=8
𝐶0 10 𝜇𝐹
(b) n capacitors, each of capacitance C, are
connected in series. Find the equivalent
capacitance of the combination.
A Answer
1
𝐶𝑆
=
1
𝐶1
𝑛
1
=
𝐶𝑆 𝐶
+
1
¥ Set 1, 1 Marks
+.........+
𝐶2
1
𝐶𝑛
𝐶
Therefore, 𝐶𝑆 =
𝑛
(c) A capacitor is charged to a potential (V) by
connecting it to a battery. After some time, the
battery is disconnected and a dielectric is
introduced between the plates. How will the
potential difference between the plates, and
the energy stored in it be affected ? Justify
your answer.
OR
Find the equivalent capacitance between
points A and B, if capacitance of each
capacitor is C.
A Answer
¥ 2 Marks
The charge on the plates does not change as the
capacitor is disconnected from the battery. The
presence of dielectric slab increases the
capacitance, which decreases the potential
difference. Thus, the energy stored is reduced.
Charge is constant.
𝑄1 = 𝑄2
𝑉1
𝑉2 =
𝐾
𝐶2 = 𝐾𝐶1
𝐶1𝑉1 = 𝐾𝐶1𝑉2
Potential difference decreases by a factor
1𝑄
2
1𝑄
2
1
2
1
𝐾
𝑄
𝑉2 =
=
=
2 𝐶2 2 𝐾𝐶1 𝐾 2𝐶2
OR
As we can see, the capacitors form a Wheatstone
bridge. The effective capacitance between M and
N is the sum of effective capacitance between P
and Q and the effective capacitance between S
and T.
𝐶 𝑀𝑁 = 𝐶 𝑃𝑄 + 𝐶𝑆𝑇
Wheatstone bridges can be divided into two
halves which have similar capacitance.
∴ 𝐶 𝑃𝑄 = 𝐶𝑆𝑇
The capacitance 𝐶 𝑃𝑄 is the reciprocal of the sums
of the reciprocal of the capacitances P and Q.
𝐶 𝑃𝑄 =
1
1
𝐶𝑃
+
1
𝐶𝑄
The capacitance of the capacitor 𝑃 is 𝐶 and
capacitance of 𝑄 is 𝐶 .
1
𝐶 𝑃𝑄 =
𝐶 𝑃 = 𝐶, 𝐶𝑄 = 𝐶
1
1
+
𝐶𝑃
𝐶 𝑃𝑄 =
1
1
𝐶
+
𝐶𝑄
1 𝐶
= =
1 2 2
𝐶
𝐶
The capacitance 𝐶 𝑃𝑄 is equal to the capacitance
𝐶𝑆𝑇
𝐶 𝑃𝑄 = 𝐶𝑆𝑇 =
𝐶
2
=⇒ 𝐶 𝑀𝑁 = 𝐶 𝑃𝑄 + 𝐶𝑆𝑇 =
=⇒ 𝐶 𝑀𝑁 = 1𝐶
𝐶
2
+
𝐶
2
The effective capacitance between the point M
and point N is 1C.
Now only two capacitors in parallel left between
A and B, so the effective capacitance between the
point A and point B is ( 1 + 1) C = 2C.
§ 2023 55/1/1 All Sets
š Question 9
(a) Twelve negative charges of same magnitude
are equally spaced and fixed on the
circumference of a circle of radius R as shown
in Fig. (i). Relative to potential being zero at
infinity, find the electric potential and electric
field at the centre C of the circle.
(b) If the charges are unequally spaced and fixed
on an arc of 120° of radius R as shown in Fig.
(ii), find electric potential at the centre C.
A Answer
¥ 3 Marks
(a) Electric potential due to point charge
𝑘𝑞
V=
𝑅
Value of each charge = −𝑞,
Total charge = −12𝑞
𝑘(−12𝑞)
Total potential V =
𝑅
−12𝑘𝑞
−12𝑞
=
V=
𝑅
4𝜋 ∈0 𝑅
By symmetry the resultant of all electric field
vectors becomes zero. So electric field is zero.
(b) Electric potential is a scalar quantity and does
not depend on placement of charges
−12𝑘𝑞
−12𝑞
Therefore V =
=
𝑅
4𝜋 ∈0 𝑅
š Question 10
A 100 𝜇 F capacitor is charged by a 12 V battery.
(a) How much electrostatic energy is stored by
the capacitor ?
(b) The capacitor is disconnected from the
battery and connected in parallel to another
uncharged 100 𝜇 F capacitor. What is the
electrostatic energy stored by the system ?
A Answer
¥ Set 2, 3 Marks
−6
C = 100 𝜇𝐹 = 100 × 10 𝐹, 𝑉 = 12 V
1
2
1
−6
(a) U = 𝐶𝑉 = × 100 × 10
2
2
1
−4
= × 10 × 144
2
= 72 × 10
−4
× ( 12)
2
J = 7.2 m J
(b) Ceq = C1 + C2 = 200 𝜇𝐹
−6
−4
𝑄 = 𝐶𝑉 = 100 × 10 × 12 = 12 × 10 C
2
−
4
2
12 × 10
𝑄
144
−4
𝑈=
=
=
×
10
−
6
2𝐶𝑒𝑞 2 × 200 × 10
4
= 36 × 10
−4
J = 3.6 m J
š Question 11
Three point charges Q1 (−15 𝜇 C) , Q2 ( 10 𝜇 C) and
Q3 ( 16 𝜇 C) are located at (0 cm, 0 cm),(0 cm, 3 cm)
and (4 cm, 3 cm) respectively. Calculate the
electrostatic potential energy of this system of
charges.
A Answer
¥ Set 3, 3 Marks
1
𝑄1𝑄2
UQ1Q2 =
4𝜋 ∈𝑜 𝑟12
−6
9
−6
× 10 × 10
9 × 10 × −15 × 10
=
−
2
3 × 10
= −45 J
1 𝑄2𝑄3
UQ2Q3 =
4𝜋 ∈𝑜 𝑟23
9
−6
−6
9 × 10 × 10 × 10
× 16 × 10
=
4 × 10−2
= 36 J
1 𝑄1𝑄3
UQ1Q3 =
4𝜋 ∈𝑜 𝑟13
9
−6
−6
9 × 10 × −15 × 10
× 16 × 10
=
5 × 10−2
= −43.2 J
Unet = UQ1Q2 + UQ2Q3 + UQ1Q3
= −45 + 36 − 43.2 = −52.2 J
§ 2023 55/2/1 All Sets
š Question 12
A point P lies at a distance 𝑥 from the mid point
of an electric dipole on its axis. The electric
potential at point P is proportional to
1
1
1
1
(a) 2
(b) 3
(c) 4
(d) 1/2
𝑥
𝑥
𝑥
A Answer
𝑥
¥ Set 1, 1 Marks
If you know the formula of electric potential due
to dipole which is
1 𝑝
𝑉 =±
4𝜋𝜖𝑜 𝑟 2
1
From formula it is clear that 𝑉 ∝ 2
𝑟
1
So 2 is the correct option.
𝑥
š Question 13
A point charge 𝑞0 is moving along a circular path
of radius a, with a point charge −𝑄 at the centre
of the circle. The kinetic energy of 𝑞0 is
𝑞0 𝑄
(a)
4𝜋𝜖0a
𝑞0 𝑄
(c)
4𝜋𝜖0a2
𝑞0 𝑄
(b)
8𝜋𝜖0a
𝑞0 𝑄
(d)
8𝜋𝜖0a2
A Answer
¥ Set 3, 1 Marks
To find the kinetic energy of the moving point
charge 𝑞0, we need to consider the electrostatic
potential energy between the charges and the
kinetic energy of the charge in motion.
The electrostatic potential energy between two
point charges 𝑞0 and 𝑄 separated by a distance 𝑟
is given by:
𝑘 · |𝑞0 · 𝑄|
𝑈=
,
𝑟
▲ 𝑘 is Coulomb’s constant, 𝑘 =
1
4𝜋𝜖0
2
2
9
approximately equal to 9 × 10 N m /C ,
▲ 𝑟 is the distance between the charges, and
▲ |𝑞0 · 𝑄| is the product of their magnitudes.
The kinetic energy 𝐾 of the moving point charge
𝑞 is given by:
1 2
𝐾 = 𝑚𝑣 ,
2
where 𝑚 is the mass of the moving point charge,
and 𝑣 is its velocity.
Since 𝑞 is moving in a circular path of radius 𝑎
around the charge −𝑄 at the center, there must
be a centripetal force acting on 𝑞 to keep it in
circular motion.
This force is provided by the electrostatic force
between the two charges.
At any point on the circular path, the
electrostatic force (𝐹𝑒) between 𝑞 and −𝑄 is given
by Coulomb’s law:
𝑘 · |𝑞 · 𝑄|
1
𝐹𝑒 =
,
where
𝑘
=
2
𝑎
4𝜋𝜖0
where 𝑎 is the radius of the circular path.
The centripetal force (𝐹𝑐) required to keep 𝑞 in
circular motion is:
2
𝑚𝑣
𝐹𝑐 =
𝑎
Since 𝐹𝑐 = 𝐹𝑒, we can equate the two expressions:
2
𝑚𝑣
𝑘 · |𝑞 · 𝑄|
=
𝑎
𝑎2
2
Solving for 𝑣 , we get:
𝑘 · |𝑞 · 𝑄|
.
𝑣 =
𝑚·𝑎
2
Now, substituting this value of 𝑣 into the
2
expression for kinetic energy, we get:
1
𝑘 · |𝑞 · 𝑄|
𝑘 · |𝑞 · 𝑄|
=
𝐾 = 𝑚·
2
𝑚·𝑎
2𝑎
1
𝑞𝑄
Replacing by 𝑘 =
, we get
4𝜋𝜖0
8𝜋𝜖0a
To explain you line by line, the answer
has become long but you don’t have
to write so long in your answer sheet.
Just write the correct option as it’s of
1 Mark only.
š Question 14
Two charged conducting spheres of radii a and b
are connected to each other by a wire. Find the
ratio of the electric fields at their surfaces.
A Answer
¥ 3 Marks
When connected by a conducting wire both
spheres will be at the same potential.
𝑞2
1
𝑞1
where 𝑘 =
∴𝑉 =𝑘 =𝑘
𝑎
𝑏
4𝜋𝜖0
1 𝑄
∵ 𝑉 (𝑟) =
4𝜋𝜖0 𝑟
𝑞1 𝑎
∴ =
𝑞2 𝑏
𝑞1
𝑘
2
𝐸1
𝑏
𝑎
= 𝑞 =
𝐸2 𝑘 2 𝑎
𝑏2
š OR
A parallel plate capacitor (A) of capacitance C is
charged by a battery to voltage V. The battery is
disconnected and an uncharged capacitor (B) of
capacitance 2C is connected across A. Find the
ratio of
(i) final charges on A and B.
(ii) total electrostatic energy stored in A and B
finally and that stored in A initially.
A Answer
¥ 3 Marks
(i) Initially 𝑄 = 𝐶𝑉
Finally 𝑞 𝐴 = 𝐶 𝐴𝑉1
𝑞𝐴 𝐶 𝐴 1
=
𝑞𝐵
=
𝐶𝐵
& 𝑞𝐵 = 𝐶 𝐵𝑉1
2 (ii) 𝑞 𝐴 + 𝑞𝑏 = 𝑄
∴ 𝑞𝐴 =
𝑄
&
3
𝑈 𝑓 𝑈 𝐴 + 𝑈𝐵
=
𝑈𝑖
𝑈 𝐴𝑖
2
2
𝑞𝐵
𝑞𝐴
+
2𝐶 𝐴 2𝐶 𝐵 1
=
=
2
3
𝑄
2𝐶 𝐴
Alternatively,
Common potential
𝑄1 + 𝑄2
𝑉1 =
𝐶1 + 𝐶2
𝑄 𝑉
=
=
3𝐶 3
1
2
𝑒𝑞 1
𝐶 𝑉
𝑈𝑓 2
=
𝑈𝑖 1
2
𝐶 𝐴𝑉
2
𝑄
∵ =𝑉
𝐶
𝑞𝐵 =
2𝑄
3
1
=
2
3𝐶 ×
1
2
2
𝑉
𝐶𝑉
3
=
2
1
3
š Question 15
(i) Consider two identical point charges located
at points ( 0, 0) and ( a, 0) .
(1) Is there a point on the line joining them at
which the electric field is zero?
(2) Is there a point on the line joining them at
which the electric potential is zero?
Justify your answers for each case.
(ii) State the significance of negative value of
electrostatic potential energy of a system of
charges.
Three charges are placed at the corners of an
equilateral triangle ABC of side 2.0 m as
shown in figure. Calculate the electric
potential energy of the system of three
charges.
A Answer
¥ 5 Marks
(i.1) Yes , electric field is zero at mid point.
Electric field being a vector quantity, its
resultant is zero.
(i.2) No, potential cannot be zero on line joining
the charges. Electric potential being a scalar
quantity, the net potential due to two
identical charges cannot be zero.
(ii) Negative value of electrostatic potential
energy of a system signifies that the system
has attractive forces and is stable.
1
𝑞1𝑞2
𝑈=
×
4𝜋𝜖0
𝑟
h
i
𝑞 𝐴𝑞 𝐵 𝑞 𝐵𝑞𝑐 𝑞𝑐 𝑞 𝐴
1
+
+
𝑈=
4𝜋𝜖0
𝑟
𝑟
𝑟
9
9 × 10
−12
[−16 − 8 + 8] × 10
=
2
−2
= −7.2 × 10 J
§ 2023 55/3/1 All Sets
š Question 16
(i) Define electric potential at a point and write
its SI unit.
(ii) Two capacitors are connected in series.
Derive an expression of the equivalent
capacitance of the combination.
(iii) Two point charges +𝑞 and −𝑞 are located at
points ( 3𝑎, 0) and ( 0, 4𝑎) respectively in 𝑥 − 𝑦
plane. A third charge 𝑄 is kept at the origin.
Find the value of 𝑄, in terms of 𝑞 and 𝑎, so
that the electrostatic potential energy of the
system is zero.
A Answer
¥ 5 Marks
(i) Electrostatic potential at any point in a region
with electrostatic field is the work done in
bringing a unit positive charge (without
acceleration) from infinity to that point.
𝑉=
Work Done
=−
∫
→
−
𝐸® · 𝑑𝑙
𝑞
S.I. unit of electrostatic potential is Volt or J/C.
(ii)
𝑄 𝑄
𝑉 = 𝑉1 + 𝑉2 = +
𝐶
𝐶
1
2
𝑄
1
1
=𝑄
+
𝐶𝑒𝑞.
𝐶1 𝐶2
1
=
𝐶𝑒𝑞.
