CHAPTER
1
Electric Charges and Fields
Year Wise Number of Questions Analysis (2022-2013)
5
4
3
2
1
0
2022
2021
2020
2020
Covid
2019
2018
2017
2016 I 2016 II
2015 2015 Re
2014
Year
Topicwise Number of Questions Analysis (2022-2013)
7
6
4
3
2
1
Coulomb’s Law
5
Topics
3
2
Dipole
8
0
Electric Flux
9
Electric Field Due to Point Charges
10
1
Gauss Law & Its Application
1
Electric Field Due to Continuous Charge
Distribution
4
Number of Questions
Number of Questions
6
2013
CHARGES AND ITS PROPERTIES
™
Electrostatics is the branch of physics that deals with
stationary electric charges.
™
Electric charge is the property associated with a body or a
particle due to which it is able to produce as well as experience
the electric and magnetic effects.
Characteristics of Electric Charges
™
Charge is a fundamental property of matter.
™
The excess or deficiency of electrons in a body is responsible
for charge on the body.
™
There are two types of charges namely positive and negative
charges.
™
The deficiency of electrons in a body is responsible for
positive charge.
™
The excess of electrons in a body is responsible for negative
charge.
™
If a body gets positive charge, its mass slightly decreases.
™
If a body is given negative charge, its mass slightly increases.
™
Charge is relativistically invariant, i.e., it does not change
with motion of charged particle.
qstatic = qdynamic
Charge is a scalar quantity. S.I. unit of charge is coulomb(C).
One electrostatic unit of charge is esu
1
coulomb.
1 esu =
3 × 109
One electromagnetic unit of charge (emu) = 10 coulomb
Charge is a derived physical quantity with dimensions [AT].
™
™
Charge is quantised. The charge on any body is an integral
multiple of the minimum charge or electron charge, i.e., if q
is the charge then q = ±ne where n is an integer, and e is the
charge of electron = 1.6 × 10–19 C.
™
The minimum charge possible is 1.6 × 10–19 C.
™
If a body possesses n1 protons and n2 electrons, then net
charge on it will be (n1 – n2) e,
i.e., n1 (e) + n2(–e) = (n1 – n2)e
™
In an isolated system charge remains conserved. It can neither
be created nor destroyed. It can only be transferred from one
object to the other.
™
Like charges repel each other and unlike charges attract each
other.
Key Note
P
P
2
Charge always resides on the outer surface of a charged
conducting body. It accumulates more at sharp points.
The total charge on a body is algebraic sum of the charges
located at different points on the body.
Train Your Brain
Example 1: Which of the following charge is not possible?
(1) 1.6 × 10–18 C
(2) 1.6 × 10–19 C
(3) 1.6 × 10–20 C
(4) None of these
–20
Sol. (3) 1.6 × 10 C, because this is 1/10 of electronic
charge and hence not an integral multiple of charge of
electron.
Example 2: What is the total charge of a system containing
five charges +1, +2, –3, +4 and –5 in some arbitrary unit?
Sol. Total charge is + 1 + 2 – 3 + 4 – 5 = –1 in the same unit.
Example 3: How many electrons are there in one coulomb
of charge?
Q
1 coulomb
n =
Sol. Q =
e 1.6 × 10−19 coulomb
= 6.25 × 1018 electrons.
Example 4: An electron at rest has a charge of 1.6 × 10–19
C. It starts moving with a velocity v = c/2, where c is the
speed of light, then the new charge on it is
(1) 1.6 × 10–19 coulomb
2
(2) 1.6 × 10–19
1
1 −   coulomb
2
(3) 1.6 × 10–19
2
  − 1 coulomb
1
2
1.6 × 10−19
coulomb
2
1
1−  
2
Sol. (1) Charge does not depend on velocity.
Example 5: Which of the following charge cannot be
present on an oil drop in Millikan’s experiment
(1) 4.0 × 10–19 C
(2) 6.0 × 10–19 C
(3) 10.0 × 10–19 C (4) None of these
Sol. (4) Charge cannot be in fraction of e.
Example 6: If in Millikan’s oil drop experiment charges
on drops are found to be 8µC, 12µC, 20µC, then quanta of
charge is
(1) 8 µC(2) 4µC
(3) 20µC(4) 12µC
Sol. (2) Quanta is smallest charge in whose integer multiple
charges remains present.
Example 7: In 1 g of a solid, there are 5 × 1021 atoms. If
one electron is removed from 0.01% atoms of the solid, the
charge gained by the solid is
(1) + 0.08 C
(2) + 0.8 C
(3) – 0.08 C
(4) – 0.8 C
(4)
Sol. (1) 0.01% of 5 × 1021 =
5 × 1021 × 0.01
100
NEET (XII) Moduel-1 PW
= 5 × 1017
\ Charge gained = 5 × 1017 e
= 5 × 1017 × 1.6 × 10–19
= +0.08 C
Concept Application
1. The weight of a body which is charged negatively by
rubbing
(1) Remain constant
(2) Decreases
(3) Increases
(4) May increase or may decrease
2. Which of the following option is correct?
(1) The total number of changed particles in the
universe remains conserved
(2) The magnitude of total positive charge of the
universe is constant
(3) The magnitude of total negative charge of the
universe is constant
(4) The total charge of the universe is constant
3. If a glass rod is rubbed with silk, it acquires a positive
charge because
(1) Protons are added to it
(2) Protons are removed from it
(3) Electrons are added to it
(4) Electrons are removed from it
4. Which one of the following statement regarding
electrostatics is wrong ?
(1) Charge is quantized
(2) Charge is conserved
(3)There is an electric field near an isolated charge
at rest
(4)A stationary charge produces both electric and
magnetic fields
5. In nature, the electric charge of any system is always
equal to
(1) Half integral multiple of the least amount of charge
(2) Zero
(3) One third of the least amount of charge
(4) Integral multiple of the least amount of charge
other acquires negative charge. For example, when a glass rod is
rubbed with a silk, the glass rod acquires positive charges and the
silk acquires negative charges.
2. By conduction
When an uncharged conducting body is made in contact with a
charged conducting body, charge flows to the uncharged body and
the body gets charged. For example, if an uncharged conducting
sphere A is brought in contact with a charged conducting sphere
B, the sphere A will be charged with the same nature of charge as
in the sphere B.
3. By induction
When a charged particle or body is brought near an uncharged
body without touching, charges are developed on the uncharged
body. This method of charging a body is called charging by
induction.
positively
positively
COULOMB’S LAW AND FORCE
DUE TO MULTIPLE CHARGES
Coulomb’s law states that the magnitude of the electrostatic force
of attraction or repulsion between two stationary point charges is
directly proportional to the product of the magnitudes of charges
and inversely proportional to the square of the distance between
them.
F=
=
and
Methods of Charging a Body
1. By rubbing
When a body is rubbed with another body, both of them get
charged. One of the two bodies acquires positive charge and the
P Electric Charges and Fields
W
(in air or vaccum)
kq1q2
d2
∈0 = 8.857 × 10–12
™
C2
Nm 2
or
farad
,
metre
1
= 9 × 109 Nm 2 / C 2 = k
4π ∈0
(say)
Key Note
P
ELECTRIFICATION
1 q1q2
4π ∈0 d 2
P
P
P
P
The electric force is independent of mass and velocity of
the charged particle. It depends upon the charge.
The electric force is conservative in nature.
Coulomb force is a central force.
Electric forces act as action reaction pair, i.e the two
charges exert equal and opposite forces on each other.
Coulomb force is much stronger than gravitational force.
3
™
Relative permitivity (∈r): The relative permitivity is the
ratio of absolute permitivity of the medium to the absolute
∈
permitivity of the free space ∈r = = K (dielectric constant)
∈0
∈r has no units and dimensional formula is
[M° L°T° A°]
And also
Force between two charges in air
∈r =
Force between the same two charges in the
medium at same distance
=
™
™
Train Your Brain
Example 8: Two charges of 3 × 10–19 C and –10–6 C are
placed at (0 , 0, 0) and (1, 1,1) respectively. Find the force
on 2nd charge in vector form.
1 q1q2
→
ˆ
Sol. F21 = 4πε r 2 r21
0 12
→
r = (1− 0) iˆ + (1− 0) ˆj + (1− 0) kˆ = iˆ + ˆj + kˆ
21
Fair
→
| r21
Fmedium
For air k = 1
k > 1 for any dielectric medium;
k = ∞ for conducting medium like metals
|=
12 + 12 + 12 =
→
r21
→
| r21 |
r̂21 =
=
3
(iˆ + ˆj + kˆ)
3
(
(Net force on either charges in a medium)
∈0 - permitivity of free space or vacuum or air.
K = ∈r - Relative permitivity or dielectric constant of the
medium in which the charges are situated.
)
ˆ ˆ ˆ
9×109 × 3 ×10−19 × (−10−6 ) . i + j + k
=
F21
3
3
= –3 × 10–16 (iˆ + ˆj + kˆ) N
Example 9: Three charges (each q) are placed at the corners
of an equilateral triangle. Find out the resultant force on any
one charge due to other two.
→
Force between two charges in a medium
q1q2
1
q1q 2
1
=
F=
2
4
π
∈
K
d2
4π ∈0 ∈r d
0
F12 + F22 + 2 F1 F2 cos 60º
Sol. F =
Coulomb’s Law in Vector Form
™

Suppose the position vector of two charges q1 and q2 are r1

and r2 respectively, then electric force on charge q1 due to
q2 is,

F12
=
q1q2  
1
r1 − r2 ...(i)


4πε0 r − r 3
1
2
(
)
Similarly, electric force on q2 due to charge q1 is

q1q2  
1
F21
r2 − r1 ...(ii)
=

4πε0 r − r 3
2
1


\ F12 = − F21
(
)
Principle of Superposition
The principle of superposition states that for a system of charges
q1, q2, ..., qn, the force on q1 due to q2 is the same as given by
Coulomb’s law, i.e., it is unaffected by the presence of the other
→
charges q3, q4, ..., qn. The total force F1 on the charge q1, due
to all other charges is given by the vector sum of the forces
→
→
F12 , F13 ,
→
..., F1n :
i.e.,
→
F1 =
→
→
→
F12 + F13 + ... + F1n =

q
F1 = 1
4πε0
4
n
qi

qq
qq
1  q1q 2
rˆ12 + 12 3 rˆ13 + ... + 1 2 n rˆln 

4πε 0  r122
r13
rln

...(iii)
∑ r 2 rˆ1i ...(iv)
But F1 = F2 =
∴ F =
1
4π ∈0
1 q2
4π ∈0 a 2
3 q2
a2
Example 10: A proton and an electron are placed 1.6 cm
apart in free space. Find the magnitude and nature of
electrostatic force between them.
Sol. Using Coulomb’s law
F=
Q1Q2
4πε0 r 2
=−9 × 10−25 N
Negative sign indicates that the force is attractive in nature.
Example 11: Nucleus 92U238 emits a-particle (2He4). At
any instant a-particle is at distance of 9 × 10–15 m from the
centre of nucleus of uranium. What is the force on a-particle
at this instant?
92U
238
→ 2He4 + 90Th234
i = 2 1i
NEET (XII) Moduel-1 PW
1 q1q2
4πε0 R 2
\ q1 = 90e, q2 = 2e, and R = 9 × 10–15 m
(90e)(2e)
\ F = 9 × 109 ×
(9 × 10−15 ) 2
Sol. F =
9
−19 2
9 × 10 × 90 × (1.6 × 10 ) × 2
=
512N
(9 × 10−15 ) 2
Example 12: Two fixed charges +4q and +q are at a distance
3 m apart. At what point between the charges a third charge
+q must be placed to keep it in equilibrium?
Q2 = +q
q
Sol. Q1 = +4q
x
3–x
kQ1q
kQ2 q
Force on q by Q1 =
and that by Q2 =
2
(3 − x) 2
x
Now,
kQ1q
=
kQ2 q
(3 − x) 2
Q
Q2
4
1
=
=
or 2
or 21
2
(3 − x)
(3 − x) 2
x
x
2
1
Take the square root; =
x (3 − x)
x
2
⇒ x = 2 m. So, q will be placed at a distance 2m from
Q1 and at 1 m from Q2.
Shortcut Method:
(a) If Q1Q2 > 0
x1 =
r0
Q2
+1
Q1
(x1 is distance from Q1 and Q2)
Example 13: Two point charges of + 2µC and +6µC repel
each other with a force of 12 N. If each is given an additional
charge of –4 µC, then force will become
(1) 4N (attractive)
(2) 60 N (attractive)
(3) 4 N (Repulsive) (4) 12 N (attractive)
9 × 2 × 2 × 10−3
= – 4N
9 × 10−3
Example 14: Two identical small conducting spheres
carry charge of Q1 and Q2 with Q1 > > Q2. The charges
are d distance apart. The force they exert on one another
is F1. The spheres are made to touch one another and then
separated to distance d apart. The force they exert on one
another now is F2. Then F1/F2 is
4Q1
Q1
(2)
Q2
4Q 2
4Q 2
Q2
(3)
(4)
Q1
4Q1
(1)
Sol. (3) F1 =
KQ1Q2
d2
On touching each other
Q1 + Q2
2
d
Q1 + Q2
2
 Q + Q2   Q1 + Q2 
K 1
2   2 