1
𝐶1
1
+
𝐶2
(iii) Potential energy of the system
2
−𝑄𝑞 𝑄𝑞 𝑞
=
+
−
4𝑎
3𝑎 5𝑎
2
−𝑄𝑞 𝑄𝑞 𝑞
∴𝐾
+
−
=0
4𝑎
3𝑎 5𝑎
−𝑄 𝑄 𝑞
𝑄 𝑞
⇒
+ − =0
⇒+ − =0
4
⇒𝑄=+
3 5
12𝑞
5
12
5
§ 2023 55/4/1 All Sets
š Question 17
Obtain an expression for electrostatic potential
energy of a system of three charges 𝑞, 2𝑞 and
−3𝑞 placed at the vertices of an equilateral
triangle of side 𝑎.
A Answer
¥ 2 Marks
1
𝑞1𝑞2
𝑈=
·
4𝜋𝜖0
𝑟
2
2
2
1
2𝑞
6𝑞
3𝑞
𝑈=
−
−
4𝜋𝜖0 𝑎
𝑎
𝑎
2
1 −7 𝑞
𝑈=
4𝜋𝜖0 𝑎
š OR
Two small conducting balls A and B of radius 𝑟1
and 𝑟2 have charges 𝑞1 and 𝑞2 respectively. They
are connected by a wire. Obtain the expression
for charges on A and B, in equilibrium.
A Answer
¥ 2 Marks
According to law of conservation of charge
𝑞𝑖 = 𝑞 𝑓
′
′
𝑞1 + 𝑞2 = 𝑞1 + 𝑞2 = 𝑄
When two balls are connected with wire
𝑉1 = 𝑉2
′
′
′
′
𝑘𝑞1 𝑘𝑞2
𝑞1 𝑞2
=
or
=
𝑟1
𝑟2
𝑟1 𝑟2
′
′
𝑞1𝑟2 = 𝑞2𝑟1
′
′
𝑞1𝑟2 = 𝑄 − 𝑞1 𝑟1
′
′
𝑞1𝑟2 = 𝑄𝑟1 − 𝑞1𝑟1
′
𝑞1 (𝑟1 + 𝑟2) = 𝑄𝑟1
(𝑞1 + 𝑞2) 𝑟1
𝑄𝑟1
=
𝑞1 =
𝑟1 + 𝑟2
𝑟1 + 𝑟2
(𝑞
)
𝑄𝑟
𝑄𝑟
+
𝑞
𝑟
1
2
1
2
2
′
′
𝑞2 = 𝑄 − 𝑞1 = 𝑄 −
=
=
𝑟1 + 𝑟2 𝑟1 + 𝑟2
𝑟1 + 𝑟2
′
š Question 18
Electrostatics deals with the study of forces,
fields and potentials arising from static charges.
Force and electric field, due to a point charge is
basically determined by Coulomb’s law. For
symmetric charge configurations, Gauss’s law,
which is also based on Coulomb’s law, helps us to
find the electric field. A charge/a system of
charges like a dipole experience a force/torque
in an electric field. Work is required to be done
to provide a specific orientation to a dipole with
respect to an electric field.
Answer the following questions based on the
above :
(a) Consider a uniformly charged thin conducting
shell of radius R. Plot a graph showing the
→
−
variation of | E | with distance 𝑟 from the
centre, for points 0 ≤ r ≤ 3R.
A Answer
¥ Set 1, 1 Marks
(b) The figure shows the variation of potential 𝑉
with 1/r for two point charges Q1 and Q2,
where V is the potential at a distance r due to a
point charge. Find Q1/Q2.
A Answer
𝑄
V=𝑘
¥ Set 1, 1 Marks
r
Slope of graph is proportional to 𝑄
𝑄1 tan 60
=
=
3
◦
𝑄2 tan 30
◦
(c) An electric dipole of dipole moment of
−7
6 × 10 C − m is kept in a uniform electric field
4
of 10 N/C such that the dipole moment and
the electric field are parallel. Calculate the
potential energy of the dipole.
A Answer
𝑈 = −𝑝𝐸 cos 𝜃
◦
𝜃=0
4
−7
𝑈 = − 6 × 10 × 10
−3
𝑈 = −6 × 10 J
¥ 2 Marks
OR
An electric dipole of dipole moment 𝑝® is initially
→
−
→
−
kept in a uniform electric field E such that p is
→
−
perpendicular to E . Find the amount of work
done in rotating the dipole to a position at which
→
−
𝑝® becomes antiparallel to E .
A Answer
¥ 2 Marks
Work done W = −pE ( cos 𝜃2 − cos 𝜃1)
where 𝜃2 = 180 , 𝜃1 = 90
◦
◦
◦
◦
⇒ W = −pE ( cos 180 − cos 90 )
W = +pE
§ 2023 55/5/1 All Sets
š Question 19
The capacitors, each of 4 𝜇 F are to be connected
in such a way that the effective capacitance of
the combination is 6 𝜇 F. This can be achieved by
connecting
(a) All three in parallel
(b) All three in series
(c) Two of them connected in series and the
combination in parallel to the third.
(d) Two of them connected in parallel and the
combination in series to the third.
A Answer
¥ Set 1, 1 Marks
Two of them connected in series and the
combination in parallel to the third
š Question 20
Depict the orientation of an electric dipole in
(a) stable
(b) unstable equilibrium
in an external uniform electric field. Write the
potential energy of the dipole in each case.
A Answer
¥ 2 Marks
(a) 𝑈 = −𝑝𝐸
(b) 𝑈 = 𝑝𝐸
𝜃 = 0°
𝜃 = 180°
š Question 21
Three point charges 𝑄, 𝑞 and −𝑞 are kept at the
vertices of an equilateral triangle of side L as
shown in figure. What is
(i) the electrostatic potential energy of the
arrangement?
(ii) the potential at point D ?
A Answer
¥ Set 3, 1 Marks
2
𝑘𝑄𝑞 𝑘𝑄𝑞 𝑘𝑞
+
−
(i) 𝑈 = −
𝐿
𝐿
𝐿
2
𝑘𝑞
𝑈=−
𝐿
(ii) 𝑉 = 𝑉𝐷𝐴 + 𝑉𝐷𝐵 + 𝑉𝐷𝐶
2 𝐾𝑄
2 𝐾𝑞
2 𝐾𝑞
+
𝑉= √ −
𝐿
𝐿
𝐿 3
2 𝐾𝑄
𝑉= √
𝐿 3
š Question 22
A capacitor is a system of two conductors
separated by an insulator. The two conductors
have equal and opposite charges with a potential
difference between them. The capacitance of a
capacitor depends on the geometrical
configuration (shape, size and separation) of the
system and also on the nature of the insulator
separating the two conductors. They are used to
store charges. Like resistors, capacitors can be
arranged in series or parallel or a combination
of both to obtain desired value of capacitance.
(i) Find the equivalent capacitance between
points A and B in the given diagram
(ii) A dielectric slab is inserted between the plates
of a parallel plate capacitor. The electric field
between the plates decreases. Explain.
(iii) A capacitor A of capacitance C, having charge
Q is connected across another uncharged
capacitor B of capacitance 2C. Find an
expression for
(a) the potential difference across the
combination
(b) the charge lost by capacitor A.
OR
Two slabs of dielectric constants 2K and K fill the
space between the plates of a parallel plate
capacitor of plate area A and plate separation 𝑑
as shown in figure. Find an expression for
capacitance of the system.
A Answer
¥ 4 Marks
(i)
𝐶net = 𝐶 + 𝐶 = 2𝐶
(ii) Within the dielectric slab, the induced electric
field due to polarization decreases the electric
field. As the dielectric slab is introduced there
is some charge distribution in the slab and
because of this the electric field between the
two plates is decreased, due to which the
capacitor can hold more charge. Thus, the
capacity to hold charge of the capacitor is
increased.
E0
E=
(iii) (a)
𝐾
𝑄
𝑄Total
=
𝑉 =
𝐶eqi
3𝐶
𝑄 𝑉
′
𝑉 =
=
3𝐶 3
′
(b)
′
𝑄𝐴
=𝐶×
𝑉
3
=
𝑄
3
𝑄 𝐴 = 𝐶𝑉 = 𝑄
Charge lost by capacitor A is
Δ𝑄 = 𝑄 −
𝑄
3
=
2𝑄
OR
Capacitance of left portion, C1 =
3
6 𝐾 ∈0 𝐴
𝑑
3 𝐾 ∈0 𝐴
Capacitance of right portion, C2 =
2𝑑
As the capacitors are in series
1
1
1
+
=
𝐶𝑒𝑞𝑖
1
𝐶1
𝐶2
𝑑
2𝑑
5𝑑
=
+
=
𝐶𝑒𝑞𝑖 6𝐾 ∈0 𝐴 3𝐾 ∈0 𝐴 6𝐾 𝐴 ∈0
6 𝐾 𝐴 ∈0
𝐶𝑒𝑞𝑖 =
5𝑑
§ 2022 Term I 55/2
š Question 23
The electric potential V at any point (𝑥, 𝑦, 𝑧) is
2
given by V = 3𝑥 where 𝑥 is in metres and V in
volts. The electric field at the point (1 m, 0, 2 m)
(a) 6 V/m along −𝑥 -axis
(b) 6 V/m along +𝑥 -axis
(c) 1.5 V/m along −𝑥 -axis
(d) 1.5 V/m along +𝑥 -axis
A Answer
−𝑑 V
𝑑
2
E=
=−
3 𝑥 = −6 𝑥
𝑑𝑥
𝑑𝑥
Putting 𝑥 = 1,
E = 6 V/m along −𝑥 direction.
¥ Set 1, 1 Marks
š Question 24
Which of the diagrams correctly represents the
electric field between two charged plates if a
neutral conductor is placed in between the
plates?
A Answer
¥ Set 1, 1 Marks
Option (d) is correct.
Upper side of the neutral conductor will be
negatively charged. Lower side of the neutral
conductor will be positively charged. Then the
field lines will be from negative to positive as
shown in the diagram.
š Question 25
A variable capacitor is connected to a 200 V
battery. If its capacitance is changed from 2 𝜇 F
to X 𝜇 F, the decrease in energy of the capacitor is
(a) 1 𝜇 F
(b) 2 𝜇 F
(c) 3 𝜇 F
(d) 4 𝜇 F
A Answer
1
¥ Set 1, 1 Marks
2
Energy = CV
2
1
−6
2
−2
E1 = × 2 × 10 × 200 = 4 × 10 J
2
1
−6
2
E2 = × (𝑋) × 10 × 200 J
2
Decrease in energy = E1 − E2
−2
−2
−2
⇒ 2 × 10 = 4 × 10 − 2 × ( X) × 10
⇒ 2 = 4 − 2𝑋
∴ X = 1𝜇 F
š Question 26
A +3.0nC charge 𝑄 is initially at rest at a distance
of 𝑟1 = 10 cm from a +5.0nC charge 𝑞 fixed at the
origin. The charge 𝑄 is moved away from 𝑞 to a
new position at 𝑟2 = 15 cm. In this process work
done by the field is
−5
5
(a) 1.29 × 10 J
(b) 3.6 × 10 J
(c) −4.5 × 10
−7
J
A Answer
(d) 4.5 × 10
−7
J
¥ Set 1, 1 Marks
Work done = 𝑈𝑖 − 𝑈 𝑓




 1

1
1
1

= 𝑘𝑞𝑄 
−
= 𝑘𝑞𝑄
−

15
10
𝑟1 𝑟2


 100 100 


1
𝑘𝑞𝑄 1
𝑘𝑞𝑄 1
−
=
=
−
2
−
2
( 10 ) 10 15
( 10 ) 30
9
−9
−9 9 × 10 × 5 × 10
× 3 × 10
1
=
−
2
( 10 )
30
−7
= 4.5 × 10 J
š Question 27
A car battery is charged by a 12 V supply, and
5
energy stored in it is 7.20 × 10 J. The charge
passed through the battery is
4
3
(a) 6.0 × 10 C
(b) 5.8 × 10 J
6
5
(c) 8.64 × 10 J
(d) 1.6 × 10 C
A Answer
𝑄 = 𝑊/𝑉
5
= 7.2 × 10 /12
4
= 6 × 10 C
¥ Set 1, 1 Marks
š Question 28
Two charges 14 mC and −4 𝜇 C are placed at
(−12 cm, 0, 0) and (12
cm, 0, 0) in an external
𝐵
electric field 𝐸 = 2
𝑟
6
2
where 𝐵 = 1.2 × 10 N/(cm ) and 𝑟 is in metres.
The electrostatic potential energy of the
configuration is
(a) 97.9 J
(b) 102.1 J
(c) 2.1 J
(d) –97.9 J
A Answer
¥ Set 1, 1 Marks
Potential energy of the system
= the total work done in assembling the
configuration
𝑞1𝑞2
𝑘𝑞1𝑞2
= 𝑞1𝑉 + 𝑞2𝑉 +
= 𝑞1𝑉 + 𝑞2𝑉 +
4𝜋𝜖0𝑟
𝑟
"
#
|𝛿𝑉 |
|𝛿𝑉 |
=+
As we know, | E | = −
𝛿𝑙
𝛿𝑙
6
𝐵
𝐵 1.2 × 10
𝑉 = 𝐸𝑟 = 2 𝑟 = =
𝑟
𝑟
𝑟
6
1.2 × 10
−6
= 𝑞1𝑉 + 𝑞2𝑉 =
(
14
−
4
)
×
10
12 × 10−2
9
−6
−6
𝑘𝑞1𝑞2
9 × 10 × 14 × 10 × 4 × 10
And,
=−
−
2
( 24 × 10 )
𝑟
Therefore, combining all
𝑘𝑞1𝑞2
= 100 − 2.1 = 97.9 J
= 𝑞1𝑉 + 𝑞2𝑉 +
𝑟
š Question 29
Equipotentials at a large distance from a
collection of charges whose total sum is not zero
are
(a) spheres
(b) planes
A Answer
(c) ellipsoids
(d) paraboloids
¥ Set 1, 1 Marks
(a) spheres is the correct answer.
The collection of charges, whose total sum is not
zero, at great distance may be considered as a
point charge. So, potential is inversely
proportional to the distance from the charge. So,
electric potentials due to point charge are the
same for all equidistant points. The locus of
these equidistant points, which are at same
potential, is a sphere.
š Question 30
Four charges −𝑞, −𝑞, +𝑞 and +𝑞 are placed at the
corners of a square of side 2L is shown in figure.
The electric potential at point A midway
between the two charges +𝑞 and +𝑞 is
(a)
1 2𝑞
1
1−√
4𝜋𝜀0 L
5
1
1 𝑞
1−√
(c)
4𝜋𝜀0 2L
5
A Answer
(b)
1 2𝑞
1+√
4𝜋𝜀0 L
5
(d) zero
¥ Set 1, 1 Marks
Electric potential due to two +𝑞 charges
1
2𝑞
=
4𝜋𝜀0
×
𝐿
1
Electric potential due to two −𝑞 charges
1
−2 𝑞
=
×√
4𝜋𝜀0
5𝐿
Total potential at A =
1
4𝜋𝜀0
×
2𝑞
1
1−√
𝐿
5
§ 2021 55/C Compart
š Question 31
(i.a) Why does the electric field inside a dielectric
slab decrease when kept in an external
electric field?
(i.b) Derive an expression for the capacitance of a
parallel plate capacitor filled with a medium
of dielectric constant K.
(ii) A charge 𝑞 = 2 𝜇 C is placed at the centre of a
sphere of radius 20 cm. What is the amount
of work done in moving 4 𝜇 C from one point
to another point on its surface?
® of a
(iii) Write a relation for polarisation P
dielectric material in the presence of an
external electric field.
A Answer
¥ 5 Marks
(i.a) A dielectric material gets polarized when it is
placed in an external electric field. The field
produced due to the polarization of material
reduces the effect of external electric field.
Hence, the electric field inside a dielectric
decreases.
(i.b) Electric field in vacuum between the plates
𝜎
= E0 =
𝜀𝑜
Electric field in dielectric between the plates,
𝐸0
E=
𝐾
Potential difference between the capacitor
plates V = Et + E0 ( d − t)
where ’ 𝑡 ’ is the thickness of dielectric slab.
𝐸0
V = t + E0 ( d − t)
𝐾
𝜎 𝑡
𝜎 𝑡 + 𝐾 (𝑑 − 𝑡)
V=
+ ( d − t) =
𝜀𝑜 𝐾
𝜀𝑜
𝐾
𝜀0 𝐴𝐾
𝑄
⇒C=
As C =
𝑉
𝑡 + 𝐾 (𝑑 − 𝑡)
(ii) The surface of the sphere is equipotential. So,
the work done in moving the charge from
one point to the other is zero.
W = 𝑞ΔV = 0 (∵ ΔV = 0)
(iii) 𝑃 = 𝜖𝑜 𝜒𝑒 𝐸
Theory Recap
Relation between Dielectric and Polarisation
The dipole moment per unit volume is called
polarisation and is denoted by P. For linear
isotropic dielectrics,
𝑃 = 𝜖𝑜 𝜒𝑒 𝐸
where 𝜒𝑒 is a constant characteristic of the
dielectric and is known as the electric
susceptibility of the dielectric medium.
š OR
(i) Obtain an expression for the potential energy
of an electric dipole placed in a uniform
electric field.
(ii) Three capacitors of capacitance C1, C2 and C3
are connected in series to a source of V volt.
Show that the total energy stored in the
combination of capacitors is equal to sum of
the energy stored in individual capacitors.
(iii) A capacitor of capacitance C is connected
across a battery. After charging, the battery is
disconnected and the separation between the
plates is doubled. How will
(a) the capacitance of the capacitor, and
(b) the electric field between the plates be
affected ? Justify your answer.
A Answer
¥ 5 Marks
(i) Consider a dipole with charges 𝑞1 = +𝑞 and
𝑞2 = −𝑞 is placed in a uniform electric field E,
as shown in Fig The dipole experiences no net
force; but experiences a torque 𝝉 given by
𝝉 = p × E which will tend to rotate it (unless p
is parallel or antiparallel to E ).
Suppose an external torque 𝝉ext is applied in
such a manner that it just neutralizes this
torque and rotates it in the plane of paper
from angle 𝜃𝑜 to angle 𝜃1 at an infinitesimal
angular speed and without angular
acceleration. The amount of work done by the
external
torque
will
be
given
by
∫
∫
𝜃1
𝜃1
𝜏𝑒𝑥𝑡 (𝜃)𝑑𝜃 =
𝑊=
𝜃0
𝑝𝐸 sin 𝜃𝑑𝜃
𝜃0
W = U = 𝑝𝐸 ( cos 𝜃0 − cos 𝜃1)
(ii) U =
1
𝐶eff
U=
1𝑄
2
2 𝐶eff
=
1
+
𝐶1
1𝑄
+
𝐶2
2
2 𝐶1
1
+
1𝑄
1
𝐶3
2
2 𝐶2
+
1𝑄
2
2 𝐶3
U = U1 + U2 + U3
(iii) When battery is disconnected then charge (𝑞)
remains constant.
(a) Capacitance is halved
𝜀𝑜 𝐴 𝐶
C =
=
2𝑑
2
′
(b) Electric field (E) is unaffected.
𝑞
𝜎
𝐸= =
𝜀 𝜀𝑜 𝐴
Alternatively for effect on electric field.
𝑉
E=
𝑑
𝑄
𝑄
′
= 2V
V = ′=
𝐶 𝐶/2
′
𝑉
2
𝑉
𝑉
′
E =
=
= =E
𝑑′ 2𝑑 𝑑
§ 2020 55/C All Sets
š Question 32
A charge particle is placed between the plates of
a charged parallel plate capacitor. It experiences
a force F. If one of the plates is removed, the
force on the charge particle becomes
F
(a) F
(b) 2F
(c)
(d) Zero
2
A Answer
¥ Set 1, 1 Marks
F
is the correct option.
2
š Question 33
An air-filled parallel plate capacitor is connected
across a battery. After it is fully charged, the
battery is disconnected. Now a dielectric slab is
inserted between the plates of the capacitor to
fill the space completely. Then the
(a) capacitance will decrease.
(b) electric field between the plates will increase.
(c) potential difference between the plates will
increase.
(d) charge on plates will remain the same.
A Answer
¥ Set 2, 1 Marks
(d) change on plates will remain the same
š Question 34
Two point charges 𝑞 and −𝑞 are located at
( 0, 0, −𝑎) and ( 0, 0, 𝑎) respectively.
(a) Depict the equipotential surfaces due to this
arrangement.
(b) Find the amount of work done in moving a
small test charge q0 from point (𝑙, 0, 0) to
( 0, 0, 0)
A Answer
¥ 2 Marks
(a)
(b) W = 𝑞0 ΔV
As a small test charge 𝑞0 is moving along
𝑥−axis which is equipotential line for a given
system,
therefore ΔV = 0, Hence W = 0
š Question 35
(a) Consider a system of two parallel metal plates
of area 𝐴, each placed at a separation 𝑑 in air.
Derive the expression for the capacitance of
this parallel plate capacitor.
(b) If the two plates of the capacitor have +𝑞 and
−𝑞 charges, respectively, find the force
experienced by the negative plate due to the
positive plate.
(c) A network of four capacitors each of
capacitance 12 𝜇 F is connected to a battery as
shown in the figure. Find the total charge
stored in the network.
A Answer
¥ 5 Marks
(a) Electric field believes the plates of parallel
plate capacitor.
𝜎
𝑄
=
E=
𝜖0 𝐴𝜖0
𝜎
We know V = Ed =
𝑑
𝐴𝜖0
𝑄 𝜖0 𝐴
As capacitance C = =
𝑉
𝑑
(b) Electric Field due to the positive plate on the
negative plate
𝜎
𝜎
E=
=
2𝜖0 2 𝐴𝜖0
Hence Force experienced by negative plate
due to positive plate
2
𝑞
𝑞
F = −qE = −𝑞 ×
=−
2 𝐴𝜖0
2 𝐴𝜖0
−𝑣𝑒 sign shows attractive force.
(c) C2, C3 and C4 are connected in series.
1
𝐶𝑠
=
1
+
1
𝐶2 𝐶3
Cs = 4 𝜇 F
+
1
𝐶4
=
1
12
+
1
12
+
1
12
Equivalent capacitance of the Network
C = Cs + C4
= 4𝜇 F + 12 𝜇 F
= 16 𝜇 F
Total charge
−16
𝑄 = 𝐶𝑉 = 16 × 10 × 100 = 1600𝜇𝐶
š Question 36
Define the SI unit of capacitance.
A Answer
¥ Set 3, 1 Marks
When a charge of one coulomb develop potential
of one volt between the plates of capacitor its
capacity is said to be one farad.
−1
1 F = 1 CV
§ 2020 55/5/1 All Sets
š Question 37
−1
The physical quantity having SI unit NC
A Answer
m is
¥ Set 1, 1 Marks
Electrostatic potential difference or Electric
potential
š Question 38
Obtain the expression for the energy stored in a
capacitor connected across a dc battery. Hence
define energy density of the capacitor.
A Answer
¥ 2 Marks
Let the charge on the capacitor plates at any
instant, during charging process be 𝑞, amount of
work done to supply further charge 𝑑𝑞 to the
capacitor
𝑑𝑊 = 𝑉 𝑑𝑞
where 𝑉 is the potential difference and
𝑞
equals to
𝐶
Total work done to charge the capacitor upto
charge 𝑄
∫
𝑄
𝑊=
𝑉 𝑑𝑞
∫0 𝑄
2
𝑄 1 2 1
𝑞
𝑑𝑞 =
𝐶𝑉 = 𝑄𝑉
=
2𝐶 2
2
0 𝐶
Since Energy stored = work done
2
𝑄 1 2 1
⇒𝑈 =
𝐶𝑉 = 𝑄𝑉
2𝐶 2
2
Energy density: Electrical energy stored per unit
volume is known as energy density.
1
2
1𝜎
2
Energy density = 𝜀0 𝐸 =
2
2 𝜀0
š Question 39
(a) Two point charges 𝑞1 and 𝑞2 are kept at a
distance of 𝑟12 in air. Deduce the expression
for the electrostatic potential energy of this
system.
(b) If an external electric field (𝐸) is applied on
the system, write the expression for the total
energy of this system.
A Answer
¥ 2 Marks
(a) Work done in bringing the charge 𝑞2, from
infinity, to a point
= 𝑞2 × potential at the point due to charge 𝑞1
= 𝑞2 ×
1 q1
4𝜋𝜀0 𝑟12
∴ potential energy of the system =
1 𝑞1𝑞2
4𝜋𝜀0 𝑟12
(b) Let the potentials, at two points, due to an
external electric field (E) be 𝑉1 and V2
respectively.
Now the total energy of the system is:
𝑞1𝑉1 + 𝑞2𝑉2 +
1 𝑞1𝑞2
4𝜋𝜀0 𝑟12
§ 2019 55/5/1 All Sets
š Question 40
Why is the electrostatic potential inside a
charged conducting shell constant throughout
the volume of the conductor ?
A Answer
¥ Set 1, 1 Marks
E = 0 inside the conductor and has no tangential
component on the surface.
No work is done in moving charge inside or on
the surface of the conductor. Therefore, Potential
is constant.
OR
Because E = 0 inside the conductor
OR
No work is done in moving a charge inside the
conductor
š Question 41
Two identical capacitors of 12 pF each are
connected in series across a 50 V battery.
Calculate the electrostatic energy stored in the
combination. If these were connected in parallel
across the same battery, find out the value of the
energy stored in this combination.
A Answer
¥ 2 Marks
𝐶 𝑠 = 6 𝑝𝐹
1
2
𝐸 𝑠 = 𝐶 𝑠𝑉 =
1
× 6 × 10
−12
× 50 × 50
2
2
−12
−9
= 7500 × 10 J = 7.5 × 10 J
𝐶 𝑝 = 24 𝑝𝐹
1
2
𝐸 𝑝 = 𝐶 𝑝𝑉 =
1
2
2
−8
= 3 × 10 J
× 24 × 50 × 50 × 10
−12
š Question 42
A charge Q is distributed over the surfaces of
two concentric hollow spheres of radii r and R
(R>>r), such that their surface charge densities
are equal. Derive the expression for the
potential at the common centre.
A Answer
¥ 3 Marks
2
𝑄 = 𝑞1 + 𝑞2 = 4𝜋𝜎 𝑟 + 𝑅
2
Potential at common centre
h
i
𝑞1 𝑞2
1
𝑉=
=
+
4𝜋𝜀0 𝑟
𝑅
2
1
4𝜋𝑟 𝜎
4𝜋𝜀0
×
2
+
4𝜋𝑅 𝜎
𝑟
𝑅
(𝑟 + 𝑅)𝜎
1
𝑄(𝑟 + 𝑅)
=
=
2
2
𝜀0
4𝜋𝜀0 𝑟 + 𝑅
š OR
Three concentric metallic shells A, B and C of
radii a, b and c ( a < b < c) have surface charge
densities +𝜎, −𝜎 and +𝜎 respectively as shown.
(a) Obtain the expressions for the potential of
three shells A, B and C.
(b) If shells A and C are at the same potential,
obtain the relation between a, b and c.
A Answer
¥ 3 Marks
𝑘𝑄 𝐴 𝑘𝑄 𝐵 𝑘𝑄𝐶
(a) 𝑉𝐴 =
+
+
𝑎
𝑏
𝑐
2
2
2
4𝜋𝑎 𝜎 4𝜋𝑏 𝜎 4𝜋𝑐 𝜎
1
−
+
𝑉𝐴 =
4𝜋𝜀0
𝑎
𝑏
𝑐
𝜎
= [𝑎 − 𝑏 + 𝑐]
𝜀0
𝑘𝑄 𝐴 𝑘𝑄 𝐵 𝑘𝑄𝐶
𝑉𝐵 =
+
+
𝑏
𝑏
𝑐
2
2
2
4𝜋𝑎 𝜎 4𝜋𝑏 𝜎 4𝜋𝑐 𝜎
1
−
+
𝑉𝐵 =
4𝜋𝜀0
𝑏
𝑏
𝑐
2
2
𝜎 𝑎 −𝑏
𝑉𝐵 =
+𝑐
𝜀0
𝑏
𝑘𝑄 𝐴 𝑘𝑄 𝐵 𝑘𝑄𝐶
+
+
𝑉𝐶 =
𝑐 𝑐
𝑐
2
2
2
1
4𝜋𝑎 𝜎 4𝜋𝑏 𝜎 4𝜋𝑐 𝜎
𝑉𝐶 =
−
+
4𝜋𝜀0
𝑐
𝑐
𝑐
2
2
2
𝜎 𝑎 −𝑏 +𝑐
𝑉𝐶 =
𝜀0
𝑐
(b) 𝑉𝐴 = 𝑉𝐶
2
2
𝑎 −𝑏
𝑎−𝑏+𝑐 =
+𝑐
𝑐
𝑐 = 𝑎+𝑏
š Question 43
Two identical capacitors of 10 pF each are
connected in turn (i) in series, and (ii) in parallel
across a 20 V battery. Calculate the potential
difference across each capacitor in the first case
and charge acquired by each capacitor in the
second case.
A Answer
(i) V = 10 V
¥ Set 2, 2 Marks
(ii) Q = CV = 200 pC
š Question 44
The figure shows a network of three capacitors
C1 = 2 𝜇 F; C2 = 6 𝜇 F and C3 = 3 𝜇 F connected
across a battery of 10 V. If a charge of 6 𝜇 C is
acquired by the capacitor C3, calculate the
charge acquired by C1.