F2 =
d2
K
(Q1 + Q2 ) 2
F2 =
2
4d
K
4d
F2 =
x1 in this case is not between the charges.
r2
= −
=
(b) If Q1 Q2 < 0
r0
x1 =
Q2
−1
Q1
9 × 109 × 2 × 10−6 × 2 × 10−6
F=–
Q2
2 1
 Q2 
1 +

 Q1 
2
KQ12
(Since Q1 >> Q2)
4d 2
F1 KQ1Q2 4d 2
=
×
F2
d2
KQ12
=
4Q2
Q1
Example 15: Four charges are arranged at the corners of
square ABCD, as shown. The force on a + ve charge kept at
the centre of the square is
Sol. (1) 2mC ------------------------6mC
12 =
9 × 109 × 2 × 10−6 × 6 × 10−6
r2
9 × 2 × 6 × 10−3
12
2
r = 9 × 10–3
On giving –4mC charge
–2mC ------------------------2mC
r2 =
P Electric Charges and Fields
W
(1) Zero
(2) Along diagonal AC
(3) Along diagonal BD
(4) Perpendicular to CD
5
Sol. (4)
B
+q
=
C
–q
=
Q
A –2q
D
3
10
Fr =
+2q
Force on charge Q kept at the centre of the square
C
F
Fr
5× 2
O
F
cm
3
4 × 9 × 109 × 2 × 10−6 × 2 × 10−6
2
Fr = 43.2 N
 10 
−4

 × 10
3


Example 17: Two pith balls with mass m are suspended
from insulating threads. When the pitch balls have equal
positive charge Q, they hang in equilibrium as shown.
D
–q
+q
Net force on Q due to –q and –2q is along OA & due to
+q and 2q is along OB.
A
\ Direction of resultant force is ⊥ to CD.
Example 16: The electric force on 2 µC charge placed at
the centre O of two equilateral triangles each of side 10 cm,
as shown in Fig. is Fr . If charge A, B, C, D, E, & F are 2 µC,
2 µC, 2 µC, –2 µC, –2 µC, –2 µC, respectively then Fr is.
(1)
(3)
(2)
(4)
(1) 21.6 N(2) 64.8 N
(3) 0(4) 43.2 N
2F
F
F
120°
x
2
2F
2F
120°
F
cos 30° =
6
5
x
5
cos 30°
Concept Application
F
–2
x=
Sol. (4) Internal force do not affect symmetry of charge only
angle increases.
–2
Sol. (4)
F
6. Find the number of electrons that should be removed
from a coin so that it acquires a charge of 10–6 coulomb
(1) 6.25 × 1011
(2) 1.6 × 106
–19
(3) 1.6 × 10
(4) 6.25 × 1012
7. Two identical small bodies each of mass m and charge
q are suspended from two strings each of length l
from a fixed point. This whole system is taken into an
orbiting artificial satellite, then tension in the strings is
(1)
(3)
Kq 2
l2
Kq 2
l2
+ 2mg
(2)
(4)
Kq 2
4l 2
Kq 2
+ 2mg
4l 2
NEET (XII) Moduel-1 PW
8. Two point charges +4 q and +q are separated by
distance r, where should a third point charge Q be
placed that the whole system remains in equilibrium.
2r
r
(1)
from q
(2)
from q
3
3
r
r
(3)
(4)
from 4q
from q
3
2
9. If two identical spheres having charges 16 mC and –8
mC are kept at a certain distance apart, the force acting
on them is F. They are touched and again kept at the
same distance, the force now becomes.
F
(1) F
(2)
18
F
F
(3)
(4)
8
2
10. Two identical small conducting spheres, having
charges of opposite sign, attract each other with a
force of 0.108 N when separated by 0.5 m. The spheres
are connected by a conducting wire, which is then
removed, and thereafter, they repel each other with a
force of 0.036 N. The initial charges on the spheres
are
(1)
(2)
(3)
(4)
± 5 × 10–6 C and ± 15 × 10–6C
± 1.0 × 10–6 C and ± 3.0 × 10–6C
± 2.0 × 10–6 C and ± 6.0 × 10–6C
± 0.5 × 10–6
11. A charge particle q, is at position (2, –1, 3). The
electrostatic force on another charged particle q2 at
(0, 0, 0) is
(1)
(2)
(3)
q1 q 2
ˆ
(2iˆ – ˆj + 3k)
56 π∈0
q1 q 2
56 14 π∈0
q1 q 2
56 14 π∈0
ˆ
(2iˆ – ˆj + 3k)
ˆ
(ˆj – 2iˆ – 3k)
(4) None of these
Electric Charge Density
™
Linear charge density (l) is defined as the charge per unit
length.
Surface charge density (σ) is defined as the charge per unit
area.
™
σ=
dq
ds
s = constant (uniform distribution)
= variable (non uniform distribution)
where dq is the charge on an infinitesimal surface area ds.
Unit of (σ) is coulomb/meter2 (C/m2).
Examples: Plane sheet of charge, conducting sphere.
Volume charge density (ρ) is defined as charge per unit
volume.
™
dq
dV
r = constant (uniform distribution)
= variable (non uniform distribution)
ρ=
where dq is the charge on an infinitesimal volume element
dv. Unit of (ρ) is coulomb/meter3 (C/m3 )
Examples: Charge on a dielectric sphere i.e., non conducting
sphere etc.
Key Note
P
In conductors having non spherical surfaces, the surface
charge density (σ) will be larger when the radius of
curvature is small.
1

σ ∝ R 


P
The working of lightening conductor is based on leakage
of charge through sharp point due to high surface charge
density.
ELECTRIC FIELD
™
The space around an electric charge where its influence is felt
is known as electric field.
™
Electric field is a conservative field.
Electric Lines of Force and its Characteristics
™
Electric line of force is an imaginary path along which a unit
+ve test charge would tend to move in an electric field.
™
Electric lines of force start from +ve charge and end at –ve
charge.
E
dq
λ=
dl
l = constant (for uniform distribution)
= variable ( for nonuniform distribution)
where dq is the charge on an infinitesimal length dl.
Unit of l is coulomb / meter (C/m)
n)
tio
c
ire
(D
+q
–q
Examples: Straight charged wire, charged ring
P Electric Charges and Fields
W
7
™
Electric field line of similar charges repels each other.
(f)
(e)
cannot be circular
cannot be circular
(g)
+10C
–10C
(h)
cannot be directed
towards +ve charge
™
™
No. of electric field lines originated from positive charge
or terminated to negative charge is directly proportional to
strength of charge.
Electric lines of force in the case of isolated +ve charge are
radially outwards and in the case of isolated –ve charge are
radially inwards.
Difference Between Electric and
Magnetic Field Lines
™
™
™
A
q
™
The difference between electric field lines and magnetic field
lines is that magnetic field lines form closed loops and electric
field lines never form closed loops.
Electric field lines do not exist inside a conductor, but
magnetic field lines may exist inside a magnetic material.
Total electric lines of force linked with a closed surface may
or may not be zero, but total magnetic field lines linked with a
closed surface is always zero (as monopoles do not exist).
Intensity of Electric Field
–q
B
cannot intersect
The tangent at any point to the curve gives the direction of
electric field at that point.
EP
ER
R
Intensity of electric field is a vector quantity. Its direction is always
away from the positive charge and towards the negative charge.
Electric field intensity at a point in space is defined as the force
experienced by a unit test charge placed at that point.


F
E = Limit
δq →0 δq
P
F
∴ E=
q
Its units are N/C or V/m
™
™
™
™
(a)
Electric lines of force do not intersect.
Electric lines of force never exist inside a conductor.
The electric lines of force are perpendicular to equipotential
surface.
The force experienced by a unit positive test charge placed at
a point in the electric field gives the intensity of electric field
at that point both in magnitude and direction.
Uniform Electric Field (b) Non-uniform Field
E
E
(Direction)
B
q
F
F = qE
(c)
A
|EB| > |EA|
Non uniform field due (d) Non uniform field due
to direction
to change in magnitude
Following pattern of electrostatics field lines cannot exist
8
A positive charge experiences force in the direction of electric
field and a negative charge experiences force opposite to field.
Electric Field Intensity Due to Point Charge
Let us consider a point charge Q placed in vacuum at the origin
 
O. We place another point charge q at a point P, where OP = r .
q

Q
F

O
P
r
The electric field produced by the charge Q at a point P is given as

 F
1 Q
E
rˆ
= =
q 4πε0 r 2
™
™
S.I unit is newton/coulomb (NC–1) or volt / metre (Vm–1).
Dimensional formula MLT–3 A–1
If instead of a single charge, let the electric field is produced
by n number of charges, using the principle of super-position
resultant electric field intensity
  

E = E 1 + E 2 + E 3 + ....
NEET (XII) Moduel-1 PW
Train Your Brain
Example 18: A metal sphere is placed in an uniform electric
field which one is a correct electric line of force?
q
A
D
–q
So according to given problem
Fe
C
–2q
+2q
Sol. Side = 0.1 m q = 1 × 10–6 C
a
Half diagonal =
Sol. Only 4 is normal to the conducting surface. So 4 is a
correct electric line of force.
Example 19: Calculate the electric field intensity which
would be just sufficient to balance the weight of a particle
of charge –10 mC and mass 10–5 kg. (take g = 10 ms2)

Sol. As force on a charge q in an electric field E is


Fq = qE
B
q
2
A
=
B
0.1
2
A
–2q
⇒
–q
D
C
E E
O
+2q
D
C+q
Electric field due to charge q,
9 109 1 106 9 103 2
18 105 N/C
2
0.01
0.1 2
At the centre of the square there are two electric fields
which are perpendicular to each other, so net electric field
E0 E 2 E 2 E 2 18 105 2 2.54 106 N/C
A q E
Example 22: Show that if a charge q is placed at each
vertex of a regular polygon, the net electric field at its centre
is zero.
Sol.
q
q
q
W


Fq = W i.e., |q|E = mg
mg
= 10 N/Cin downward direction.
|q|
Example 20: Two charges 2 mC and 8 mC are placed 12 cm
apart. Find the position of the point where the electric field
intensity will be zero.
Sol. Let the E field be zero at point P, at a distance x from
2mC as shown
i.e.,=
E
2mC
8mC
P
12 – x
x
So, at P, the electric Field due to both the charges must be
equal and opposite.
⇒
k2
x
2
k8
(12 x)
2
Taking square root of both sides gives
1
2
x (12 x)
\ x = 4 cm
Example 21: Four charges are placed at the corners of
a square as shown in figure having side 10 cm. If q is
1 mC then what will be electric field intensity at the centre
of the square?
P Electric Charges and Fields
W
B
–q
E3
q
E2
E1
q
(a)
q
q
E2
q
E3
q
E1
E4
q
E2 q q E1
q E3 E4
(b)
q
q
E5
(c)
q
The distance of the centre of a regular polygon from
each vertex will be same. Therefore,
|E
=
=
1| |E
2 | | E3 | in figure (a)
in figure (b)
|E
=
=
1| |E
2 | | E3 |=| E4 |
E1 | | E
=
E3 | | E
=
and | =
2 | |=
4 | | E5 | in figure (c)
\ Net field is zero in all cases.
9
Example 23: An electric dipole is placed at some angle in
an electric field generated by a point charge
(1) The net electric force on the dipole must be zero
(2) The net electric force on the dipole may be zero
(3)The torque on the dipole due to the field must be
zero
(4)The torque on the dipole due to the field may be
zero
Sol. (4) Electric field produced by a point charge is nonuniform in nature.
Example 24: Abhishek, Hritik, John and Amir are assigned
the tasks of moving equal positive charges slowly through
an electric field, along assigned path (shown as dotted line).
In each case the charge is at rest at the beginning. They all
have paths of exactly equal lengths. Who must do the most
positive work?
Sol. (2) Displacement = −aiˆ − bjˆ and force F = qEiˆ
 
Work done W = F .S = −qEa
Example 26: Two identical particle of mass m carry a charge
Q each. Initially one is at rest on a smooth horizontal plane
and the other is projected along the plane directly towards
first particle from a large distance with speed v. The closest
distance of approach be
(1)
1 Q2
1 2Q 2
(2)
4πε0 mv
4πε0 mv 2
(3)
1 Q2
1 4Q 2
(4)
4πε0 mv
4πε0 mv 2
Sol. (2)
Q
v
Q
∞
Q
Q
r0
From mechanical energy conservation
KI + UI = Kf + Uf
1 2 KQ 2
mv =
2
r0
r0 =
(1) Abhishek (3) Amir (2) Hritik
(4) John
2KQ 2
mv 2
1 2Q 2
=
4πε0 mv 2
Sol. (4) John is moving opposite to direction of most strong
electric field. So, work done by john is most positive
Example 25: Point charge q moves from point P to point
S along the path PQRS (figure) in a uniform electric
field E pointing parallel to the positive direction of the
x-axis. The co-ordinates of the point P, Q, R and S are
(a, b, 0), (2a, 0, 0), (a, –b, 0) and (0, 0, 0) respectively. The
work done by the field in the above process is
Concept Application
12. The magnitude of electric field intensity E is such that
an electron placed in it would experience an electrical
force equal to its weight is given by
(1) mge
(2)
e
mg
(4)
(3)
(1)
qEa
qEa 2
(2) −qEa
2
2
(3) qEa 2 (4) aE (2a) + b
10
mg
e
e2
m2
g
13. The distance between the two charges 25mC and 36mC
is 11 cm At what point on the line joining the two, the
intensity will be zero.
(1) At a distance of 5 cm from 25 mC
(2) At a distance of 5 cm from 36 mC
(3) At a distance of 10 cm from 25 mC
(4) At a distance of 11 cm from 36 mC
NEET (XII) Moduel-1 PW
™
14. A charge produces an electric field of 1N/C at a point
distance 0.1 m from it. The magnitude of charge is
(1) 1.11 × 10–12 C
(2) 9.11 × 10–12 C
(3) 7.11 × 10–6 C
(4) None of these
At any instant of time t, horizontal component of velocity,
vx = u and vertical component of velocity
 Eq 
v y= at= 
t
 m 
v x2 + v y2 =
∴v = v =
Motion of a Charged Particle in a
Uniform Electric Field
(a) A charged body of mass ‘m’ and charge ‘q’ is initially at rest
in a uniform electric field of intensity E. The force acting
on it is given by, F = Eq.
™
K .E
=
E 2 q 2t 2
m2
Null Points (Or) Neutral Points
™
Two charges q1 and q2 are separated by a distance ‘d’. Then
the point of zero intensity (null point) lies at a distance of
x=
The body travels in a straight line path with uniform
F Eq
acceleration, =
.
a =
m m
At any instant of time t,
 Eq 
Its final velocity, v =u + at =
 t [Q u = 0]
 m 
1 2 1  Eq  2
Displacement s =+
ut
at =
t
2
2 m 
Momentum, P = mv = (Eq)t
Kinetic energy,
u2 +
d
q2
±1
q1
from q1
+ve sign for like charges –ve sign for unlike charges.
Some Important Considerations about
Electric Field
™
1 2 1  E 2q2  2
mv
=

t
2
2  m 
Two charges +Q each are placed at the two vertices of an
equilateral triangle of side a. The intensity of electric field at
the third vertex is
E′ = 2E cos
θ
=
2
3E (Q q = 60°)
(b) When a charged particle enters perpendicularly into a
uniform electric field of intensity E with a velocity v then it
describes parabolic path as shown in figure.
u
E′ =
™
Along the horizontal direction, there is no acceleration and
hence x = ut.
3
1 Q
.
4π ∈0 a 2
Two charges +Q, –Q are placed at the two vertices of an
equilateral triangle of side ‘a’, then the intensity of electric
field at the third vertex is
E′ = 2E cos
θ
= E (Q q = 120°)
2
Along the vertical direction, acceleration
F Eq
=
(here gravitational force is not considered)
m m
Hence vertical displacement,
a
=
1  Eq  2
y= 
t
2  m 
2
 qE  2
1  qE   x 
=
y =
x


 
2
2  m  u 
2
mu


P Electric Charges and Fields
W
E′ =
1 Q
.
4π ∈0 a 2
11
™
If three charges +Q each are placed at the three vertices of
an equilateral triangle of side ‘a’ then the intensity of electric
field at the centroid is zero.
Electric Field Due to Finite and Infinite
Line Charge
(1) Electric field due to finite line charge at a point along
the line of charge.
q charge in uniformly distributed over its length.
x
™
If three charges +Q each are placed at the three corners of a
square of side ‘a’ as shown in figure.
P
++++++++++++++++++
dx
L
dE =
r
dq
4πε0 x 2
To find field at O, take a point charge dq at distance x.
dq
dE =
4πε0 x 2
As the field due to all such point charges is along the same
direction E = ∫dE
q
q dx
dx ⇒ dE =
L
4πε0 L x 2
As, dq =
Intensity of electric field at the fourth corner
=
1 Q
1
Q
E
and E ' =
Where E =
.
=
2
2
4π ∈0 a
4π ∈0 2a
2
2E + E ' .
Hence the intensity of electric field at the fourth corner
1

= E 2 + 
2

™
=
⇒E
l
g−
EQ
m
+Q
q dx
1
q
=
4πε0 L x 2 4πε0 (r )(r + L)
r+L
r
(2) Electric field due to finite line charge at perpendicular
distance
P
q2 q1
r
The bob of a simple pendulum is given a +ve charge and it is
made to oscillate in a vertically upward electric field, then the
time period of oscillation = 2π
∫
+++++++++++++++
l
=
E⊥
1 λ
(sin θ1 + sin θ2 ) (Perpendicular to rod)
4πε0 r
=
E
1 λ
(cos θ2 − cos θ1 )
4πε0 r
(Parallel to rod)
(3) Electric field due to circular arc
Y
In the above case, if the bob is given a –ve charge then the time
period is = 2π
l
g+
+
EQ
m
+ + + +
+
q
O
X
–Q
E0 =
12
λ
θ
sin   along negative y-axis
2πε0 R
2
NEET (XII) Moduel-1 PW
4. Electric Field at the Axis of a Uniformly Charged Ring
1. For points on the axis
Let the point P be at distance r from the centre of the dipole
on the side of the charge q, as shown in Fig.(a). Then
q
pˆ E − q = −
...(i)
2
4πε 0 ( r + a )
R
R
Intensity of electric field at a point P that lies on the axis of the
1
qx
ring at a distance x from its centre is E =
3
4πε0 2
x + R2 2
(
R
2
and Emax =
2  1 q 


3 3  4πε0 R 2 
E
x= −
R
2a
2
R
2
x
P
+q
E–q
r
Fig.: (a) a point on the axis,
–q
E+q
The total field at P is
=
E E + q + E −q
=
Emax
x=
where p̂ is the unit vector along the dipole axis (from –q
to q). Also
q
E+q = +
pˆ ...(ii)
2
4πε 0 ( r − a )
)
Where R is the radius of the ring. From the above expression E
= 0 at the centre of the ring.
dE
E will be maximum when
= 0.
dx
Differentiating E w.r.t x and putting it equal to zero we get
x= ±
Electric Field Intensity due to an Electric Dipole
q
4ar
4πε 0 r 2 − a 2
(
)
2
p̂ ...(iii)
For r >> a
4qa
pˆ
=
( r >> a ) E
4πε0 r 3

2Kp
= r 3
2. For points on the equatorial plane
...(iv)
+q
DIPOLE
An electric dipole is a pair of equal and opposite point charges q
and –q separated by a small distance 2a.
Dipole Moment ( p )
It is defined as the product of magnitude of either charge and the
distance between the two charges.


p = q 2a

Dipole moment p always points from –q to +q.
–q
( )
→
Fig.: (b) a point on the equatorial plane of the dipole. p
is the dipole moment vector of magnitude p = q × 2a and
directed from –q to q
The magnitudes of the electric fields due to the two charges
+q and –q are given by
q
1
E + q =
...(v)
2
4πε0 r + a 2
q
1
E − q =
...(vi)
2
4πε0 r + a 2
The directions of E+q and E–q are as shown in Fig.(b). Clearly,
the components normal to the dipole axis cancel away. The
components along the dipole axis add up.
P Electric Charges and Fields
W
13
(
)
E=
− E + q + E − q cos θpˆ
= −
(
2qa
4πε 0 r 2 + a 2
)
3/2
p̂
or tan
=
α
1
tan θ
2
⸫ α can be calculated
...(vii)
Particular Cases:
At large distances (r >> a)
1. When the point K lies on axial line of dipole.
2qa
θ = 0°, cos θ = cos 0° = 1
E = −
pˆ
( r >> a ) ...(viii)
4πε0 r 3

p
2p

3cos 2 0° + 1
=
∴| E |
=
3
−Kp
4
π
∈
r
4
π
∈0 r 3
0
= 3
r
1
tan
0° 0, ∴ α = 0°
and tan
=
α
=
2
Electric Field due to a Short Dipole at any Point
i.e., resultant intensity is along the axial line.
K(r, q)

2.
When the point K lies on equatorial line of dipole.
Figure AB represents a short electric dipole of moment p along

θ = 90°, cos θ = cos 90° = 0
AB and O is the centre of the dipole. We have to calculate electric

p
p
3cos 2 90° + 1
=
∴E
=
field E intensity at any point K, where
4π ∈0 r 3
4π ∈0 r 3
1
tan 90° = ∞, α = 90°
2
i.e., direction of resultant field intensity is perpendicular to the
equatorial line (and hence anti-parallel to axial line of dipole).
and tan α =
Torque on a Dipole Placed in a Uniform
Electric Field
OK = r, ∠BOK = θ