A Answer
C23 = ( 6 + 3)𝜇 F = 9 𝜇 F
𝑞
𝑞
⇒ V1 =
C1 =
𝑉1
2 𝜇𝐹
𝑞
𝑞
C23 =
⇒ V2 =
𝑉2
9 𝜇𝐹
V = V1 + V2
1
1
+
⇒ 10 = 𝑞
2 𝜇𝐹 9 𝜇𝐹
10 × 18
⇒𝑞=
𝜇 C = 16.4𝜇 C
11
¥ Set 3, 2 Marks
§ 2019 55(B)
š Question 45
Define the term dielectric constant of a material.
A Answer
¥ Set 1, 1 Marks
The dielectric constant is the ratio of the
capacitance of a capacitor filled with the given
material (permittivity of the medium) to the
capacitance of an identical capacitor in a
vacuum without the dielectric material
(permittivity of the vacuum).
𝜖
𝐶
𝜅=
OR 𝜅 =
𝜖0
𝐶0
š Question 46
A 600 pF capacitor is charged by a 100 V battery.
(i) Calculate the electrostatic energy stored by
the capacitor.
(ii) If the capacitor is disconnected from the
battery and connected to another 600 pF
capacitor, calculate the electrostatic energy
stored by the system.
A Answer
1
2
¥ 3 Marks
1
−12
U = CV = × 600 × 10
2
2
−6
U = 3 × 10 J
× 100 × 100
When connected across another capacitor of
capacitance 600pF, energy stored by the system
1
2
𝑞
′
𝑈 =
2 (𝐶1 + 𝐶2)
2
1 (𝐶1𝑉 )
=
2 (𝐶1 + 𝐶2)
2
−12
600 × 10
× 100
1
= ×
2
( 600 + 600) 10−12
′
−6
𝑈 = 1.5 × 10 J.
Å
Click to YouTube icon to see the full
question bank video
Click the photo to see the full question bank video
*
Unit II
Current Electricity
www.cbse.page
www.cuet.pw
’ Chapter 3
™ Current Electricity
Weightage for CBSE Board 2024
Unit I and II
Combined
Weightage is
16 Marks
À Chapter 1
À Chapter 2
À Chapter 3
It’s not fixed which chapters will
have more weightage, but overall weightage
will be 16 Marks. It may vary too in your
question paper.
` Syllabus
Electric current, flow of electric charges in a
metallic conductor, drift velocity, mobility and
their relation with electric current;
Ohm’s law, V−I characteristics (linear and
non-linear), electrical energy and power,
electrical resistivity and conductivity,
temperature dependence of resistance,
Internal resistance of a cell, potential difference
and emf of a cell, combination of cells in series
and in parallel, Kirchhoff’s rules,
Wheatstone bridge.
Deleted Topics
Carbon resistors, colour code for carbon
resistors;
Metre bridge, Potentiometer - principle and its
applications to measure potential difference and
for comparing EMF of two cells;
§ 2024 SQP
š Question 1
An ammeter of resistance 0.81 ohm reads up to 1
A. The value of the required shunt to increase
the range to 10 A is
(a) 0.9 ohm
(b) 0.09 ohm
(c) 0.03 ohm
(d) 0.3 ohm
¥ Set 1, 1 Marks
A Answer
9 × S = 1 × 0.81
S=
0.81
9
= 0.09 ohm
š Question 2
A heating element using nichrome connected to
a 230 V supply draws an initial current of 3.2 A
which settles after a few seconds to a steady
value of 2.8 A. What is the steady temperature of
the heating element if the room temperature is
27 °C and the temperature coefficient of
−4
−1
resistance of nichrome is 1.70 × 10 ℃ ?
¥ 2 Marks
A Answer
Given: 𝑉 = 230 V, I0 = 3.2 A, I = 2.8 A,
◦
−4◦ −1
𝑇0 = 27 C, 𝛼 = 1.70 × 10 C
Using equation R = R0 ( 1 + 𝛼ΔT)
V
I
=
V
I◦
[ 1 + 𝛼ΔT]
and solving ΔT = 840, i.e. T = 840 + 27 = 867 C
◦
§ 2023 55/C Compart
š Question 3
Assertion (A) : The temperature coefficient of
resistance is positive for metals and negative for
semi-conductors.
Reason (R) : The charge carriers in metals are
negatively charged whereas in semiconductors
they are positively charged.
(a) Both Assertion (A) and Reason (R) are true and
Reason (R) is the correct explanation of the
Assertion (A).
(b) Both Assertion (A) and Reason (R) are true, but
Reason (R) is not the correct explanation of the
Assertion (A).
(c) Assertion (A) is true, but Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is also
false.
A Answer
¥ Set 1, 1 Marks
Assertion (A) is true, but Reason (R) is false.
In intrinsic semiconductors, the free electrons
(𝑛𝑒) and holes (𝑛ℎ) are the charge carriers.
š Question 4
(i) Derive the relation between the current and
the drift velocity of free electrons in a
conductor. Briefly explain the variation of
resistance of a conductor with rise in
temperature.
(ii) An ammeter, together with an unknown
resistance in series is connected across two
identical batteries, each of emf 1.5 V,
connected (i) in series, and (ii) in parallel.
If
the
current
recorded in the two cases be
1
1
A and
A respectively, calculate the
2
3
internal resistance of each battery.
A Answer
¥ 5 Marks
(i) Total charge transported across the area A in
time Δt is ΔQ = −𝑛𝑒Avd Δt
....(1)
Also the amount of charge crossing area A in
time Δt is ΔQ = IΔt
....(2)
Comparing equation (1) and (2) ⇒ I = neAvd
With increase in temperature, average speed
of electrons increases resulting in more
frequent collisions.
𝑚𝑙
As R = 2
𝑛𝑒 𝜏𝐴
Hence relaxation time 𝜏 decreases, resistance
increases.
𝐸
(ii) I =
𝑅+𝑟
For series
For parallel
1
=
3
2 𝑅 + 2𝑟
1
1.5
3
=
𝑅+
𝑟
R + 2r = 6
2R + r = 9
2
After solving both equation 𝑟 = 1Ω
§ 2023 55/2/1 All Sets
š Question 5
A current of 0.8 A flows in a conductor of 40Ω
for 1 minute. The heat produced in the
conductor will be
(a) 1445 J
(b) 1536 J
(c) 1569 J
(d) 1640 J
¥ Set 1, 1 Marks
A Answer
From Ohm’s law, we know that 𝑉 = 𝐼𝑅
Given: 𝐼 = 0.8 A, R = 40Ω, t = 1 min ( 60sec)
Now, 𝑉 = 𝐼𝑅 = 0.8 × 40 = 32 volt
Power, 𝑃 = 𝑉 × 𝐼 = 32 × 0.8 = 25.6 watt
Heat = 𝑃 × 𝑡 = 25.6 × 60 = 1536 J
š Question 6
A steady current of 8 mA flows through a wire.
The number of electrons passing through a
cross-section of the wire in 10 s is
16
(a) 4.0 × 10
17
(b) 5.0 × 10
16
(c) 1.6 × 10
17
(d) 1.0 × 10
¥ Set 2, 1 Marks
A Answer
Given: I = 8 mA = 8 × 10
𝑡 = 10 s
−3
A
𝑒 = 1.6 × 10
−19
We know that,
𝑄 = 𝑛𝑒
𝑄 = 𝐼𝑡
Equating both the above equations, we get
𝑛𝑒 = 𝐼𝑡
−19
−3
𝑛 × 1.6 × 10
= 8 × 10 × 10
−2
−2
80 10
8 × 10
17
=
×
=
5
×
10
𝑛=
−
19
−
19
1.6 × 10
16 10
Theory Recap
Graph for temperature dependence of resistivity
for a typical semiconductor.
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*
Unit III
Magnetic Effects of
Current and Magnetism
www.cuet.pw
’ Chapter 4
™ Moving Charges
and Magnetism
Weightage for CBSE Board 2024
Unit III and IV
Combined
Weightage
is 17 Marks
À Chapter 4
À Chapter 5
À Chapter 6
À Chapter 7
It’s not fixed which chapters will
have more weightage, but overall weightage
will be 17 Marks. It may vary too in your
question paper.
§ 2023 55/C Compart
š Question 7
An electron with velocity 𝑣® = 𝑣𝑥 𝑖ˆ + 𝑣 𝑦 𝑗ˆ moves
through a magnetic field 𝐵® = 𝐵𝑥 𝑖ˆ − 𝐵 𝑦 𝑗ˆ .
The force 𝐹® on the electron is
(a) −𝑒 𝑣𝑥 𝐵 𝑦 − 𝑣 𝑦 𝐵𝑥 𝑘ˆ
(c) −𝑒 𝑣𝑥 𝐵 𝑦 + 𝑣 𝑦 𝐵𝑥 𝑘ˆ
(b) 𝑒 𝑣𝑥 𝐵 𝑦 − 𝑣 𝑦 𝐵𝑥 𝑘ˆ
(d) 𝑒 𝑣𝑥 𝐵 𝑦 + 𝑣 𝑦 𝐵𝑥 𝑘ˆ
¥ Set 1, 1 Marks
A Answer
𝑒 𝑣𝑥 𝐵 𝑦 + 𝑣 𝑦 𝐵𝑥 𝑘ˆ
Lorentz Force
F = 𝑞[ E ( r) + 𝑣 × B ( r)] ≡ Felectric + Fmagnetic
§ 2023 55(B)
š Question 8
An electron is revolving around a nucleus in a
circular orbit of radius r with a constant speed 𝑣.
The magnetic moment associated with the
circulating current is
evr
evr
(a)
(b) 4 evr
(c) 2 evr
(d)
4
2
A Answer
¥ Set 1, 1 Marks
evr
(d)
2
Explanation
Current flowing in the circular orbit:
e
𝐼 = = ev
𝑇
where, 𝜈 is the frequency of revolution of
electron
v
∴ 𝐼 =e×
2𝜋 r
2
Area of the circular orbit: A = 𝜋r
Magnetic moment of an electron :
ev
evr
2
𝜇 = IA =
× 𝜋r =
2𝜋 r
2
§ 2022 Term I
š Question 9
Two parallel conductors carrying current of
4.0 A and 10.0 A are placed 2.5 cm apart in
vacuum. The force per unit length between them
is
−5
−2
(a) 6.4 × 10 N/m
(b) 6.4 × 10 N/m
−4
−4
(c) 4.6 × 10 N/m
(d) 3.2 × 10 N/m
A Answer
¥ Set 1, 1 Marks
F 𝜇0𝑖1𝑖2
=
𝑙
2𝜋𝑟
−7
4𝜋 × 10 × 4 × 10
−4
=
=
3
.
2
×
10
N
/
m
2𝜋 × 2.5 × 10−2
Theory Recap
Formula for force between two parallel currents,
𝜇0𝐼 𝑎𝐼𝑏
𝐹𝑏𝑎 = 𝐼𝑏𝐿𝐵𝑎 =
𝐿
2𝜋𝑑
Formula for force per unit length
F
𝜇0𝑖1𝑖2
=
𝑙
2𝜋𝑟
Parallel currents attract, and antiparallel
currents repel. This rule is the opposite of what
we find in electrostatics. Like (same sign)
charges repel each other, but like (parallel)
currents attract each other.
Define the ampere (A), which is one of the
seven SI base units.
The ampere is the value of that steady current
which, when maintained in each of the two very
long, straight, parallel conductors of negligible
cross-section, and placed one metre apart in
vacuum, would produce on each of these
−7
conductors a force equal to 2 × 10 newtons per
metre of length.
š Question 10
Assertion (A) : When radius of a current
carrying loop is doubled, its magnetic moment
becomes four times.
Reason (R) : The magnetic moment of a current
carrying loop is directly proportional to the area
of the loop.
(a) Both Assertion (A) and Reason (R) are true and
Reason (R) is the correct explanation of the
Assertion (A).
(b) Both Assertion (A) and Reason (R) are true, but
Reason (R) is not the correct explanation of
the Assertion (A).
(c) Assertion (A) is true, but Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is also
false.
¥ Set 1, 1 Marks
A Answer
Both Assertion (A) and Reason (R) are true and
Reason (R) is the correct explanation of the
Assertion (A).
Theory Recap
Relation between current and magnetic moment
of a current carrying loop is 𝑀 = 𝐼 𝐴
𝐴 = 𝜋𝑅
2
𝑀∝𝑅
2
So, If the radius is doubled, the moment becomes
four times. The assertion is true.
Since 𝑀 = 𝐼 𝐴, ∴ 𝑀 ∝ 𝐴. The reason is true.
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*
Unit III
Magnetic Effects of
Current and Magnetism
www.cbse.page
www.cuet.pw
’ Chapter 5
™ Magnetism and
Matter
Weightage for CBSE Board 2024
Unit III and IV
Combined
Weightage
is 17 Marks
À Chapter 4
À Chapter 5
À Chapter 6
À Chapter 7
It’s not fixed which chapters will
have more weightage, but overall weightage
will be 17 Marks. It may vary too in your
question paper.
` Syllabus
Bar magnet, bar magnet as an equivalent
solenoid (qualitative treatment only), magnetic
field intensity due to a magnetic dipole (bar
magnet) along its axis and perpendicular to its
axis (qualitative treatment only), torque on a
magnetic dipole (bar magnet) in a uniform
magnetic field (qualitative treatment only),
magnetic field lines.
Magnetic properties of materialsPara-, dia- and ferro - magnetic substances with
examples, Magnetization of materials, effect of
temperature on magnetic properties.
§ 2024 Additional Questions
¥ Released by CBSE on 6th
September, 2023
š Question 11
A rod when suspended in a uniform magnetic
field aligns itself perpendicular to the magnetic
field as shown below.
Which of the following statements is/are true for
the rod?
(P) Every atom in the rod, has a zero magnetic
moment.
(Q) The rod is attracted when taken near the
poles of a strong magnet.
(R) The relative permeability of the material of
the rod is slightly less than 1.
(S) The susceptibility of the material of the rod is
directly proportional to temperature.
(a) only Q
(c) only Q and S
A Answer
only P and R
(b) only P and R
(d) only R and S
¥ Set 1, 1 Marks
Click the photo to see the full question bank video
*
Unit IV
Electromagnetic
Induction and
Alternating Currents
www.cuet.pw
’ Chapter 6
™ Electromagnetic
Induction
Weightage for CBSE Board 2024
Unit III and IV
Combined
Weightage is
17 Marks
À Chapter 4
À Chapter 5
À Chapter 6
À Chapter 7
It’s not fixed which chapters will
have more weightage, but overall weightage
will be 17 Marks. It may vary too in your
question paper.
` Syllabus
Electromagnetic induction; Faraday’s laws,
induced EMF and current; Lenz’s Law, Self and
mutual induction.