The dipole moment p can be resolved into two rectangular
components:
(p cosθ) along A1B1 and (p sinθ) along A2B2 ⊥ A1B1. Electric
field intensity at K on the axial line of A1B1,


2p cos
E1
along KL
3
4π ∈0 r
The torque due to the force on the positive charge about a point O
is given by Fa sin q. The torque on the negative charge about O
is also Fa sin q
t = 2 Fa sin q
⇒ t = 2aq E sin q
⇒ t = pE sin q
a
Electric field intensity at K on equatorial line of A2B2,


p sin θ
E2 =
along
KM
4π ∈0 r 3

∴ | E |=
2
E12 + E 22 =
p
3
 2p cos θ   p sin θ 
+

3 
3 
 

 4π ∈0 r   4π ∈0 r 
(
4π ∈0 r 3
  
τ= p × E
)
3cos 2 θ + 1
α
In ∆KLN, tan=
Key Note
Let electric force between two dipoles be F, then
P
 p 3cos 2 θ + 1
i.e., | E |=
4π ∈0 r 3
Let ∠LKN = α
14
a
2
4 cos 2 θ + sin 2 θ
4π ∈0 r

p
=
|E|
3cos 2 θ + cos 2 θ + sin 2 θ
3
4π ∈0 r
p
LN KM
p sin θ 4π ∈0 r 3
=
=
.
KL KL 4π ∈0 r 3 2 p cos θ
P
1 6 p1 p2
when dipoles are placed coaxially to
4πε0 r 4
each other.
F=
1 3 p1 p2
when dipoles are placed perpendicular
4πε0 r 4
to each other.
F=
NEET (XII) Moduel-1 PW
Work Done in Rotating a Dipole in a Uniform
Electric Field
Sol. (3)
L
2
When an electric dipole is placed in a uniform electric field E,
→
→
dW = τdθ
⇒
dW = pE sinθ dθ
θ = θ1 to θ = θ2 will be given by
=
W
∫
θ2
θ1
M
–q
pEsin θ dθ
Wexternal forces = pE (cosθ1 – cosθ2)
and work done by electric forces,
Welectric force = –Wexternal force = pE(cosθ2 – cosθ1)
If Taking θ1 = θ and θ2 = 90°
We have,
Welectric dipole = p. E (cos90° – cosθ) = – pE cosθ
Train Your Brain
Example 27: The electric field due to a short dipole at a
distance r, on the axial line, from its mid point is the same
as that of electric field at a distance r′, on the equatorial line,
r
from its mid-point. Determine the ratio .
r´
1 2p
1 p
Sol.
=
3
4πε0 r
4πε0 r ′3
r
=
1
r3
or 3 = 2
3
r′
r′
r
or
= 21/3
r′
Example 28: A point particle of mass M is attached to
one end of a massless rigid non-conducting rod of length
L. Another point particle of same mass is attached to the
other end of the rod. The two particles carry charges +q
and –q respectively. This arrangement is held in a region
of uniform electric field E such that the rod makes a small
angle (< 5°) with the field direction. The minimum time
needed for the rod to become parallel to the field after it is
set free. (rod rotates about centre of mass)
(1) 2π
(3)
ML
2qE
π ML
2 2qE
P Electric Charges and Fields
W
(2) π
(4) 4π
+q
M
qE
L
ML2
sin
=
θ
×2
2
4
qE L sin θ
4
×
a= −
2
ML2
2qE
θ [For small q]
a= −
ML
= –w2q.
ML
2qE
ML
2qE
2qE
ML
w2 =
2qE 2π
=
ML
T
w=
ML
2qE
T = 2π
→ →
= − p. E
3
q
q
2 × qE
Total work done by external forces in rotating a dipole from,
2
L
2
→
a torque, τ = p × E acts on it. If we rotate the dipole through a
small angle dθ the work done by the torque is
qE
t=
T
4
t=
π
2
ML
2qE
Example 29: The force on a charge situated on the axis of
a dipole is F, if the charge is shifted to double the distance,
the acting force will be
(1) Zero
F
4
Sol. (4) F = qE
At axial point
(3)
E=
E∝
F
2
F
(4)
8
(2)
2KP
r3
1
r3
When distance is doubled
E
E'=
8
F
\ F′ =
8
Example 30: An electrical dipole with dipole moment

p = 3iˆ + 4 ˆj × 10−30 C − m is placed in an electric field

E = 4000iˆ(N/C) . An external agent turns the dipole slowly
until its electric dipole moment becomes −4iˆ + 3 ˆj × 10−30
C – m. The work done by the external agent is equal to
(1) 4 × 10–28 J
(2) –4 × 10–28 J
–26
(3) 2.8 × 10 J
(4) –2.8 × 10–26 J
(
)
(
)
15
Sol. (3)
4 ˆj
53°
q = 53°
3iˆ
3 ˆj
−4iˆ
Concept Application
15. Find the magnitude of dipole moment of the following
system.
+2q
W = PE(cosq1 – cosq2)
W = 5 × 4000[cos 53° – cos 143°] × 10–30
3 4
−30
= 20000  +  × 10
5
5


7
−30
= 20000 × × 10
5
= 4000 × 7 × 10–30
= 28 × 103 × 10–30
= 2.8 × 10–26 J
Example 31: Two charges, each of 5μC but opposite in sign,
are placed 4 cm apart. Calculate the electric field intensity
of a point that is at a distance 4 cm from the mid point on
the axial line of the dipole.
Sol. We can no use formula of short dipole here because
distance of the point is comparable to the distance between
the two point charges.
q = 5 × 10–6 C, a = 4 × 10–2 m, r = 4 × 10–2 m
+3q
60°
a
–5mC
4cm
Eres = E+ + E–
=
k (5µC )
2
−
k ( 5µC )
144
=
=
NC−1 108 NC−1
2
−8
(6 cm)
144 × 10
(2 cm)
Example 32: Two charges ±10 mC are placed 5 × 10–3 m
apart. Determine the electric field at a point Q which is 0.15
m away from O, on the equatorial line.
Sol. In the given problem, r >> a
Q
–5q
(1) qa 21
(2) qa 13
(3) qa 19
(4) Zero
16. Four charges are placed each at a distance a from
origin. The dipole moment of configuration is
Y
3q
–2q
O
+2q
X
q
4cm
5mC
60°
(2) 3qajˆ
(1) 2qajˆ
(3) 2aq(i + j)
(4) 3aq(i + j)
17. An electric dipole is placed perpendicular to an infinite
line of charge at some distance as shown in figure.
identify the correct statement.
+
+
+
+
+
+

P
(1) The dipole is attracted towards the line charge
(2) The dipole is repelled away from the line charge
0.15 m
(3) The dipole does not experience a force
(4) The dipole experience a force as well as a torque
–10mC
=
\ E
O
+10mC
1 p
1 q(a)
=
4πε0 r 3 4πε0 r 3
ELECTRIC FLUX AND ITS APPLICATION

Electric flux Df through an area element ∆S is defined as
 
10 × 10 × 5 × 10
or E = 9 × 109
NC−1 =
1.33 × 105 NC-1
Df = E ⋅ ∆ S = E ∆S cos θ . The angle q here is the angle between
0.15 × 0.15 × 0.15


electric field E and are element vector ∆S .
−6
16
−3
NEET (XII) Moduel-1 PW

∆S
Applications of Gauss’s Law
1. Electric field due to an infinitely long charged wire.

E
q

For a closed surface, q is angle between E and the outward normal
to the area element. The SI unit of electric flux is Nm2C–1. It is a
scalar quantity. Mathematically electric flux Df is proportional to
the number of electric field lines cutting the area element.
GAUSS’S LAW
According to Gauss’s law, “the net electric flux through any
closed surface is equal to the net charge enclosed by it divided by
ɛ0”. Mathematically, it can be written as
φ E=
→
→
∫ E⋅ dS=
q
ε0
Derivation of Gauss’s Law
Let a point charge +q be placed at centre O of a sphere S. Then S
is a Gaussian surface.
Electric field at any point on S is given by
Electric flux through the curved part of the Gaussian surface
=
φ
E
→ →
= ∫ E.dS cos0
∫ E.ds
1
0
= E × 2πrl
(Electric flux through other surfaces becomes zero as
θ = 90°)
q
ε0
Using Gauss law, φE =
E.2πrl =
λl
λ
⇒ E=
ε0
2πε0 r
2. Electric field intensity due to a uniformly charged infinite
plane sheet.
ds = A
1 .q
E=
4πε0 r 2
The electric field and area element points radially outwards. So
θ = 0°.

Flux through area dS is
→ →
=
dφ E=
.dS EdScos
=
0° EdS
Total flux through surface S is
φ=
∫ dφ=
S
∫ EdS= E ∫ dS= E × Area of Sphere
S
→ →
ϕE =
∫ E.ds = 2EA
(due to both circular surface)
q = σA
q
ε0
Using Gauss’s law, φE =
2EA =
σA
σ
⇒ E=
2ε 0
ε0
S
1 q
q
=
. 2 .4πr 2=
or, φ
which proves Gauss 's theorem.
ε0
4πε0 r
3. Electric field intensity due to two equally and oppositely
charged parallel plane sheets of charge.
Continuous charge distribution
Linear charge distribution: q = λl
Where λ = linear charge density.
Surface charge distribution: q = σA
where σ = surface charge density.
Volume charge distribution: q = ρA
where ρ = volume charge density.
P Electric Charges and Fields
W
E=
σ
(between two plates)
ε0
–
–
–
–
–
–
E = 0 (outside the plates)
17
4. Electric field intensity due to two positively charged
parallel plane sheets of charge.
Consider, σ1 > σ2 > 0;
In region I: E net = E1 + E 2

We have to calculate electric field intensity E at any point
P, where OP = r. With O as centre and r as radius, imagine a
sphere S, which acts as a Gaussian surface.
If q´ is the charge enclosed by the sphere S, then according to
Gauss’s law
( σ + σ2 )
= 1
2ε0
In region II: E net = E1 − E 2 =
 
( towards right )
( σ1 − σ2 )
2ε 0
( σ + σ2 )
= 1
In region III: E net = E1 + E 2
2ε0
(towards right)
(towards left)
5. Electric field due to uniformly charged thin spherical
shell.
q′
∫ E. ds = ∈
S
where ∈ is electrical permitivity of the material of the
insulating sphere.
∴
(
)
E 4=
πr 2
q'
q'
or E
...(i)
=
∈
4π ∈ r 2
Now, charge inside S is q′ = volume of S × volume density of
charge
4 3
πr × ρ
3
Putting the value of q′ in eqn (i)
q '=
=
E
Inside the sphere (r < R)
By Gauss’s law,

q
∫ E.ds = ε
0
E = 0 (as q = 0 inside)
Outside the sphere (r > R)
By Gauss’s law,

q
∫ E.ds = ε
0
q
E × 4πr =
ε0
σR 2
1 q
 q =σ4πR 2 
E=
. 2 ⇒ E=
4πε0 r
∈0 r 2
2
1
⇒ E∝ 2
r
On the surface (r = R)
E=
q
4πε0 R 2
4 πr 3ρ
rρ
...(ii)
=
3 4π ∈ r 2 3 ∈
⇒E∝r
i.e., electric intensity at any point inside a non-conducting
charged solid sphere varies directly as the distance of the
point from the centre of the sphere.
At the centre of the sphere, r = 0.
⸫
E=0
Also,
=
E
Rρ
⇒ maximum at surface of sphere
3∈
We have already proved that outside the sphere,
E ∝ 1/r2.
Figure below represents the variation of electric field intensity
E with distance (r) from the centre of a non conducting
uniformly charged solid sphere.
σ
⇒ E=
ε0
6. Electric Field Intensity Due to a Non-conducting Charged
Solid Sphere
Suppose a non conducting solid sphere of radius R and centre
O has uniform volume density of charge ρ.
18
NEET (XII) Moduel-1 PW
Train Your Brain
Example 33: A metal sphere a of radius r1 charged to a
potential f1 is enveloped by a thin walled conducting
spherical shell B of radius r2. Then potential (f2) of the
sphere A after it is connected by a thin conducting wire to
the shell B will be
Example 35: A non-conducting solid sphere of radius R is
uniformly charged. The magnitude of the electric field due
to the sphere at a distance r from its centre
(1) A, C(2) C, D
(3) A, B(4) B, D
Sol. (1)
E∝r
E
E∝
1
r2
r=R
r
Example 36: Three concentric metallic spherical shells of
radii R, 2R, 3R, are given charges Q1, Q2, Q3 respectively.
It is found that the surface charge densities on the surfaces
of the shells are equal. Then, the ratio of the charges given
to the shells, Q1 : Q2 : Q3, is
(1) 1 : 2 : 3
(2) 1 : 3 : 5
(3) 1 : 4 : 9
(4) None of these
Sol. (2)
O
(1) φ1
 r2 
rl
(2) φ1  
r2
 r1 
 r2 
(3) φ1 1 − 
 r1 
 r1r2 
(4) φ1 

 r1 + r2 
q
Kq
Sol. (1) φ1 = r1
2R
R Q1 Q 2
s
3R
q will be transfered to the surface of outer conductor
Kq φ1r1
=
r2
r2
Example 34: A sphere of radius R and charge Q is placed
inside an imaginary sphere of radius 2R whose centre
coincides with the given sphere. The flux related to
imaginary sphere is
φ2=
(1)
Q
Q
(2)
∈0
2 ∈0
(3)
2Q
4Q
(4)
∈0
∈0
Sol. (1)
2R
Q
R
Flux depends on charge enclosed. It does not depend
on shape & size of Gaussian surface.
A. increases as r increase, for r < R
B. decreases as r increase, for 0 < r < ∞
C. decrease as r increases, for R < r < ∞
D. is discontinuous at r = R
P Electric Charges and Fields
W
Q3
Q1+Q2+Q3
–(Q1+Q2)
Q1+Q2
–Q1
Q1
s
s
Q1
:
Q1 + Q2
2
:
Q1 + Q2 + Q3
= 1:1:1
4πR 4π(2 R )
4π(3R ) 2
Q + Q2 Q1 + Q2 + Q3
Q1 : 1
:
4
9
Q1 + Q2 + Q3
Q1 + Q2
= Q1
= Q1
4
9
Q2 = 3Q1 Q1 + 3Q1 + Q3 = 9Q1 ⇒ Q3 = 5Q1
Q1 : Q2 : Q3 = 1 : 3 : 5
Example 37: A rectangular surface of sides 10 cm and
15 cm is placed inside a uniform electric field of 25 N/C,
such that normal to the surface makes an angle of 60° with
the direction of electric field. Find the flux of electric field
through the rectangular surface.
Sol. Flux = EAS cosq
= (25 N/C)(0.15 × 0.10)m2(cos60°)
= 0.1875 N m2/C
Example 38: A hemispherical surface of radius R is kept in
a uniform electric field E such that E is parallel to the axis
of hemisphere, Net flux from the surface will be
2
19
Sol. f =
→ →
Concept Application
∫ E . ds
s
= (E) (Area of surface perpendicular to E) = E. pR2.
Example 39: In a region of space, the electric field is

in the x-direction and proportional to x, i.e., E = E0 xiˆ.
Consider an imaginary cubical volume of edge a, with its
edges parallel to the axes of coordinates. What is the charge
enclosed by this volume?
Sol. Using the Gauss’s Law, Net flux
Qnet
ε0

E
therefore Qnet enclosed = e0 (Net flux) = e0E0a3
y
O
Surface 1
C
E
B
d
D
A
F
H
G
h
Surface
2
R
x
ERh
2
(3) ERh
(1)
z
Example 40: A long string with a charge of l per unit
length passes through an imaginary cube of edge a. What
is the maximum flux of the electric field through the cube?
1
Sol. As per Gauss’s law, the maximum flux is
times the
ε0
maximum charge included in the cube.
Maximum length in a cube is the largest diagonal =
So, the maximum flux =
18. Consider two concentric spherical surface S1 with
radius R and S2 with radius 2R both centered at the
origin. There is a charge +2q at the origin and there
are no other charges. Compare the flux f1 through S1
with the flux f2 through S2.
(1) f1 = 2f2
(2) f1 = 4f2
φ
(3) f1 = f2
(4) φ1 = 2
2
19. In the given figure a cone lies in a uniform electric
field E. The electric flux entering the cone is
(2) 2ERh
(4) Zero
20. Eight point charges (can be assumed as small spheres
uniformly charged and their centres at the corner of the
cube) having values q each are fixed at vertices of a
cube. The electric flux through square surface ABCD
of the cube is
3a
3aλ
ε0
Example 41: If a charge Q is kept at the mouth centre
of a hemisphere of radius R, what is flux through the
hemispherical surface?
Sol. Total flux a point charge Q happens to have is Q/e0
through an sphere so the flux through one hemisphere is
Q/2e0.
Example 42: A charge Q is placed at the centre of the edge
of a cube. What is the flux through the cube?
Sol. Once again you have to cover the charge completely
and symmetrically.
(1)
q
12 ∈0
q
(4)
8 ∈0
q
24 ∈0
(2)
q
6 ∈0
21. Consider the charge configuration and a spherical
Gaussian surface as shown in the figure. When
calculating the flux of the electric field over the
spherical surface, the electric field will be due to
(3)
Q
Q
In this case you require total four cubes to cover the Q
completely as shown in the other figure. So the flux through
one of the cubes is Q/4e0.
20
(1)
(2)
(3)
(4)
q2
Only the positive charges
All the charge
None of these
NEET (XII) Moduel-1 PW
Short Notes
Coulomb’s Law

q1q 2 
1
Force between two charges F =
r, ∈r = dielectric
4π ∈0 ∈r r 2
constant
q1
Equilibrium of Suspended Point Charge System
For equilibrium position
q2
r
Tcos
Principle of Superposition
Fe
Force on a point charge due to many charges is given by
   
F = F1 + F2 + F3 + ........
⇒ tan=
θ


F
1 Q
=
E Lim
=
r
q0 → 0 q
4π ∈0 r 2
0
Fe
kQ 2
= 2
mg x mg
(Fe ) 2 + (mg) 2
q
⇒ T = Fe =
Null Point for Two Charges
q
180º
kq 2
4 2
Electric Dipole
r
Q2
If |Q1| > |Q2|
™
™
™
⇒ Null point near Q2
Q1 ± Q 2
Q
If whole set up is taken into an artificial satellite (geff  0)
2
Electric Field Due to Charge Q
Q1 r
Q
mg
T cosq = mg & T sinq = Fe
=
T
Q1
x
Electric Field or Electric Field Intensity
(Vector Quantity)
x=
T
Tsin
Notes: The force due to one charge is not affected by the presence
of other charges.

 F
E = , unit is N/C or V/m.
q
; x → distance of null point from Q1 charge
Electric dipole moment p = qd
  
Torque on dipole placed in uniform electric field τ = p × E
At a point which is at a distance r from dipole midpoint and
making angle q with dipole axis.
y
E
Er
E
(+) for like charges
(–) for unlike charges
P
r
Equilibrium of Three Point Charges
–q
+q
x
x
q
r
Q1
(i) Two charges must be of like nature.
(ii) Third charge should be of unlike nature.
x=
Q1
Q1 + Q 2
r and q =
(
P Electric Charges and Fields
W
Electric field E =
Q2
−Q1Q 2
Q1 + Q 2
)
2
Direction
™
™
1 p 1 + 3cos 2 θ
4π ∈0
r3
tan =
α
Eθ 1
=
tan θ
Er 2


1 2p
Electric field at axial point (or End-on) E =
of
4π ∈0 r 3
dipole
Electric field at equatorial position (Broad-on) of dipole

E=

1 (−p)
4π ∈0 r 3
21
 
Electric flux: φ = ∫ E.ds
Gauss’s Law:
surface)
 
∫ E.ds =
+
∑q
q
R
+
(Applicable only on closed
∈0
q
Net flux emerging out of a closed surface is en
ε0
  q en
where qen = net charge enclosed by the
=
φ ∫=
E.dA
ε0
+
+
For r < R : E = 0
For a Charged Circular Ring
E
closed surface.
φ does not depend on the
(i) Shape and size of the closed surface
(ii) The charges located outside the closed surface.
V
x
For a Conducting Sphere
+
q
+
q
R
+
+
+
+ C
x
P
R
+
For r ≥ R : E =
1 q
and For r < R : E = 0
4π ∈0 r 2
+
EP =
E
1
qx
2
4π ∈0 (x + R 2 )3/ 2
Electric field will be maximum at x = ±
r
R
+
™
q
For r ≥ R : E =
2π ∈0 r
™
For r < R : E = 0
q
R
+
+
Electric Field Intensity at a Point near a Charged
Conductor
+
1 q
For r ≥ R : E =
4π ∈0 r 2
E=
σ2
2 ∈0
P=
R
r
1 qr
For r < R : E =
4π ∈0 R 3
For a Conducting/Non-conducting Spherical Shell
22
1 q
4π ∈0 r 2
σ
∈0
Mechanical Pressure on a Charged Conductor
E
For r ≥ R : E =
2
For a Charged Long Conducting Cylinder
For a Non-conducting Sphere
+
R
For Non-conducting Infinite Sheet of Surface
Charged Density s
E=
σ
2 ∈0
For Conducting Infinite Sheet of Surface Charge
Density s
E=
σ
∈0
NEET (XII) Moduel-1 PW
x
Solved Examples
q
(1)
q
+
++++++
++
q
q
+
++
++
++++++
++
+ + + + + ++
+
+
+
++++++
++
+
++
q
+
+++ ++++
q
+
∴
θ= 27°
From (i), =
T
qE
3 × 10 −8 × 2 × 104
=
= 8.8 × 10 −4 N
sin θ
sin 27°
2. Two infinite plane parallel sheets, separated by a distance d
have equal and opposite uniform charge densities σ. Electric
field at a point between the sheet is
σ
(1)
2ε0
σ
(2)
ε0
(3) Zero
(4) Depends on the location of the point
Sol. (2)
At every point between the plates.
3. A metallic shell has a point charge q kept inside its cavity.
Which one of the following diagrams correctly represents
the electric lines of forces?
P Electric Charges and Fields
W
+
3 × 10 −8 × 2 × 104
=
0.5102
80 × 10 −6 × 9.8
+ + + + + ++
+
=
tan θ
...(i)
++
In equilibrium, T sinθ = qE
and
T cosθ = mg
Dividing, we get
qE
tan θ =
mg
++
–ve
++
+
Sol. (3)
+ve
+
+
++
++
q
q
(4)
++
++
++++++++
++
+
+++ ++++
(3)
++
++
+
q
(2)
+
1. A pendulum bob of mass 80 milligram and carrying a charge
of 3 × 10–8 C is at rest in a horizontal uniform electric
field of 2 × 104 Vm–1. Find the tension in the thread of the
pendulum and the angle it makes with the vertical.
(1) 8.8 × 10–4 N, 27°
(2) 4.2 × 10–6 N, 28°
–10
(3) 9.8 × 10 N, 25°
(4) 7.2 × 10–4 N, 30°
Sol. (1) Here, m = 80 mg = 80 × 10–6 kg, q = 3 × 10–8 C,
E = 2 × 104 V/m
Let T be the tension in the string and θ be the angle it
makes with the vertical, Fig.
No field lines inside metal
→
E is ⊥ to surface
4. Four point + ve charges of same magnitude (Q) are placed
at four corners of a rigid square frame as shown in figure.
The plane of the frame is perpendicular to z-axis. If a – ve
point charge is placed at a distance z away from the above
frame, the
Z
–q
+Q
z << L
+Q
L
O
+Q
L
L
+Q
(1) – q charge oscillates along the Z-axis
(2) It moves away from the frame
(3) It moves slowly towards the frame and stays in the plane
of the frame
(4) It passes through the frame only once
Sol. (1) Since, the net force on the –ve charge will act
downwards due to force of attraction it goes down the
frame, till it reaches the plane of frame the charge
continues to experience downward force but as it goes
down it starts experiencing an upward pull. Similarly,
it will always experience an attractive force in the
z-axis. Hence, it will oscillate along z-axis.
23
5. The spatial distribution of the electric field due to two
charges (A, B) is as shown in figure
Fnet = O
KQ 2
KQq
2 0
2