§ 2024 SQP
š Question 12
In a coil of resistance 100Ω a current is induced
by changing the magnetic flux through it. The
variation of current with time is as shown in the
figure. The magnitude of change in flux through
coil is
(a) 200 Wb
(b) 275 Wb
(c) 225 Wb
(d) 250 Wb
A Answer
¥ Set 1, 1 Marks
ΔΦ
𝐼Δ𝑡 =
= Area under 𝐼 − 𝑡 graph, 𝑅 = 100 ohm
𝑅
ΔΦ
1 ΔΦ
𝑒=
,𝐼 =
Δ𝑡
𝑅 Δ𝑡
∴
1
ΔΦ = 100 × × 10 × 0.5 = 250 Wb.
2
§ 2023 55/C Compart
š Question 13
Answer the following, giving reasons :
(a) Induced emf does not always produce induced
current.
(b) The motion of a copper plate is damped when
it is allowed to oscillate between pole pieces of
a strong magnet.
(c) No power is consumed in an ac circuit
containing a pure inductor.
A Answer
¥ Set 3, 3 Marks
(a) Even in the open loop emf can be induced but
since circuit is open no current can flow.
(b) When copper plate oscillates between poles of
a strong magnet eddy currents are generated
in the plate opposing the change in flux.
(c) Power consumed by a pure inductor oscillates.
For first half of the cycle it is positive and for
the next half it is negative. So for a complete
cycle it is zero.
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*
Unit IV
Electromagnetic
Induction and
Alternating Currents
www.cbse.page
’ Chapter 7
™ Alternating
Current
Weightage for CBSE Board 2024
Unit III and IV
Combined
Weightage is
17 Marks
À Chapter 4
À Chapter 5
À Chapter 6
À Chapter 7
It’s not fixed which chapters will
have more weightage, but overall weightage
will be 17 Marks. It may vary too in your
question paper.
` Syllabus
Alternating currents, peak and RMS value of
alternating current/voltage; reactance and
impedance; LCR series circuit (phasors only),
resonance, power in AC circuits, power factor,
wattless current. AC generator, Transformer.
§ 2023 55/4/1 All Sets
š Question 14
The figure shows variation of current (I) with
time (𝑡 ) in four devices P, Q, R and S. The device
in which an alternating current flows is :
(a) P
(b) Q
A Answer
(d) S
(c) R
(d) S
¥ Set 1, 1 Marks
Theory Recap
AC Current moves sinusoidally and has
corresponding positive and negative values
during each cycle.
Click the photo to see the full question bank video
*
Unit V
Electromagnetic waves
www.cbse.page
’ Chapter 8
™ Electromagnetic
Waves
Weightage for CBSE Board 2024
Unit V and VI
Combined
Weightage is
18 Marks
À Chapters
2 Chapter 8
2 Chapter 9
2 Chapter 10
It’s not fixed which chapters will
have more weightage, but overall weightage
will be 18 Marks. It may vary too in your
question paper.
` Syllabus
Basic idea of displacement current,
Electromagnetic waves, their characteristics,
their transverse nature (qualitative idea only).
Electromagnetic spectrum (radio waves,
microwaves, infrared, visible, ultraviolet, X-rays,
gamma rays) including elementary facts about
their uses.
§ 2024 SQP
š Question 15
Identify the part of the electromagnetic
spectrum which:
(a) produces heating effect,
(b) is absorbed by the ozone layer in the
atmosphere,
(c) is used for studying crystal structure.
Write any one method of the production of each
of the above radiations.
A Answer
¥ 3 Marks
(a) Infrared
Infrared waves are produced by vibration of
atoms and molecules hot bodies and molecules
(b) Ultraviolet
UV Rays is produced and released when inner
shell electrons in atoms moving from one energy
level to a lower level.
(c) X rays
One common way to generate X-rays is to
bombard a metal target by high energy
electrons.
§ 2023 55/C Compart
š Question 16
Which of the following physical quantities
remain the same for X-ray, red light and radio
waves when travelling through a medium ?
(a) Wavelength
(c) Frequency
A Answer
(b) Speed
(d) Momentum
¥ Set 1, 1 Marks
Speed
š Question 17
Which of the following radiations has the
highest frequency ?
(a) Visible light
(b) Infrared rays
(c) Microwaves
(d) X-rays
A Answer
X-rays
¥ Set 2, 1 Marks
š Question 18
(a) The electric field of an electromagnetic wave
passing through vacuum is represented as
Ex = E0 sin (𝑘 z − 𝜔 t) . Identify the parameter
which is related to the (i) wavelength, and (ii)
the frequency of the wave in the above
equation.
(b) Write two properties of a medium that
determine the velocity of light in that medium.
¥ 2 Marks
2𝜋
(i) Parameter relating wavelength is 𝑘 =
𝜆
Parameter relating frequency is 𝜔(= 2𝜋𝜈)
A Answer
(ii) Electric properties of the medium
Magnetic properties of the medium
Alternatively:
−→ Permittivity (𝜀) of the medium
−→ Permeability (𝜇) of the medium
§ 2023 55(B) Compart
š Question 19
Speed of electromagnetic waves in vacuum is
1
1
√
(a) √
(c) 𝜇0𝜀0
(d)
(b) 𝜇0𝜀0
𝜇0𝜀0
𝜇0𝜀0
A Answer
√
¥ Set 1, 1 Marks
1
𝜇0𝜀0
š Question 20
Identify the part of electromagnetic spectrum
which is :
(i) suitable for radar systems.
(ii) sometimes referred to as heat waves.
Write their wavelength range.
A Answer
¥ Set 1, 2 Marks
(i) Microwave
Wavelength range (1mm − 0.1m)
(ii) Infrared wave
Wavelength range (700nm-1mm)
š OR
Write two characteristics of electromagnetic
waves. Name the radiation used to kill germs in
water purifiers. Write the range of their
frequency.
A Answer
¥ 2 Marks
Any two
▲ EMW travel with speed of light in vaccum.
▲ EMW carries energy and momentum.
▲ Speed of EMW in vaccum is given by 𝑐 = √
▲ EMW are transverse in nature
▲ In EMW electric and magnetic field are
1
𝜇0𝜀0
perpendicular to each other and to the direct
propagation.
Ultraviolet(UV)
15
17
Frequency range 10 − 10 Hz
§ 2023 55/5/1 All Sets
š Question 21
Which one of the following electromagnetic
radiation has the least wavelength?
(a) Gamma rays
(b) Microwaves
(c) Visible light
(d) X-rays
A Answer
¥ Set 1, 1 Marks
(a) Gamma rays
š Question 22
Identify the electromagnetic wave whose
wavelengths range is from about
−12
−8
(a) 10
m to about 10 m.
−3
−1
(b) 10 m to about 10 m.
Write one use of each.
A Answer
¥ Set 1, 2 Marks
(a) X-rays
Used as diagnostic tool in medicine
Treatment for certain forms of cancer
To study crystal structure
Alternatively
Gamma rays
Used in medicine to destroy the cancer cell.
(b) Microwaves
Used in radar system for aircraft navigation
š Question 23
Identify the electromagnetic radiation and write
its wavelength range, which is used to kill germs
in water purifier. Name the two sources of these
radiations.
A Answer
¥ Set 2, 2 Marks
Ultraviolet rays. It covers wavelengths ranging
−7
from about 4 × 10 m ( 400 nm) to
−10
6 × 10
m ( 0.6 nm) .
Sources :
(1) The Sun
(2) Any device in which inner shell electrons in
atoms move from one energy level to a lower
level.
š Question 24
Choose the correct option related to wavelengths
(𝜆) of different parts of electromagnetic
spectrum.
(a) 𝜆 x-rays < 𝜆 micro waves < 𝜆 radio waves < 𝜆 visible
(b) 𝜆 visible > 𝜆 x-rays > 𝜆 radio waves > 𝜆 micro waves
(c) 𝜆 radio wave > 𝜆 micro waves > 𝜆 visible > 𝜆 x-rays
(d) 𝜆 visible < 𝜆 micro waves < 𝜆 radio waves < 𝜆 x-rays
A Answer
¥ Set 2, 1 Marks
(c) 𝜆 radio wave > 𝜆 micro waves > 𝜆 visible > 𝜆 x-rays
š Question 25
How are electromagnetic waves produced?
Write their two characteristics.
A Answer
¥ Set 3, 2 Marks
Electromagnetic waves can be produced by
1. Charged particle moving with varying speed.
2. Oscillating charge
3. By using LC circuit
Characteristics of em waves (Any two)
(1) Travel with speed of light in vacuum
(2) Transverse in nature
(3) No medium required for propagation
(4) Can be polarized
(5) Exert pressure
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Unit VIII
Atoms and Nuclei
www.cbse.page
’ Chapter 12
™ Atoms
Weightage for CBSE Board 2024
Unit VII and VIII
Combined
Weightage is
12 Marks
À Chapters
2 Chapter 11
2 Chapter 12
2 Chapter 13
It’s not fixed which chapters will
have more weightage, but overall weightage
will be 12 Marks. It may vary too in your
question paper.
` Syllabus
Alpha-particle scattering experiment;
Rutherford’s model of atom;
Bohr model of hydrogen atom,
th
Expression for radius of 𝑛 possible orbit,
velocity and energy of electron in nth orbit,
hydrogen line spectra (qualitative treatment
only).
§ 2023 55/1/1 All Sets
š Question 26
th
The radius of the 𝑛 orbit in Bohr model of
hydrogen atom is proportional to :
1
1
2
(b)
(c) 𝑛
(d) 𝑛
(a) 2
𝑛
𝑛
¥ Set 1, 1 Marks
A Answer
2
𝑛
Theory Recap
According to Bohr’s postulates
2 2
𝑛
ℎ 4𝜋𝜀0
𝑟𝑛 =
2
𝑚 2𝜋
𝑒
From the formula we can directly say 𝑟𝑛 ∝ 𝑛
2
2
▶ Don’t get confused by 1/𝑛
th
▶ If the question was asked energy of the 𝑛
2
orbit then the answer would be 1/𝑛 because
−18
2.18 × 10
13.6
𝐸𝑛 = −
J
=
−
𝑒𝑉
2
2
𝑛
𝑛
▶ The negative sign of the total energy of an
electron moving in an orbit means that the
electron is bound with the nucleus.
š Question 27
Hydrogen atom initially in the ground state,
absorbs a photon which excites it to 𝑛 = 5 level.
The wavelength of the photon is
(a) 975 nm
(b) 740 nm
(c) 523 nm
(d) 95 nm
¥ Set 1, 1 Marks
A Answer
The energy of ground state (𝑛 = 1) will be
𝐸1 = −
13.6
= −13.6eV
12
The energy of fifth energy level (𝑛 = 5) will be
𝐸5 = −
13.6
2
5
=−
13.6
eV
25
Thus, energy of a photon will be 𝐸 = 𝐸5 − 𝐸1
=−
=
13.6
25
− −
13.6 × 24
25
13.6
1
=
13.6 × 24
× 1.6 × 10
25
−19
eV
= 20.9 × 10
−19
J
ℎ𝑐
Also the energy will be equal to 𝐸 =
𝜆
−34
8
6.6 × 10
× 3 × 10
−7
𝜆=
=
0
.
95
×
10
m
−
19
20.9 × 10
= 95 nm
š Question 28
The ground state energy of hydrogen atom is
−13.6 eV. What is the potential energy and
kinetic energy of an electron in the third excited
state ?
¥ Set 1, 2 Marks
A Answer
−13.6
En =
eV
=
Total
energy
𝑛2
For third excited state n = 4
−13.6 −13.6
E4 =
=
=
−
0
.
85eV
2
4
16
Potential Energy = 2 × Total Energy = 2 × 𝐸4
= 2 × (−0.85) eV
= −1.70eV
Kinetic energy = − (Total Energy) = −𝐸4 = 0.85 eV
š Question 29
Calculate the wavelength of the second line of
Lyman series in a spectrum of hydrogen atom.
7 −1
(Take Rydberg constant, 𝑅 = 1.1 × 10 m )
A Answer
¥ Set 2, 2 Marks
1
"
1
1
#
−
2
2
𝑛1 𝑛2
=𝑅
𝜆
For second line of Lyman series n1 = 1, n2 = 3
1
𝜆
= 1.1 × 10
7
7
1
1
1
32
1
−
2
= 1.1 × 10 1 −
7
= 1.1 × 10 ×
𝜆=
9
8.8
× 10
−7
8
9
9
=
8.8
9
m = 1.023 × 10
−7
7
× 10 m
−1
m = 1023Å
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*
Unit VIII
Atoms and Nuclei
www.cbse.page
’ Chapter 13
™ Nuclei
Weightage for CBSE Board 2024
Unit VII and VIII
Combined
Weightage is
12 Marks
À Chapters
2 Chapter 11
2 Chapter 12
2 Chapter 13
It’s not fixed which chapters will
have more weightage, but overall weightage
will be 12 Marks. It may vary too in your
question paper.
` Syllabus
Composition and size of nucleus, nuclear force
Mass-energy relation, mass defect; binding
energy per nucleon and its variation with mass
number; nuclear fission, nuclear fusion.
Deleted Topics
Radioactivity; Alpha, beta and gamma
particles/rays and their properties;
Radioactive decay law, half life and mean life.
§ 2023 55(B) Compart
š Question 30
A nucleus with A = 280 splits into two nuclei X1
and X2 whose mass numbers are in the
ratio
r1
27 : 8. Find the ratio of their radii,
r2
¥ 2 Marks
A Answer
𝑅 = 𝑅0 𝐴
1/3
1
1
𝑅1
𝐴1 3
27 3
=
=
𝑅2
𝐴2
8
𝑅1 3
=
𝑅2 2
š Question 31
(a) Obtain the relationship between atomic mass
unit (u) and electron volt (eV).
(b) The mass of a ball is 0.5 kg. It is totally
converted into energy. Calculate the energy
output in eV
A Answer
¥ 3 Marks
(a) 𝐸 = Δ𝑚𝑐
2
−27
Δ𝑚 = 1.66 × 10 kg
−27
8 2
1.66 × 10
× 3 × 10
6
𝐸=
eV
=
931
.
5
×
10
eV
1.6 × 10−19
2
(b) 𝐸 = 𝑚𝑐
𝑚 = 0.5 kg
8 2
𝐸 = ( 0.5) × 3 × 10
J
16
𝐸 = 4.5 × 10 J
𝐸=
4.5 × 10
16
23
eV
=
2
.
8
×
10
−
19
1.6 × 10
eV
§ 2022 Term II 55/6 (C)
š Question 32
Determine from the given data, whether the
following reaction is exothermic or endothermic
1
1H
+
3
1H
−→
2
1H
2
1H
+
Atomic masses
2
𝑚 1H = 2.014102u
3
𝑚 1H = 3.016049u
1
𝑚 1H = 1.007825u
What is the ratio of the nuclear density of the
197
107
gold isotope 79 Au and the silver isotope 47 Ag.