/ 2
q Q / 4
Which one of the following statements is correct?
(1) A is + ve and B – ve and |A| > |B|
(2) A is – ve and B +ve and |A| = |B|
(3) Both are +ve but A > B
(4) Both are –ve but A > B
r
Sol. (1) E lines are coming out of A and reaching to B this
shows that A must be +ve and B should be a –ve
charge. Also, more will be the electric lines of forces
more will be its magnitude.
8. Two charge +q and –q are placed at a distance of 2r as
shown in the figure. Electric field at the centre of the line
joining 2 charges is
+q
–q
2r
Kq
r2
2Kq
(3)
r2
4Kq
r2
Kq
(4)
2r 2
(1)
(2)
Kq
r2
E1 + E2
Sol. (3) Enet = E1 + E2 = 2E = 2
6. 3 charges are placed as shown in the figure. The net force
on charge q is
q
+Q (0,a)
(x,0)
+q
+Q (0,–a)
(1)
(3)
2KQqx
(a
2
+x
2
2
)y
(2)
2
KQqx
(a + x
2
(4)
)
2 3/ 2
2KQqx
(a
2
+ x2 )
3/ 2
3 KQqx
2 ( a 2 + x 2 )3/ 2
9. An electric dipole is placed at an angle of 60° to a nonuniform electric field. The dipole will experience
(1) A translational force only in the direction of field
(2) A torque only
(3) A translational force only in direction normal to the
direction of field
(4) A translational force as well as a torque
Sol. (4)
F1 = qE1
0
60
qE2 = F2
+Q
F
Sol. (2)
θ
θ
θ
θ
Fnet
F
+Q
Fnet = 2F cosθ = 2 ×
KQq
x
2
2
2
a x a x2
Electric field at the location of 2 charges will be different.
Hence, the 2 forces will be unequal. The net force will be
non-zero. Hence, it will experience a translational as well
as rotational motion.
10. In a uniform charged thin spherical shell of total charge
Q and radius R, the electric field is plotted as function of
distance, from the center. The corresponding graph is
E
2KQqx
a
2
x2 3
2
7. Two equal charges ‘Q’ are placed at a distance l apart. A
third charge ‘q’ is placed between them such that entire
system is in equilibrium. The value of charge q is
(1) –Q/4
(2) Q/4
(3) Q/2
(4) –Q/2
q
Sol. (1) Q
Q
/2
24
–q
/2
(1)
(2)
r
R
E
E
(3)
(4)
R
r
R
r
NEET (XII) Moduel-1 PW

Sol. (2) E inside = 0

Kq
E on the surface = 2

Kq
E outside the shell = 2
R
r
11. Charge Q is placed at each of the two diagonally opposite
corner of a square. Also charge q is placed at each of the
other two corners. If net force on charge Q is zero, then Q/q
is
(1) −2 2
(2) –2
(3) − 2
(4)
Sol. (1) Q
1
2
q
2 F2
F2
q
∴ qE= mg ⇒ q =
mg 9 × 10 −15 × 10
=
= 4.5 × 10–19 C
E
2 × 10 5
14. Charge q is uniformly distributed over a thin half ring of
radius R. The electric field at the centre of the ring is
(1)
q
2π ε 0 R 2
(3)
q
4πεR 2
q
4π ε 0 R 2
q
(4)
2πε0 R 2
(2)
2
2
0
45
F2
Sol. (1)
F1
∴ 2 F2 =
−F1 ⇒ 2
⇒
Q
=
−2 2
q
+++ +
dq + +
++ q
+
+
+
+
θ +
d
+
+
+
+
+
+
θ
+
+
θ
dEcos θ
dEcos θ
r
dE
dE
2dEsinθ
Q
Let F1 be the force b/w Q and Q. For the net force on Q
to be zero, the force between Q and q should be attractive
Let F2 be the force between Q and q.
∴ net force on Q is zero.
2F2 cos450 = –F1
KQq
KQ 2
=
−
2
l2
2l
( )
dEnet
12. The flux through the hemispherical shell as shown in the
figure is
q
.
x
(1)
Sol. (4) At equilibrium, electric force on charged oil drop will
balance the weight
dθ
13. A charged oil drop is suspended in a uniform field of
2 × 105 V/m so that it neither fall nor rises. The charge on
the drop will be
(mass of the charge is 9 × 10–15 kg)
(1) 9 × 10–18 C
(2) 9 × 10–19 C
(3) 4.5 × 10–20 C
(4) 4.5 × 10–19 C
q
ε0
(2)
q
2ε0
(4) 2q
ε0
(3) 0
Sol. (2)
q
kdq
R2
2kdq
q
qd 2dE sin 2 sin dq Rd R
R
Electric field due to small charge dq is dE =
dE
net
l 2
o
2k
sin d
q
R2

1 
q
 k=
 ⇒ Enet =
2
4
πε
2
π
ε0 R 2
0 

∴ Horizontal components gets cancelled, and only vertical

component of E field gets added due to the field of a charge
element in opposite half.
15. Considers the following system of two charges q and 4q,
separated by a distance of 1 m. Charge ‘Q’ is placed near
them such that entire system comes in equilibrium. The
value of charge Q is
Q
q
4q
1m
q
ε0
q
∴ for hemisphere, φ =
2ε0
For complete sphere, φ =
P Electric Charges and Fields
W
9
q
4
(3) 2 q
9
(1)
(2)
4
q
9
(4) 5 q
9
25
Sol. (2) Let the charge ‘Q’ is placed at a distance of ‘r’ from
charge ‘q’. For the system to be in equilibrium
KQq KQ 4q
1
r
2
r2
3
1 r Force on ‘q’ due to ‘4q’
F1 Kq 4q
12
Force on ‘q’ due to Q
9 KqQ
F2 =
12
18. Find out electric field intensity at point A(0, 1m, 2m) due to
a point charge –20mC situated at point B 2m, 0, 1m .
(
KQ
Sol. E =  2 rˆ
r
r = rA − rB
9 × 109 × (−20 × 10−6 )
=
− 2iˆ + ˆj + kˆ
E
8
=−22.5 × 103 − 2iˆ + ˆj + kˆ N/C
(
(
4 Kq 2 9 KqQ
4
Q q
2
2
1
1
9
16. Consider a neutral conducting sphere. A positive point
charge is placed outside the sphere. The net charge on the
sphere is
(1) Zero
(2) Negative and distributed uniformly over the surface of
the sphere
(3) Negative and distributed non-uniformly over the surface
of the sphere
(4) Negative and appears only at the point on the sphere
closest to the point charge
F1 F2 Sol. (1) When a positive point charge is placed outside a
conducting surface, redistribution of charges takes
place on the surface. But the total charge is zero as no
charge enters or loses the surface.
17. A spherical portion has been removed from a solid sphere
having a charge distributed uniformly in its volume as
shown in the figure. Electric field inside the emptied space
is
)
)
)
19. Six equal point charges are placed at the corners of a regular
hexagon of side ‘a’. Calculate electric field intensity at the
centre of hexagon?
+q
+q
r
E
r
E
+q
r
E
+q
+q
+q
Sol. Zero, due to symmetry, electric field cancel out each other.
20. Find out electric field intensity at the centre of uniformly
charged semicircular ring of radius R and linear charge
density l.
Sol. l = linear charge density.
dEx
q
dq
x
y
dEy
dE
The arc is the collection of large no. of point charges.
Consider a part of ring as an element of length Rdq which
subtends an angle dq at centre of ring and it lies between q
and q + dq

=
dE dEx iˆ + dE y ˆj =
Ex =
dEx 0 (due to symmetry)
∫
Kλ π
2K λ
sin θ=
⋅ dθ
0
R 0
R
21. A point charge q is placed at a distance r from a very long
charge thread of uniform linear charge density l. Find out
total electric force experienced by the line charged due to
the point charge. (Neglect gravity).
=
Ey
(1) Non- uniform
(3) Non-Zero and uniform
(2) Zero everywhere
(4) Zero only at its center
Sol. (3)
O
r
A
y
π
dE sin
=
θ
∫
Sol. Force on charge q due to the thread.
O1
ρ Electric field due to charge on big sphere E1 =
OA
3ε 0

ρ 
O1 A
Electric field due to small sphere E2 =
3ε0
ρ ρ (OA − O1 A) OO1
∴ Enet = E1 − E2 =
3ε 0
3ε 0
26
dE
∫=
∫
 2K λ 
=
F 
⋅q
 r 
F
l
q
F
r
NEET (XII) Moduel-1 PW
By Newton’s III law, every action has equal and opposite
2K λ
⋅q
reaction so force on the thread =
r
(away from point charge)
22. Three large conducting parallel sheets are placed at a finite
distance from each other as shown in figure. Find out electric
field intensity at point A, B, C & D.
Q –2Q 3Q
A
B
=
−2Q ˆ
i
Aε 0
For point D
Q − 2Q + 3Q ˆ
Q ˆ
Enet =
i=
i
ε0 A
2ε 0 A
23. A spherical shell having charge +Q (uniformly
distributed) and a point charge + q0 are placed as shown.
Find the force between shell and the point charge
(r >> R).
+Q, R
y
x
Sol. For point A

Q ˆ 3Q ˆ 2Q ˆ
Q ˆ
Enet =
−
i−
i+
i=
−
i
Aε0
2 Aε0
2 Aε0
2 Aε0
For point B

3Q ˆ 2Q ˆ
Q ˆ
Enet =
−
i+
i+
i=
0
2 Aε0
2 Aε0
2 Aε0
For point C
Q ˆ 2Q ˆ 3Q ˆ
Enet =
i−
i−
i
2ε 0 A
2ε 0 A
2ε 0 A
P Electric Charges and Fields
W
r
O
+q0
P
Sol. (i) Force on the point charge + q0 due to the shell

KQq0
 KQ 
= q=
(q0 )  =
rˆ
rˆ
0 Eshell
2 
r2
 r 
where r̂ is unit vector along OP.
From action-reaction principle, force on the shell due to the
KQq0
point charge will also be=
(−rˆ)
Fshell
r2
27
Exercise-1 (Topicwise)
COULOMB’S LAW
1. One million electrons are added to a glass rod. The total
charge on the rod is
(1) 10–13 C
(2) –1.6 × 10–13 C
(3) +1.6 × 10–12 C
(4) 10–12 C
2. A body has a charge of 9.6 × 10–20 coulomb. It is
(1) Possible
10. Two charged spheres separated at a distance R exert a force
F on each other. If they are immersed in a liquid of dielectric
constant 5 then what is the new force between them
(2) Not possible
(1) F
5
(2) F
(3) May (or) may not possible
(3) 5F
(4)
(4) Data not sufficient
3. A force of 4N is acting between two charges in air. If
the space between them is completely filled with glass
(εr = 8), then the new force will be
(1) 2 N
(2) 5 N
(3) 0.2 N
(4) 0.5 N
4. Two identical metal spheres possess +60C and –20C of
charges. They are brought in contact and then separated by
10 cm. The force between them is
(1) 36 × 1013 N
(2) 36 × 1014 N
(3) 36 ×
1012
N
(4) 3.6 ×
1012
N
5. Which of the following law gives existence of force between
two stationary charged particles?
(1) Coulomb’s law
(2) Biot-savart’s law
(3) Ohm’s law
(2) Half integral multiple of the least amount of charge
(3) Integral multiple of least amount of charge
(4) One third the least amount of charge
7. Two charged particles having charge 2 × 10-8C each are
joined by an insulating string of length 1m and the system
is kept on a smooth horizontal table, what is the tension in
the string?
(1) 3.6 × 10–6N
(2) 3.4 × 10–6N
(3) 4 × 10–7N
(1) 18F
(3)
(1) Q
4
(2) −Q
4
(3) Q
−Q
(4)
8
(2)
18F
25
F
25
(4) 25 F
18
12. Three charges q1, q2, q3 each equal to q placed at the vertices
of an equilateral triangle of side l. What will be the force on
a charge Q placed at the centroid of triangle?
1 3q 2
4π ∈0 l 2
(1)
1 q2
4π ∈0 l 2
(2)
(3)
1 2q 2
4π ∈0 l 2
(4) Zero
13. Total charge Q is broken in two parts Q1 & Q2 and they are
placed at a distance R from each other. The maximum force
of repulsion between them will occur when
Q
Q
(1) Q 2=
, Q1= Q −
R
R
(2) Q 2=
Q
2Q
, Q1= Q −
4
R
(3) Q 2=
Q
3Q
, Q1= Q −
4
4
(4) Q 2=
Q
Q
, Q1= Q −
2
2
(4) 3 × 10–4N
8. A charge q is placed at the center of the line joining two
equal charges of magnitude equal charges Q. Three charges
are in equilibrium then what will be the value of ‘q’?
F
2
11. Charges on two spheres are +10μC and –5μC respectively.
They experience a force F. If each of them is given an
additional charge +2μC then new force between them
keeping the same distance is
(4) All of these
6. Electric charge of any system is
(1) Zero or neutral
28
9. How many electrons must be removed from a piece of metal
to give it a positive charge of 1 × 10–7C ?
(1) 6.25 × 10–11
(2) 6.25 × 10–12
(3) 6.25 × 1011
(4) 6.25 × 1013
14. Two charges of +200µC and –200µC are placed at the corners
B and C of an equilateral triangle ABC of side 0.1 m. The
force on a charge of 5µC placed at A is
(1) 1800 N
(2) 1200 3N
(3) 600 3N
(4) 900 N
NEET (XII) Moduel-1 PW
(
)
15. Two charges each of 1µC are at P 2i + 3 j + kˆ m and
Q i + j − kˆ m . Then the force between them is
(
)
(1) 100N
(3) 104 dyne
(2) 10N
(4) 100 dyne
16. A charge Q is divided into two parts q1 and q2 such that they
experience maximum force of repulsion when separated by
certain distance. The ratio of Q, q1 and q2 is
(1) 1 : 1 : 2
(2) 1 : 2 : 2
(3) 2 : 2 : 1
(4) 2 : 1 : 1
17. Three charges –q, +q and –q are placed at the corners of an
equilateral triangle of side ‘a’. The resultant electric force
on a charge +q placed at the centroid O of the triangle is
(1)
3q 2
4πε0 a 2
(2)
q2
4πε0 a 2
(3)
q2
2πε0 a 2
(4)
3q 2
2πε0 a 2
ELECTRIC FIELD DUE TO POINT CHARGES
18. The distance between a proton and an electron both having
a charge of magnitude 1.6 × 10–19 C is 10–12 metre. The
value of intensity of electric field produced on electron due
to proton will be
(1) 14.4 × 1014 N/C
(2) 12.2 × 1014 N/C
(3) 11.4 × 1014 N/C
(4) 13.2 × 1014 N/C
19. Two point charges – q and q/2 are situated at the origin and
at the point (a, 0, 0) respectively. The point along the x-axis
where the electric field vanishes is
a
(1) x =
(2) x = 2 a
2
(3) x =
2a
2 −1
(4) x =
2a
2 +1
20. The magnitude of force exerted by a uniform electric field
on an electron having mass me and proton of mass mp are
represented as Fe and Fp respectively are related as
Fe m e
=
Fp m p
(1) Fp = Fe
(2)
m
(3) Fe = p
Fp m c
2
(4) Fe = m e
2
Fp m p
21. Two point charges of 30 μC and 40 μC are 20 cm apart from
each other. Where will be the electric field zero on the line
joining the charges from 30 μC charge?
(1)
20 3
2 +1
cm
(3) 20 3 cm
5
P Electric Charges and Fields
W
(2)
20 3
3+2
(4) 5 cm
cm
22. In which of the following cases the electric field at the centre
is not zero?
–
–
–
(2)
(1)
–
–
–
(3) –
–
–
(4)
–
23. Two charges of 50mC and 100mC are separated by a distance
of 0.6m. The intensity of electric field at a point midway
between them is
(1) 50 × 106 V
(2) 5 × 106 V
m
m
−6 V
6V
(3) 10 × 10
(4) 10 × 10
m
m
24. Two point charges Q and –3Q are placed some distance apart.
If the electric field at the location of Q is E , the field at the
location of –3Q is
(2) 2 E
(1) E
E
(3) +
(4) − E
3
3
25. A mass m carrying a charge q is suspended from a string
and placed in a uniform horizontal electric field of intensity
E. The angle made by the string with the vertical in the
equilibrium position is
(1) θ = tan −1
mg
Eq
(2) θ = tan −1
m
Eq
(3) θ = tan −1
Eq
m
(4) θ = tan −1
Eq
mg
26. Two charges of 10 µ C and –90 µ C are separated by a
distance of 24 cm. Electrostatic field strength from the
smaller charge is zero at a distance of
(1) 12 cm
(2) 24 cm
(3) 36 cm
(4) 48 cm
27. A proton of mass ‘m’ charge ‘e’ is released from rest in a
uniform electric field of strength ‘E’. The time taken by it
to travel a distance ‘d’ in the field is
(1)
2de
mE
(2)
2dm
Ee
(3)
2dE
me
(4)
2Ee
dm
28. A particle of mass m and charge q is thrown at a speed u
against a uniform electric field E. How much distance will
it travel before coming to rest?
(1) mu2/2qE
(2) 2mu2/qE
2
(3) mqE/2u
(4) mu/qE
29
29. Two equal and opposite charges of masses m 1 & m 2
are accelerated in a uniform electric field through the
same distance. What is the ratio of magnitudes of their
m
accelerations if their ratio of masses is 1 = 0.5 .
m2
a
(1) 1 = 0.5
(2) a1 = 1
a2
a2
a1
(3)
(4) a1 = 3
=2
a2
a2
30. There is an electric field E in x-direction. If the work done
in moving a charge of 0.2 C through a distance of 2m along
a line making an angle of 60° with x-axis is 4J, the value of
E is
31.
(1) 2 3 N / C
(2) 5 N/C
(3) 4 N/C
(4) 20 N/C
36. A charged oil drop is suspended in uniform field of
3 ×104 V/m. So that it neither falls nor rises. The charge on
the drop will be (m = 9.9 × 10–15 kg) (g = 10 m/s2)
(1) 3.3 × 10–18C
(2) 3.2 × 10–18C
(3) 1.6 × 10–18C
(4) 4.8 × 10–18C
37. The insulation of air vanishes when the electric field is
5 × 105 V/m. The maximum charge that can be given to a
sphere of radius 3m is approximately
(1) 5 μC
(2) 500 μC
(3) 10 μC
(4) 1 μC
38. The electric field inside a spherical shell of uniform surface
charge density is
(1) Zero
A charged particle of mass 3 × 10–3 kg and charge 6 × 10–6 C
(2) Constant less than zero
enters in an electric field of 7 V/m. Then its kinetic energy
after 1m is
(1) 10–4 J
(2) 9 × 10–2 J
–3
(3) 1.05 × 10 J
(4) 4.2 × 10–5 J
(3) Directly proportional to the distance from the centre
ELECTRIC FIELD DUE TO CONTINUOUS
CHARGE DISTRIBUTION
(4) None of these
ELECTRIC FLUX
39. A circular disc of radius ‘r’ is placed along the plane of paper.

A uniform electric field E is also present in the plane of
paper. What amount of electric flux is associated with it?
32. Which one of the following is not a property of electrostatic
field lines?
(1) Two lines cannot cross each other
(2) Field line start from negative and ends at positive charge
(3) They cannot form closed loop
(4) They never exist inside a conductor
33. A charge of 4 ×10–9C is distributed uniformly over the
circumference of a conducting ring of radius 0.3 m. Calculate
the field intensity at a point on the axis of the ring at 0.4m
from its center and also at the center?
(1) 112N/C, 2N/C
(2) 112N/C, 3N/C
(3) 115.2N/C, Zero
(4) 113.2N/C, Zero
34. An oil drop having a mass of 4.8 × 10­–10g and charge of
30 ×10–18C stands still between two charged horizontal
plates separated by a distance of 1cm . If now polarity of the
plates is changed, instantaneous acceleration of the drop is
(g = 10 m/s2)
(1) 10 m/s2
(2) 15 m/s2
2
(3) 25 m/s
(4) 20 m/s2
35. Two spheres of radii R1 and R2 respectively are charged and
joined by a wire. The ratio of electric field at the surface of
the spheres is
R 
(1)  2 
 R1 
(3) R1
R2
30
2
R 
(2)  1 
 R2 
(4) R 2
R1
2
(1) Eπr2
(2) Zero
(3) 2Eπr
(4)
πr 2 q
E
40. In a region, the intensity of an electric field is given by