A Answer
¥ Set 1, 3 Marks
Mass of reactants = ( 1.007825 + 3.016049)𝑢
= 4.023874𝑢
Mass of products = ( 2 × 2.014102)𝑢 = 4.028204𝑢
Mass of reactants < Mass of products
Hence, reaction is Endothermic
Ratio of the nuclear density of the gold isotope
197
107
Au
and
the
silver
isotope
Ag
is
1
:
1
47
79
š Question 33
Determine from the given data, whether the
following reaction is exothermic or endothermic
12
6 C
+
12
6 C
−→
20
10Ne
+
4
2He
Atomic masses
m
m
m
4
He
2
12
C
6
20
Ne
10
=
4.002603u
=
12.000000u
= 19.992439u
What is the ratio of the nuclear density of the
30
14
phosphorus 15P and nitrogen 7 N ?
A Answer
¥ Set 2, 3 Marks
Mass of the products = 19.992439 + 4.002603
= 23.995042𝑢
Mass of the reactant = ( 2 × 12.000000)𝑢
= 24.00000𝑢
As the mass of the reactants is more than the
mass of the products, the reaction is exothermic.
Ratio of the nuclear density of the phosphorus
30
14
P
and
nitrogen
N
is
1
:
1
7
15
š Question 34
Determine from the given data, whether the
following reaction is exothermic or endothermic
226
88 Ra
−→
222
86 Rn
+
4
2He
Atomic masses :
226
m 88 Ra = 226.02540u
222
m 86 Rn = 222.01750u
4
m 2He = 4.002603u
What is the ratio of the nuclear density of
14
and 7 N ?
4
He
2
¥ Set 3, 3 Marks
A Answer
Mass of reactant = 226.02540 u
Mass of the products = ( 222.01750 + 4.002603) u
= ( 226.020103) u
Mass of reactants > Mass of products
Hence the reaction is exothermic
Ratio of the nuclear density of
4
He
2
and
14
N
7
is 1 : 1
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Å
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question bank video
*
Unit IX
Electronics Devices
Published by :
www.cbse.page
www.cuet.pw
’ Chapter 14
™ Semiconductor
Electronics:
Materials, Devices
and Simple Circuits
Weightage for CBSE Board 2024
Unit IX Electronic Devices
Chapter 14 : § 7 Marks
` Syllabus
Energy bands in conductors, semiconductors
and insulators (qualitative ideas only).
Intrinsic and extrinsic semiconductors p and n
type, p-n junction.
Semiconductor diode, I-V characteristics in
forward and reverse bias, application of
junction diode, diode as a rectifier.
Removed Topics from 2024 Syllabus
Special purpose p-n junction diodes: LED,
photodiode, Solar cell and Zener diode and their
characteristics;
Zener diode as a voltage regulator. Digital
Electronics and Logic gates
§ 2024 SQP
š Question 1
Assertion (A) : Putting 𝑝 − type semiconductor
slab directly in physical contact with 𝑛 − type
semiconductor slab cannot form the 𝑝 − 𝑛
junction.
Reason (R) : The roughness at contact will be
much more than inter atomic crystal spacing
and continuous flow of charge carriers is not
possible.
(a) Both Assertion (A) and Reason (R) are true
and Reason (R) is the correct explanation of
the Assertion (A).
(b) Both Assertion (A) and Reason (R) are true,
but Reason (R) is not the correct explanation
of the Assertion (A).
(c) Assertion (A) is true, but Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is also
false.
A Answer
¥ Set 1, 1 Marks
Both Assertion (A) and Reason (R) are true and
Reason (R) is the correct explanation of the
Assertion (A).
š Question 2
(a) Name the device which utilizes unilateral
action of a pn diode to convert ac into dc.
(b) Draw the circuit diagram of full wave
rectifier.
A Answer
¥ 2 Marks
(a) Rectifier
(b) Circuit diagram of full wave rectifier
š Question 3
Read the following paragraph and answer the
questions that follow.
A semiconductor diode is basically a pn junction
with metallic contacts provided at the ends for
the application of an external voltage. It is a two
terminal device. When an external voltage is
applied across a semiconductor diode such that
p-side is connected to the positive terminal of
the battery and n-side to the negative terminal, it
is said to be forward biased. When an external
voltage is applied across the diode such that
n-side is positive and p-side is negative, it is said
to be reverse biased. An ideal diode is one whose
resistance in forward biasing is zero and the
resistance is infinite in reverse biasing. When
the diode is forward biased, it is found that
beyond forward voltage called knee voltage, the
conductivity is very high. When the biasing
voltage is more than the knee voltage the
potential barrier is overcome and the current
increases rapidly with increase in forward
voltage. When the diode is reverse biased, the
reverse bias voltage produces a very small
current about a few microamperes which almost
remains constant with bias. This small current is
reverse saturation current.
(i) In the given figure, a diode D is connected to
an external resistance
R = 100Ω and an emf of 3.5 V. If the barrier
potential developed across the diode is 0.5 V,
the current in the circuit will be
(a) 40 mA
(c) 35 mA
(b) 20 mA
(d) 30 mA
A Answer
¥ 1 Mark
Net potential across the resistor is difference of
the emf and the barrier potential
∴ iR = 𝜀 − Vbarrier = 3.5 − 0.5 = 3
⇒i=
3
100
A = 30 × 10
−3
A = 30 mA
(ii) In which of the following figures, the 𝑝 − 𝑛
diode is reverse biased?
(a)
(b)
(c)
(d)
A Answer
¥ 1 Mark
Correct Option is (c)
(iii) Based on the V-I characteristics of the diode,
we can classify diode as
(a) bilateral device
(b) ohmic device
(c) non-ohmic device
(d) passive element
A Answer
Correct Option is (c)
¥ 1 Mark
OR
Two identical PN junctions can be connected in
series by three different methods as shown in the
figure. If the potential difference in the junctions
is the same, then the correct connections will be
(a) in the circuits (1) and (2)
(b) in the circuits (2) and (3)
(c) in the circuits (1) and (3)
(d) only in the circuit (1)
A Answer
¥ 1 Mark
Correct Option is (b)
(iv) The V-I characteristic of a diode is shown in
the figure. The ratio of the resistance of the
diode at 𝐼 = 15𝑚𝐴 to the resistance at
𝑉 = −10𝑉 is
6
(a) 100
(b) 10
−6
(c) 10
(d) 10
A Answer
¥ 1 Mark
Considering the diode characteristics as a
straight line between 𝐼 = 10 mA to 𝐼 = 20 mA
passing through the origin, we can calculate the
resistance using Ohm’s law.
From the curve, at
𝐼 = 20 mA, 𝑉 = 0.8 V;
𝐼 = 10 mA, 𝑉 = 0.7 V;
Resistance under forward bias:
Δ𝑉
( 0 . 8 − 0 . 7) V
𝑟𝑓 𝑏 =
=
= 10Ω
Δ𝐼 ( 20 − 10) mA
From the curve at V = −10 V, 𝐼 = −1 𝜇 A,
Resistance under reverse bias
10 V
7
= 10 Ω
𝑟𝑟𝑏 =
1 𝜇A
𝑟 𝑓 𝑏 10
−6
⇒𝑟=
= 7 = 10 Ω
𝑟𝑟𝑏 10
§ 2023 55(B) Compart
š Question 4
Which of the following dopants in silicon will
make it a p-type semiconductor ?
(a) As
(b) Al
A Answer
(c) Sb
(d) P
¥ Set 1, 1 Marks
Aluminium (Al) is the correct option.
𝑝−type is obtained when Si or Ge is doped with a
trivalent impurity like Al, B, In, etc. The dopant
has one valence electron less than Si or Ge.
š Question 5
When a 𝑝 − 𝑛 junction diode is forward biased :
(a) the depletion layer width increases and the
barrier height is reduced.
(b) the depletion layer width decreases and the
barrier height is reduced.
(c) both the depletion layer width and the
barrier height increase.
(d) the depletion layer width decreases and the
barrier height increases.
A Answer
¥ Set 1, 1 Marks
The depletion layer width decreases and barrier
height is reduced.
š Question 6
Briefly explain how the valence band and
conduction band are formed in a crystal.
A Answer
¥ 2 Marks
In a crystal, the atoms are close to each other
and therefore the electrons interact with each
other and also with neighboring atomic cores.
This interaction of electrons causes the
formation of energy bands. The energy band
which includes the energy level of electrons is
called valence band. The energy band above the
valence band is called conduction band.
š Question 7
Explain how a depletion region is formed in a
𝑝 − 𝑛 junction.
A Answer
¥ 2 Marks
When a hole diffuses from 𝑝 → 𝑛 due to the
concentration gradient, it leaves behind an
ionized acceptor (negative charge) which is
immobile. As the holes continue to diffuse,a
layer of negative charge(or negative space
charge region) on the p-side of the junction is
developed. The space charge region on either
side of the junction together is known as
depletion region.
§ 2023 55/C Compart
š Question 8
In which of the following diagrams is the
capacitor C connected correctly to provide
smooth output of a half-wave rectifier?
(a)
(b)
(c)
(d)
A Answer
¥ Set 1, 1 Marks
Diagram C is correct
š Question 9
Assertion (A) : Silicon is preferred over
germanium for making semiconductor devices.
Reason (R) : The energy gap for germanium is
more than the energy gap for silicon.
(a) Both Assertion (A) and Reason (R) are true
and Reason (R) is the correct explanation of
the Assertion (A).
(b) Both Assertion (A) and Reason (R) are true,
but Reason (R) is not the correct explanation
of the Assertion (A).
(c) Assertion (A) is true, but Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is also
false.
A Answer
¥ Set 1, 1 Marks
Assertion (A) is true, but Reason (R) is false
Theory Recap
Energy required to jump the forbidden band
(about 0.72 eV for germanium and about 1.1 eV
for silicon) at room temperature in the intrinsic
semiconductor.
š Question 10
(i) Draw the circuit diagram used to study I−V
characteristics of a 𝑝 − 𝑛 junction diode in
conducting mode. Mark on the graph the
threshold voltage of the diode. Explain the
significance of this voltage.
(ii) In the circuit shown in the figure, the forward
voltage drop across the diode is 0.3 V. Find the
voltage difference between A and B.
A Answer
¥ 5 Marks
(i)
Beyond threshold voltage in forward bias diode
current increases significantly even for very
small increases in diode bias voltage.
(ii) Let potential across A and B is 𝑉 , so by
Kirchhoff’s loop law,
−3
𝑉𝐴𝐵 = 0.2 × 10 × ( 5000 + 5000) + 0.3
−3
3
= 0.2 × 10 × 10 × 10 + 0.3
𝑉𝐴𝐵 = 2 + 0.3 = 2.3 V
š OR
(i) Briefly describe the classification of solids into
metals, insulators and semi-conductors on the
basis of energy level diagrams.
(ii) In a silicon diode, the current increases from
10 mA to 20 mA when the voltage changes
from 0.6 V to 0.7 V. Calculate the dynamic
resistance of the diode.
A Answer
(i) Energy bands of metals,
¥ 5 Marks
Energy bands of insulators,
Energy bands of semiconductors,
▶ For E𝑔 > 3eV material is insulator
▶ For E𝑔 < 3eV material is semiconductor
▶ For E𝑔 = 0 or overlapping of conduction and
valence band material is conductor.
Δ𝑉
0.7 − 0.6
=
=
10
Ω
(ii) 𝑟𝑑 =
−
3
Δ𝐼 ( 20 − 10) × 10
Theory Recap
Dynamic Resistance
Dynamic Resistance is the the ratio of small
change in voltage Δ𝑉 to a small change in
current Δ𝐼 in diodes.
Δ𝑉
𝑟𝑑 =
Δ𝐼
§ 2023 55/1/1 All Sets
š Question 11
An ac source of voltage is connected in series
with a p-n junction diode and a load resistor. The
correct option for output voltage across load
resistance will be :
(a)
(b)
(c)
(d)
A Answer
¥ Set 1, 1 Marks
Theory Recap
Half-Wave Rectifier
If an alternating voltage is applied across a diode
which is forward biased in series with a load, a
pulsating voltage will appear across the load
only during the half cycles of the ac input
during. Such rectifier circuit is called a
half-wave rectifier.
▲ When the voltage at A is positive, the diode is
forward biased and it conducts.
▲ When A is negative, the diode is
reverse-biased and it does not conduct. The
reverse saturation current of a diode is
negligible and can be considered equal to
zero for practical purposes.
▲ The reverse breakdown voltage of the diode
must be sufficiently higher than the peak ac
voltage at the secondary of the transformer to
protect the diode from reverse breakdown.
Full Wave Rectifier
The circuit using two diodes, gives output
rectified voltage corresponding to both the
positive as well as negative half of the ac cycle is
known as full-wave rectifier.
Here the p-side of the two diodes are connected
to the ends of the secondary of the transformer.
The n-side of the diodes are connected together
and the output is taken between this common
point of diodes and the midpoint of the
secondary of the transformer.
So for a full-wave rectifier the secondary of the
transformer is provided with a centre tapping
and so it is called centre-tap transformer.
š Question 12
When an intrinsic semiconductor is doped with
a small amount of trivalent impurity, then :
(a) its resistance increases.
(b) it becomes a p-type semiconductor.
(c) there will be more free electrons than holes
in the semiconductor.
(d) dopant atoms become donor atoms.
A Answer
¥ Set 1, 1 Marks
(b) it becomes a p-type semiconductor
š Question 13
In the energy-band diagram of n-type Si, the gap
between the bottom of the conduction band EC
and the donor energy level ED is of the order of
(a) 10eV
(b) 1eV
(c) 0.1eV
(d) 0.01eV
A Answer
¥ Set 1, 1 Marks
(d) 0.01 eV is correct option.
▲ This question was directly asked from
diagram.
j Students be aware because, CBSE
Examination Department is increasing their
level of asking questions, so that students can
also be ready for CUET UG Exam.
š Question 14
Differentiate between intrinsic and extrinsic
semiconductors.
A Answer
¥ 2 Marks
Intrinsic semiconductor
▲ This type of semiconductors are in pure form.
▲ Low conductivity at room temperature.
▲ Intrinsic charge carriers are electrons and
holes with equal concentration [𝑛𝑒 = 𝑛ℎ]
Extrinsic semiconductor
▲ These type of semiconductors are doped with
trivalent or pentavalent impurity atoms.
▲ High conductivity at room temperature.
▲ The two concentrations are unequal in it.
There is excess of electrons in n-type and
excess of holes in p-type semiconductors [
𝑛𝑒 ≠ 𝑛ℎ]
j This Question was repeated in CBSE 2022
Term II 55/1/1, 2015 and 2017
š OR
Draw the circuit arrangement for studying the
V−I characteristics of a 𝑝 − 𝑛 junction diode in
forward bias and reverse bias. Show the plot of
V−I characteristic of a silicon diode.
A Answer
¥ 2 Marks
Forward bias
Reverse bias
Plot of V−I characteristic of a silicon diode
š Question 15
Briefly explain how the diffusion and drift
currents contribute to the formation of potential
barrier in a 𝑝 − 𝑛 junction diode.
A Answer
¥ 2 Marks
The diffusion current due to concentration
gradient at the junction forms a space charge
region consisting of immobile charge carriers.
Due to this an electric field is generated at the
junction giving rise to drift current in a direction
opposite to diffusion current. The potential at
which diffusion current becomes equal to drift
current is called potential barrier.
§ 2023 55/2/1 All Sets
š Question 16
The formation of depletion region in a p − n
junction diode is due to
(a) movement of dopant atoms
(b) diffusion of both electrons and holes
(c) drift of electrons only
(d) drift of holes only
A Answer
¥ Set 1, 1 Marks
Diffusion of both electrons and holes.
š Question 17
Which one of the following elements will require
the highest energy to take out an electron from
them?
Pb
(a) Ge
Ge
(b) C
A Answer
Carbon (C)
(c) Si
C
Si
(d) Pb
¥ Set 2, 1 Marks
Theory Recap
Number of free electrons for conduction in Pb,
Ge and Si are significant but negligibly small for
C. So carbon require the highest energy to take
out an electron. Ionisation energy of Carbon(C)
is more in comparison to Ge, Si and Pb.
š Question 18
A semiconductor device is connected in series
with a battery, an ammeter and a resisitor. A
current flows in the circuit. If the polarity of the
battery is reversed, the current in the circuit
almost becomes zero. The device is a/an
(a) intrinsic semiconductor
(b) 𝑝−type semiconductor
(c) 𝑛−type semiconductor
(d) 𝑝 − 𝑛 junction diode
A Answer
¥ Set 3, 1 Marks
(d) 𝑝 − 𝑛 junction diode
š Question 19
In an extrinsic semiconductor, the number
20
−3
density of holes is 4 × 10 m . If the number
15
−3
density of intrinsic carriers is 1.2 × 10 m , the
number density of electrons in it is
9
−3
(a) 1.8 × 10 m
(b) 2.4 × 10
10
m
9
−3
−3
(c) 3.6 × 10 m
(d) 3.2 × 10
10
m
−3
A Answer
¥ Set 1, 1 Marks
2
𝑛𝑖
Number of density of electrons =
𝑛ℎ
Where, 𝑛𝑖 is density of intrinsic carrier, 𝑛ℎ is the
density of holes.
15
𝑛𝑖 = 1.2 × 10 m
20
−3
−3
𝑛ℎ = 4 × 10 m
2
𝑛𝑖
𝑛𝑒 =
𝑛ℎ
𝑛𝑒 =
1.2 × 10
15 2
20
10
4×
9
−3
𝑛𝑒 = 3.6 × 10 m
š Question 20
With the help of a circuit diagram, explain how a
full wave rectifier gives output rectified voltage
corresponding to both halves of the input ac
voltage.
A Answer
¥ Set 3, 2 Marks
Suppose the input voltage to A with respect to
the centre tap at any instant is positive and B
being out of phase will be negative so , diode D1,
gets forward biased and conducts while D2 being
reverse biased is not conducting. For the next
half cycle of input voltage, the polarities are
reversed and the diode D2 conducts being
forward biased.
š Question 21
Draw energy band diagram for an n-type and
p-type semiconductor at T > 0 K.
A Answer
n-type
¥ 2 Marks
p-type
š Question 22
Answer the following giving reasons :
(i) A 𝑝 − 𝑛 junction diode is damaged by a strong
current.
(ii) Impurities are added in intrinsic
semiconductors.
A Answer
¥ 2 Marks
(i) Due to strong current, a junction diode gets
heated, consequently large number of
covalent bonds are broken and the junction is
damaged.
(ii) Deliberate addition of impurity atoms in
intrinsic semiconductor increases its
conductivity and is suitable for making
electronic devices.
OR
No electronic device can be developed using
intrinsic semiconductor because of their low
conductivity.
§ 2023 55/3/1 All Sets
š Question 23
At a certain temperature in an intrinsic
semiconductor, the electrons and holes
16
−3
concentration is 1.5 × 10 m . When it is doped
with a trivalent dopant, hole concentration
22
−3
increases to 4.5 × 10 m . In the doped
semiconductor, the concentration of electrons
( ne) will be :
6
−3
7
−3
(a) 3 × 10 m
(b) 5 × 10 m
9
−3
38
−3
(c) 5 × 10 m
(d) 6.75 × 10 m
A Answer
9
(c) 5 × 10 m
¥ Set 1, 1 Marks
−3
is the correct option
š Question 24
If a 𝑝 − 𝑛 junction diode is reverse biased,
(a) the potential barrier is lowered.
(b) the potential barrier remains unaffected.
(c) the potential barrier is raised.
(d) the current is mainly due to majority carriers.
A Answer
the potential barrier is raised
¥ Set 1, 1 Marks
š Question 25
At 0 K, the resistivity of an intrinsic
semiconductor is :
(a) same as that at 0°C
(b) same as that at 300 K
(c) zero
(d) infinite
A Answer
¥ Set 2, 1 Marks
infinite
š Question 26
For the forward biasing of a 𝑝 − 𝑛 junction diode,
which of the following statements is not correct?
(a) The potential barrier decreases.
(b) Minority carrier injection occurs.
(c) Width of depletion layer increases.
(d) Forward current is due to the diffusion of
both holes and electrons.
A Answer
¥ Set 3, 1 Marks
Width of depletion layer increases
š Question 27
Explain the roles of diffusion current and drift
current in the formation of the depletion layer in
a p-n junction diode.
A Answer
¥ 2 Marks
During the formation of p−n junction ,and due to
the concentration gradient across p-,and n-sides,
holes diffuse from p-side to n-side (p→n) and
electrons diffuse from n-side to p-side(n→p).
When an electron diffuses from (n → p), it leaves
behind an ionized donor(positive charge) on
n-side which is immobile. Similarly, when a hole
diffuses from (p→n) due to the concentration
gradient, it leaves behind an ionised acceptor
(negative charge) which is immobile. This
space–charge region on either side of the
junction together is known as depletion region.
As a result, an electric field is developed across
the junction. Due to this field, an electron on
p-side of the junction moves to n-side and a hole
on n-side of the junction moves to p-side. The
motion of charge carriers due to the electric field
is called drift.
Initially, diffusion current is large and drift
current is small. As the diffusion process
continues, the electric field strength and hence
drift current increases. This process continues
till diffusion and drift current becomes equal.
š Question 28
Explain the property of a 𝑝 − 𝑛 junction which
makes it suitable for rectifying alternating
voltages. Differentiate between a half-wave and
a full-wave rectifier.
A Answer
¥ 2 Marks
p−n junction is a uni-directional device which
conducts in forward bias.
The half-wave rectifier gives output only for half
of the input cycle. In half-wave rectifier only 1
diode is used.
The full-wave rectifier gives output for both the
halves of the input cycles. In full-wave rectifier 2
diodes and a centre tap transformer is used .
š Question 29
How is the width of depletion layer of a 𝑝 − 𝑛
junction diode affected when it is
(i) forward biased,
(ii) reverse biased
Justify your answers.
A Answer
¥ Set 2, 1 Marks
(i) In Forward Bias −→ Decreases
The direction of the applied voltage (V) is
opposite to the built-in potential (V0)
(ii) In Reverse Bias −→ Increases
The direction of the applied voltage (V) is
same as the built-in potential (V0)
š Question 30
Draw a circuit diagram of a p-n junction diode in
(a) forward biasing, and (b) reverse biasing.
Draw the V I characteristics for each case.
A Answer
¥ Set 3, 1 Marks
This question was repeated by CBSE in
question paper 2023 55/1
§ 2023 55/4/1 All Sets
š Question 31
The threshold voltage for a p-n junction diode
used in the circuit is 0·7 V. The type of biasing
and current in the circuit are
(a) Forward biasing, 0 A
(b) Reverse biasing, 0 A
(c) Forward biasing, 5 mA
(d) Reverse biasing, 2 mA
A Answer
¥ Set 1, 1 Marks
(a) Forward biasing, 0 A is correct option.
š Question 32
Assertion (A) : In n type semiconductor, number
density of electrons is greater than the number
density of holes but the crystal maintains an
overall charge neutrality.
Reason (R) : The charge of electrons donated by
donor atoms is just equal and opposite to that of
the ionised donor.
(a) Both Assertion (A) and Reason (R) are true
and Reason (R) is the correct explanation of
the Assertion (A).
(b) Both Assertion (A) and Reason (R) are true,
but Reason (R) is not the correct explanation
of the Assertion (A).
(c) Assertion (A) is true, but Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is also
false.
A Answer
¥ Set 1, 1 Marks
Both Assertion (A) and Reason (R) are true, and
Reason (R) is the correct explanation of the
Assertion (A).
š Question 33
(i) A germanium crystal is doped with antimony.
With the help of energy-band diagram,
explain how the conductivity of the doped
crystal is affected.
(ii) Briefly explain the two processes involved in
the formation of a p-n junction.
(iii) What will the effect of (1) forward biasing,
and (2) reverse biasing be on the width of
depletion layer in a p-n junction diode ?
A Answer
¥ 5 Marks
(i) With proper level of doping, the number of
conduction electrons can be made much
larger than the number of holes. Due to this
conductivity of the doped crystal increases.
(ii) Two processes (a) Diffusion (b) drift
Diffusion: Due to concentration gradient
majority charge carrier that is electron moves
from n → p side and holes to p → n side. This
movement of charges is called diffusion.
Drift: Due to the junction field, an electron on
p-side of the junction moves to n − side and a
hole on n−side of the junction moves to
p−side. The motion of the charge carrier due
to electric field is called drift.
(iii) (1) decreases
(2) increases
š OR
(i) With the help of a circuit diagram, briefly
explain the working of a full-wave rectifier
using p-n junction diodes.
(ii) Draw V I characteristics of a p-n junction
diode. Explain how these characteristics
make a diode suitable for rectification.
(iii) Carbon and silicon have the same lattice
structure. Then why is carbon an insulator
but silicon a semiconductor ?
A Answer
¥ 5 Marks
(i) Working: Suppose the input voltage to A with
respect to the centre tap at any instant is
positive. At that instant voltage at B, being out
of phase will be negative. So diode D1 gets
forward biased and conducts, while D2 being
reverse biased does not conduct. Similarly
during second half of the cycle polarity get
reversed so only D2 will conduct.
(ii) V−I characteristics
This diagram shows that the diode conducts
when forward biased and does not conduct
when reverse biased. This characteristics
makes it suitable for use for rectification.
(iii) The 4 bonding of electrons of C and Si lie
respectively, in the second and third orbit.
Hence energy required to take out an electron
from their atoms will be much less than that
−
for C. Hence number of free electrons (𝑒 ) for
conduction in Si significant but negligibly
small for C.
§ 2023 55/5/1 All Sets
š Question 34
During the formation of a p − n junction
(a) diffusion current keeps increasing.
(b) drift current remains constant.
(c) both the diffusion current and drift current
remain constant.
(d) diffusion current remains almost constant
but drift current increases till both currents
become equal.
A Answer
¥ Set 1, 1 Marks
Diffusion current remains almost constant but
drift current increases till both currents become
equal.
Theory Recap
Initially, diffusion current is large and drift
current is small. As the diffusion process
continues, the space-charge regions on either
side of the junction extend, thus increasing the
electric field strength and hence drift current.
This process continues until the diffusion
current equals the drift current. Thus a p − n
junction is formed. In a p − n junction under
equilibrium there is no net current.
š Question 35
The energy required by an electron to jump the
forbidden band in silicon at room temperature is
about
(a) 0.01eV
(b) 0.05eV
A Answer
(c) 0.7eV
(d) 1.1eV
¥ Set 3, 1 Marks
Energy required to jump the forbidden band is
about 1.1 eV for silicon.
Theory Recap
The energy required to jump the forbidden band
is about 0.72 eV for germanium and about 1.1 eV
for silicon at room temperature in the intrinsic
semiconductor.
š Question 36
Assertion (A) : The resistance of an intrinsic
semiconductor decreases with increase in its
temperature.
Reason (R) : The number of conduction
electrons as well as hole increase in an intrinsic
semiconductor with rise in its temperature.
(a) Both Assertion (A) and Reason (R) are true
and Reason (R) is the correct explanation of
the Assertion (A).
(b) Both Assertion (A) and Reason (R) are true,
but Reason (R) is not the correct explanation
of the Assertion (A).
(c) Assertion (A) is true, but Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is also
false.
A Answer
¥ Set 1, 1 Marks
Both Assertion (A) and Reason (R) are true and
Reason (R) is the correct explanation of the
Assertion (A).
Theory Recap
At higher temperatures (T > 0K), which excites
some electrons from the valence band to the
conduction band. These thermally excited
electrons at T > 0 K, partially occupy the
conduction band. These have come from the
valence band leaving equal number of holes
there.
š Question 37
Draw the circuit arrangement for studying V-I
characteristics of a p − n junction diode in
(i) forward biasing
(ii) reverse biasing
Draw the typical V−I characteristics of a silicon
diode.
Describe briefly the following terms
(i) minority carrier injection in forward biasing
(ii) breakdown voltage in reverse biasing.
A Answer
¥ 5 Marks
Minority carrier injection : Under forward bias
electrons from n-side cross the depletion region
and reach p−side. Similarly, holes from p−side
cross the junction and reach the n-side.
Breakdown voltage : It is the voltage under
reverse bias for which reverse current increases
sharply.
š OR
Name two important processes involved in the
formation of a p − n junction diode.
With the help of a circuit diagram, explain the
working of junction diode as a full wave rectifier.
Draw its input and output waveforms.
State the characteristic property of a junction
diode that makes it suitable for rectification.
A Answer
¥ 5 Marks
Suppose the input voltage to A with respect to
the centre-tap at any instant is positive. At that
instant, voltage at B being out of phase will be
negative. So, diode 𝐷1 gets forward biased and
conducts (while 𝐷2 being reverse biased is not
conducting). Hence, during this positive half
cycle we get an output current (and a output
voltage across the load resistor 𝑅 𝐿). In the course
of ac cycle when the voltage at A becomes
negative with respect to centre tap, the voltage
at B would be positive. In this part of the cycle
diode 𝐷1 would not conduct but diode 𝐷2 would,
giving an output current and output voltage
(across 𝑅 𝐿) during the negative half cycle of the
input ac.
Characteristic Property :Junction diode allows
current to pass only when it is forward biased.
§ 2023 55(B)
š Question 38
Which of the following is not a semiconductor ?
(a) Ge
(b) Si
(c) Sn
A Answer
(d) CdS
¥ Set 1, 1 Marks
(c) Sn
š Question 39
In an intrinsic semiconductor, the intrinsic
carrier concentration ( n𝑖 ) , electron
concentration ( n𝑒) and hole concentration ( nh)
are related as :
(a) ne + nh = ni
(b) ni =
√
nenh
(c) ne = nh = ni
ni
(d) ne + nh =
2
A Answer
¥ Set 1, 1 Marks
ne = nh = ni
In intrinsic semiconductors, the number of free
electrons, n𝑒 is equal to the number of holes, nℎ.
š Question 40
Assertion (A) : Silicon is preferred over
Germanium for making semiconductor devices.
Reason (R) : Silicon can be used at a higher
temperature as compared to Germanium.
(a) Both Assertion (A) and Reason (R) are true
and Reason (R) is the correct explanation of
the Assertion (A).
(b) Both Assertion (A) and Reason (R) are true,
but Reason (R) is not the correct explanation
of the Assertion (A).
(c) Assertion (A) is true, but Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is also
false.
A Answer
¥ Set 1, 1 Marks
Both Assertion (A) and Reason (R) are true and
Reason (R) is the correct explanation of the
Assertion (A).
š Question 41
Define the following terms in relation to the
working of a 𝑝 − 𝑛 junction diode :
(i) Knee voltage
(ii) Reverse saturation current
A Answer
¥ 2 Marks
(i) In forward bias of a 𝑝 − 𝑛 junction diode the
external applied voltage beyond which the
current begins to flow rapidly is called Knee
Voltage.
(ii) In reverse bias of a 𝑝 − 𝑛 junction diode the
current is due to minority carriers which is
not very much dependent on the applied
voltage. This current is called reverse
saturation current.
š Question 42
How are the potential barrier and width of the
depletion region affected when a 𝑝 − 𝑛 junction
diode is
(i) forward-biased
(ii) reverse-biased
A Answer
¥ 2 Marks
(i) In forward bias
▶ Potential barrier decreases
▶ Width of depletion layer decreases
(ii) In reverse bias
▶ Potential barrier increases
▶ Width of depletion layer increases
š OR
Briefly explain doping of a semiconductor and
its necessity.
A Answer
¥ 2 Marks
Doping is the process of adding impurity atom in
an intrinsic semiconductor deliberately. An
intrinsic semiconductor has low conductivity. By
doping, conductivity of semiconductor is
increased.
§ 2022 Term II 55/B
š Question 43
How is the forward biasing different from the
reverse biasing in a 𝑝 − 𝑛 junction diode ? Name
one device each which works under forward
and reverse biasing.
A Answer
¥ 2 Marks
In forward biasing the p-side of the junction is
connected to the +ve terminal and n-side to the
–ve terminal of the battery.
In reverse biasing the p-side of the junction is
connected to the –ve terminal and n-side to the
+ve terminal of the battery.
Alternatively
In forward biasing the width of the depletion
layer decreases / resistance of the junction
decreases / current increases.
In reverse biasing the width of the depletion
layer increases / resistance of the junction
increases / current decreases.
A device for forward biasing → LED
A device for reverse biasing → Zener diode
LED, Zener diode is not in your 2024 syllabus.
š Question 44
How are (a) n-type semiconductor, and
(b) p-type semiconductor obtained from an
intrinsic semiconductor ? Name the majority
charge carriers in each of the semiconductors.
A Answer
¥ 2 Marks
▶ 𝑛− type semiconductors are obtained when
the intrinsic semiconductor are dopped with
a pentavalent impurity.
▶ 𝑝− type semiconductors are obtained when
the intrinsic semiconductors are dopped with
a trivalent impurity.
Majority charge carriers in
(i) 𝑛−type −→ electrons
(ii) 𝑝−type −→ holes
Very Important
§ 2022 55/6/1 Compart
š Question 45
A 𝑝−type semiconductor is electrically neutral
although it has holes as the majority carriers.
Justify.
A Answer
¥ Set 1, 1 Marks
The crystal maintains an overall charge
neutrality as the charge of additional charge
carriers is just equal and opposite to that of the
ionised cores in the lattice.
š Question 46
In 𝑛−type semiconductor, explain how the
crystal is electrically neutral although electrons
are the majority carriers in it.
A Answer
¥ Set 2, 1 Marks
Crystal is electrically neutral as the charge of
additional electrons provided by donor impurity
is just equal and opposite to that of the ionised
cores in the lattice.
š Question 47
Draw V−I characteristics of a 𝑝 − 𝑛 junction
diode. Answer the following giving reasons
(a) Why is the reverse bias current almost
independent of applied voltage up to
breakdown voltage ?
(b) Why does the reverse current show a sudden
increase at breakdown voltage ?
A Answer
¥ 3 Marks
(a) The current is not limited by the magnitude of
the applied voltage but is limited due to the
concentration of the minority carriers on
either side of the junction.
(b) At the breakdown voltage, a large number of
covalent bonds break, resulting in the
increase of large number of charge carriers.
Hence current increases suddenly.
š Question 48
(a) Explain the formation of a 𝑝 − 𝑛 junction.
(b) Can we take one slab of 𝑝−type
semiconductor and physically join it to
another 𝑛−type semiconductor to get a 𝑝 − 𝑛
junction ? Explain.
A Answer
¥ 3 Marks
(a) During the formation of 𝑝 − 𝑛 junction due to
concentration gradient across 𝑝 and 𝑛 sides,
holes diffuse from 𝑝 side to 𝑛 side and
electrons diffuse from n side to 𝑝 side This
motion of charge carriers gives rise to
diffusion current across the junction. Diffusion
of electrons develops a layer of positive charge
on 𝑛 side of the junction and diffusion of holes
develops a layer of negative charge on 𝑝 side
of the junction. Due to this space charge region
on either side of the junction an electric field is
developed. This electric field drifts charge
carriers across the junction and sets up drift
current in a direction opposite to diffusion
current. This process continues until the
diffusion current is equal to drift current. Thus
𝑝 − 𝑛 junction is formed.
(b) No, Any slab, howsoever flat, will have
roughness much larger than the inter-atomic
spacing (≈ 2 to 3 Å) and hence continuous
contact at the atomic level will not be possible.
The junction will behave as a discontinuity for
the flowing charge carriers.
š Question 49
Explain how the barrier potential is formed in a
𝑝 − 𝑛 junction. How is it affected in
(a) forward bias, and
A Answer
(b) reverse bias ?
¥ Set 2, 3 Marks
In a 𝑝 − 𝑛 junction diode, the loss of electrons
from the 𝑛−region and the gain of electrons by
the 𝑝−region causes a difference of potential
across the junction of the two regions. This
potential tends to prevent the movement of
electrons from the n region to the 𝑝 region.
Hence a potential barrier is created across the
junction.
(a) In forward bias, the barrier potential is
reduced
(b) In reverse bias, the barrier potential increases
š Question 50
C and Si both have the same lattice structure, but
C is an insulator while Si is a semiconductor.
Justify.
A Answer
¥ Set 3, 1 Marks
Energy required to liberate an electron (i.e
ionisation energy) of C is greater than Si. Hence
number of free electrons for conduction in Si is
more than C.
š Question 51
Draw the energy band diagrams (at T > 0 K) for
𝑛−type and 𝑝−type semiconductors. Using
diagram, explain why in 𝑛−type semiconductor
the conduction band has most electrons from the
donor impurities.
A Answer
¥ Set 3, 3 Marks
Energy Band Diagram of 𝑛−type
Energy Band Diagram of 𝑝−type
In the energy band of 𝑛−type semiconductors,
donor energy level E𝐷 is formed slightly below
the bottom of E𝐶 of the conduction band. Hence
electrons from this level move into the
conduction band easily.
§ 2022 Term II 55/1/1
š Question 52
Write the characteristics of a 𝑝 − 𝑛 junction
which makes its suitable for rectification.
A Answer
¥ 2 Marks
The unidirectional property of a diode makes it
suitable for rectification.
Alternatively
The diode conducts when forward biased and
does not conduct when reverse biased.
š Question 53
Explain the formation of depletion layer and
barrier potential in a 𝑝 − 𝑛 junction diode.
A Answer
¥ Set 2, 2 Marks
Formation of depletion region: During
formation of 𝑝 − 𝑛 junction a hole diffuses from
𝑝 to 𝑛 region and electron diffuses from 𝑛 to 𝑝
region due to concentration gradient. It leaves
behind immobile ions at the junction on both
sides. This space charge region at the junction is
together called depletion region.
Potential Barrier: The immobile ions on 𝑝 side
are negative and on 𝑛 side are positive. They
create a potential difference preventing the
further flow of majority charge carriers. This
potential is called as barrier potential.
š Question 54
Draw energy band diagrams of 𝑛−type and
𝑝−type semiconductors at temperature T > 0K,
depicting the donor and acceptor energy levels.
Mention the significance of these levels.
A Answer
¥ Set 1, 1 Marks
Significance
𝑛−type semiconductors: small energy gap
between donor level and conduction band which
can be easily covered by thermally excited
electrons.
𝑝−type semiconductors: small energy gap
between acceptor level and valence band which
can be easily covered by thermally excited
electrons.
Alternatively
The conductivity of semiconductor is improved
with the creation of donor and acceptor levels.
š Question 55
Distinguish between intrinsic and extrinsic
semiconductors. Although in an extrinsic
semiconductor ne ≠ nh, yet it is electrically
neutral. Why?
A Answer
¥ Set 3, 2 Marks
Intrinsic semiconductors are pure
semiconductors while extrinsic semiconductors
are doped with either trivalent or pentavalent
impurities.
Extrinsic semiconductor maintains an overall
charge neutrality as the charge of additional
charge carriers is just equal and opposite to that
of the ionised cores in the lattice.
§ 2021 55/C Compart
š Question 56
How does the energy gap of an intrinsic
semiconductor change when doped with a
trivalent impurity?
A Answer
¥ Set 1, 1 Marks
Energy gap decreases.
š Question 57
In a 𝑝 − 𝑛 junction under equilibrium, there is no
net current. Why?
A Answer
¥ Set 1, 1 Marks
It is because the diffusion current becomes equal
to drift current.
š OR
On what factor does the wavelength of the light
emitted by an LED depend?
A Answer
¥ Set 1, 1 Marks
Depends on energy gap/band gap of the
semiconductor.
š Question 58
State whether the given ideal diodes are forward
or reverse biased
A Answer
¥ 2 Marks
(i) Reverse Biased
(ii) Reverse Biased
š Question 59
With the help of a circuit diagram, explain the
working of a diode as a half-wave rectifier.
A Answer
¥ 2 Marks
When the voltage at A is positive, the diode is
forward biased and it conducts. When A is
negative, the diode is reverse-biased and it does
not conduct. The reverse saturation current of a
diode is negligible and can be considered equal
to zero for practical purposes. Thus output is
obtain only during positive half cycle.
§ 2020 55/C Compart
š Question 60
When the temperature of an 𝑛−type
semiconductor is increased, then the
(a) number of free electrons increases while that
of the holes decreases.
(b) number of holes increases while that of the
free electrons decreases.
(c) number of free electrons and holes remains
unchanged.
(d) number of both the free electrons and the
holes increase equally.
A Answer
¥ Set 1, 1 Marks
Number of both the free electrons and holes
increases equally.
š Question 61
In an unbiased 𝑝 − 𝑛 junction diode, the p-side of
the junction is at
potential as compared to
that on the 𝑛−side of the junction.
A Answer
Lower
¥ Set 1, 1 Marks
š Question 62
In an unbiased 𝑝 − 𝑛 junction diode, the n-side of
potential as compared to
the junction is at
that on the 𝑝−side of the junction.
A Answer
¥ Set 2, 1 Marks
Higher
š Question 63
The potential barrier of a 𝑝 − 𝑛 junction tends to
prevent the movement of
from 𝑛−side to
𝑝−side.
A Answer
¥ Set 3, 1 Marks
electrons
š Question 64
Explain energy bands in a crystalline silicon
semiconductor.
(a) Explain energy bands in a crystalline silicon
semiconductor.
(b) How is an intrinsic semiconductor converted
into
(i) 𝑝−type extrinsic semiconductor ?
(ii) 𝑛−type extrinsic semiconductor ?
→ Draw their energy band diagrams showing
impurity level.
A Answer
¥ Set 3, 3 Marks
(a) In silicon semiconductor there are two
energy bands,conduction band and valence
band separated by energy band gap balance
band is filled completely or partially with
electron while conduction band is empty
(b.i) by adding trivalent / 13th Group element
(b.ii) by adding pentavalent / 15th Group element
š Question 65
With the help of the circuit diagram, explain the
working of a silicon 𝑝 − 𝑛 junction diode in
forward biasing and draw its I−V characteristics.
A Answer
¥ 2 Marks
In the forward bias the width of depletion layer
decreases and barrier height is reduced. It
supports the movement of majority charge
carriers across the junction. As soon as supply
voltage exceeds barrier potential
instantaneously current begins to flow through
junction and increases exponentially with
forward biasing voltage.
š Question 66
Draw the circuit diagram of a full-wave rectifier
using two 𝑝 − 𝑛 junction diodes. Explain its
working and draw its input and output
waveforms.
A Answer
¥ 3 Marks
This question was repeated in 2023, 2020 55/5/1
Check 2023 PYQs above to see the answer.
§ 2020 55/5/1 All Sets
š Question 67
The
, a property of materials C, Si and Ge
depends upon the energy gap between their
conduction and valence bands.
A Answer
¥ Set 1, 1 Marks
Conductivity or Resistivity
š Question 68
an
The ability of a junction diode to
alternating voltage, is based on the fact that it
allows current to pass only when it is forward
biased.
A Answer
¥ Set 1, 1 Marks
Rectify
š Question 69
Draw the circuit diagram of a full wave rectifier.
Explain its working showing its input and output
waveforms.
A Answer
¥ 3 Marks
During one half cycle of the input a.c. signal,
only diode 1 is forward biased and conducts.
During the next half cycle of the input ac signal
only diode 2 is forward biased and conducts.
However, due to the use of the centre tapped
transformer, the current in the load flows in the
same direction during both these half cycles. The
current through the load is therefore
unidirectional.
§ 2019 55/5/1 All Sets
š Question 70
Draw the output signal in a 𝑝 − 𝑛 junction diode
when a square input signal of 10 V as shown in
the figure is applied across it.
A Answer
¥ Set 1, 1 Marks
š Question 71
When a 𝑝 − 𝑛 junction diode is forward biased,
how will its barrier potential be affected ?
A Answer
¥ Set 1, 1 Marks
Its barrier potential gets reduced.
§ 2019 55(B)
š Question 72
Explain briefly how a potential barrier is formed
in a 𝑝 − 𝑛 junction diode.
A Answer
¥ 2 Marks
The loss of electron from 𝑛 region and the gain
of electron by the 𝑝−region causes a difference
of potential across the junction of two regions.
The polarity of this potential is such as to oppose
further flow of carriers so that a condition of
equilibrium exists.
The 𝑛 material has lost electrons and 𝑝 material
has acquired electrons. The 𝑛 material is thus
positive relative to the 𝑝 material. Since this
potential tends to prevent the movement of
electron from the 𝑛−region to 𝑝−region. It is
called a barrier potential.
*
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