E = 8iˆ + 8jˆ + kˆ in NC–1. The electric flux through a surface

S = 10iˆ m 2 in the region is
(1) 10 Nm2C–1
(2) 80 Nm2C–1
(3) 8 Nm2C–1
(4) None of these
41. Electric flux emanating through a surface element


ds = 5 ˆi placed in an electric field E = 4iˆ + 4ˆj + 4kˆ is
(1) 10 units
(2) 20 units
(3) 4 units
(4) 16 units
GAUSS LAW & ITS APPLICATION
42. A ball having a charge –100e is placed at the centre of a
metallic hollow spherical shell which has a net charge of
–20e. What is the charge on the shell’s outer surface?
(1) + 80e
(2) –20e
(3) –100e
(4) –120e
NEET (XII) Moduel-1 PW
43. Electric field intensity at a point in between two parallel
sheets with like charges of same surface charge densities
(σ) is
σ
(1)
(2) 3σ
2ε0
ε
(3) 2σ
ε0
0
(4) Zero
44. If the uniform surface charge density of the infinite plane
sheet is σ , electric field near the surface will be
(1)
σ
2ε0
(2)
(3)
σ
ε0
(4) None of these
3σ
ε0
45. An electric charge q is placed at the center of a cube of side
l. The electric flux through one of its faces will be
6q
q
(1)
(2)
ε0
ε0
2 q
(3) q
(4)
3 ε0
6ε0
46. The flux of electric field due to these charges through the
surface S is
s
–q
+q
+q
–q
–q
+q
+q
–q
–q –q –q –q –q
2q
ε0
−5q
(3)
ε0
(1)
(2)
3q
ε0
(4) qε0
47. What is the flux linked through a cube of side ‘l’ if a charge
2q is placed at one vertices of cube?
(1)
q
ε0
(2)
q
8ε0
(3)
2q
ε0
(4)
q
4ε0
48. A sphere S1 of radius R encloses a total charge q. If there is
another concentric sphere S2 of radius 2R and there be no
additional charges between S1 and S2. What is the ratio of
electric flux through S1 and S2?
(1) 1.5
(2) 2
(3) 1
(4) 0.5
49. Two parallel line charges +λ and –λ are placed with a
separation distance R in free space. The net electric field
exactly mid way between the two line charge is
2λ
(1) Zero
(2)
πε0 R
1
λ
(3)
(4)
2
πε
πε0 R
0 R
P Electric Charges and Fields
W
50. A charge q is located at the center of a cube. The electric
flux through any two consecutive faces will be
q
πq
(1)
(2)
6
4πε
( 0)
6 ( 4πε0 )
(3)
2πq
6 ( 4πε0 )
(4)
4πq
3 ( 4πε 0 )
51. Five electric dipoles of charge ‘q’ each are placed into the
shell. What will be the amount of electric flux associated
with the shell?
10q
(1) 5qε0
(2)
ε0
(4) 5q
ε0
(3) Zero
DIPOLE
52. An electric dipole is along a uniform electric field. If it is
deflected by 60°, work done by an agent is 2 × 10–19 J. Then
the work done by an agent if it is deflected by 30° further
is
(1) 2.5 × 10–19 J
(2) 2 × 10–19 J
(3) 4 × 10–19 J
(4) 2 × 10–16 J
53. The dipole moment of the given system is
2q
l
l
q
l
q
(1) 3ql along perpendicular bisector of q – q line
(2) 2 ql along perpendicular bisector of q – q line
(3) ql 2 along perpendicular bisector of q – q line
(4) 0

54. An electric dipole of moment p is placed normal to the
lines of force of electric intensity E , then the work done in
deflecting it through an angle of 180° is
(1) pE
(2) +2 pE
(3) –2 pE
(4) Zero
55. An electric dipole consists of a positive charge and negative
charge of magnitude 4μC each placed at a distance of 5mm.
What is the dipole moment?
(1) 2 × 10–6 C–m
(2) 2 × 10–8 C–m
(3) 4 × 10–5 C–m
(4) 5 × 10–8 C–m
56. Electric dipole consists of two charges of magnitude 0.1 μC
separated by a distance of 2 cm. The dipole is in an external
field of 105 N/C What maximum torque does the field exert
on the dipole?
–4 N–m
(1) 10­
(2) 2 × 10–8 N–m
–4
(3) 2 × 10 N–m
(4) 3 ×10–4 N–m
31
57. An electric dipole is placed along the x - axis at the origin
O. A point P is at a distance of 20 cm from this origin such
that OP makes an angle π/3 with the x - axis. If the electric
field at point P makes an angle θ with the x - axis, the value
of θ would be.
(1) π/3
(2) 2π/3
 3
 3
−1
(3) tan −1 
(4) π 3 + tan 


 2 
 2 
58. An electric dipole of dipole moment 2 × 10–12 C–m is
oriented at an angle of 60° with electric field of intensity
3NC–1. What work need to be done to turn the dipole so that
it becomes anti-parallel to the field?
(1) 12 × 10–12 J
(2) –9 × 10–12 J
(3) 9 × 10–12 J
(4) –12 × 10–12 J
59. An electric dipole consist of a positive and negative charge of
8μC each placed at a distance of 5mm. What is the magnitude
of dipole moment?
(1) 10–8 C–m
(2) 2 × 10–8 C–m
–8
(3) 4 × 10 C–m
(4) None of these
60. What is the magnitude of electric field intensity due to a
dipole of moment 4 ×10–8 C-m at a point distant 0.5m from
the center of dipole when line joining the point to the center
of dipole makes an angle of 60o with dipole axis?
(1) 4 × 10–5 N/C
(2) 5 × 104 N/C
(3) 2 × 107 N/C
(4) 3.8 × 103 N/C
61. An electric dipole, when held at 60o with respect to a uniform
electric field of 107 NC–1 experience a torque of 12 × 10–20
N–m. What is the value of dipole moment?
(1) 1.4 × 10–26 C–m
(2) 8 × 10–27 C–m
(3) 3 × 10–26 C–m
(4) 1.3 × 10–27 C–m
62. An electric dipole is placed at an angle of 45o with an electric
field of intensity 107 NC–1. It experiences a torque equal to
15 × 102 N – m. What will be magnitude of charge on dipole
if it is 5cm long?
(1) 1 mC
(2) 3.18 mC
(3) 5.35 mC
(4) 4.24 mC
63. An electric dipole of moment p is lying along a uniform
electric field E. The work done in rotating the dipole by
180o is
(1)
2pE
(3) 2 pE
(2) pE
(4)
pE
2
Exercise-2 (Learning Plus)
1. Consider a non-spherical conductor shown in the figure
which is given a certain amount of positive charge. The
charge distributes itself on the surface such that the charge
densities are σ1, σ2 and σ3 at the region 1, 2 and 3 respectively.
Then
3
(1) σ1 > σ2 > σ3
(2) σ2 > σ3 > σ1
(3) σ3 > σ1 > σ2
(4) σ2 > σ1 > σ3
2. A pendulum bob carriers a negative charge –q. A positive
charge +q is held at the point of support. Then, the time
period of the bob is
(3) Equal to 2π
32
L
g
L
g
13 m
(2) qE0t
(3)
2
(1) Greater than 2π
(1)
qE 0 t
(4) Zero
m
4. Two identical charged spheres of material density ρ,
suspended from the same point by inextensible strings of
equal length make an angle θ between the strings. When
suspended in a liquid of density σ the angle θ remains the
same. The dielectric constant k of the liquid is
3
1
3. A particle having charge q and mass m is projected with velocity


v= 2iˆ − 3jˆ in a uniform electric field E = E 0 ˆj. Change in
momentum Dp during any time interval t is given by?
(2) Less than 2π
(4) Equal to 2π
L
g
2L
g
ρ
ρ−σ
ρ
(3)
ρ+σ
(1)
ρ−σ
ρ
ρ+σ
(4)
ρ
(2)
5. The electric intensity outside a charged sphere of radius R
at a distance r(r > R) is
(1)
σR 2
ε0r 2
(2)
σr 2
ε0 R 2
(3)
σr
ε0 R
(4)
σR
ε0r
NEET (XII) Moduel-1 PW
6. An electron of mass m and charge q is accelerated from rest
in a uniform electric field of strength E. The velocity acquired
by it as it travels a distance l is
(1)
2Eql
m
(2)
2Eq
ml
(3)
2Em
ql
(4)
Eq
ml
(3)
7. The figure shows some of the electric field lines corresponding
to an electric field. The figure suggests
B
A
2E 2 t 2
mq
(4)
Eq 2 m
2t 2
12. Two point charges 2µC and – 2µC are placed at point A and
B as shown in figure. Find out electric field intensity at point
(C) All the distances are measured in meter.
C
(2) EA = EB = EC
(4) EA > EC > EB
A
(1)

1 q
 − φ
2  ε0

(2)
q
2ε0
(3)
φ
3
(4)
q
−φ
ε0
9. A charged particle q 1 is at position (2, – 1, 3). The
electrostatic force on another charged particle q 2 at
(0, 0, 0) is
(1)
q1 q 2
ˆ
(2 ˆi − ˆj + 3k)
56 π ∈0
(2)
1 q1 q 2
ˆ
( − 2 ˆi + ˆj − 3 k)
4π∈0 14 3
(1) −8000i N / C
(3) 800i N / C
13. Eight charges, 1µC, –7µC, – 4µC, 10µC, 2µC, –5µC, –3µC
and 6µC are situated at the eight corners of a cube of side
20 cm. A spherical surface of radius 80 cm encloses this
cube. The centre of the sphere coincides with the centre of
the cube. Then the total outgoing flux from the spherical
surface (in unit of volt meter) is
(1) 36π × 103
(2) 684π × 103
(3) Zero
(4) None of these
14. A positive charge Q is uniformly distributed along a circular
ring of radius R. A small test charge q is placed at the centre
of the ring as shown in figure. Then
10. A conductor PQ as shown in figure is charged σA, σB, σC
and σD are surface charge densities of points A, B, C and D
respectively
A
C D
P
Q
B
(1) σA < σB < σC < σD
(3) σA > σC > σB > σD
P Electric Charges and Fields
W
(2) σA = σB = σC > σD
(4) σA = σB = σC = σD
+++++
q
+
+++ Q
+
R
Z
+++
(3)
++
++
( )
q1 q 2 ˆ
ˆ
( j − 2 ˆi − 3k)
56 π ∈0
q1 q 2
ˆ
(4)
(2 ˆi − ˆj + 3k)
56 14 π ∈0
(2) −800i N / C
(4) +8000i N / C
+++
B
C
++
8. A hollow cylinder has a charge q within it. If ϕ is the electric
flux in unit of volt meter associated with the curved surface
B, the flux linked with the plane surface A in unit of volt
meter will be
+
(1) EA > EB > EC
(3) EA = EC > EB
11. A charged particle of charge q and mass m is released from
rest in an uniform electric field E. Neglecting the effect of
gravity, the kinetic energy of the charged particle after time
‘t’ seconds is
Eqm
E2q2 t 2
(1)
(2)
t
2m
(1) If q > 0, and is displaced away from the centre in the
plane of the ring, it will be pushed back towards the
centre
(2) If q < 0 and is displaced away from the centre in the
plane of the ring, it will never return to the centre and
will continue moving till it hits the ring
(3) If q > 0 it will perform SHM for small displacement
along the axis
(4) All of the above
33
15. A metallic spherical shell has an inner radius R1 and outer
radius R2. A charge is placed at the centre of the spherical
cavity. The surface charge density on the inner surface is
R2
+q R1
(1)
(3)
q
–q
4πR12
q
(4)
4πR 22
(2)
4πR12
2
q
4πR 22
16. The electric field components in the given figures are
Ex = αx1/2, Ey = Ez = 0 in which α = 800N C–1 m–1/2. The
charge within the cube is [if net flux through the cube is 1.05
N m2C–1 (assume a = 0.1 m)]
19. Two metal spheres of same mass are suspended from a
common point by a light insulating string. The length of
each string is same. The spheres are given electric charges
+q on one of the sphere and +4q on the other. Which of the
following diagrams best represents the resulting positions
of the spheres?
(1)
(2)
(3)
(4)
Y
̌
nL
20. Two charges q and –3q are placed fixed on x-axis separated
by a distance ‘d’. Where should a third charge 2q be placed
such that it will not experience any force?
̌
a
nR
X
a
a
Z
(1) 9.27 × 10–12 C
(3) 6.97 × 10–12 C
(2) 9.27 × 1012 C
(4) 6.97 × 1012 C
17. A nonconducting ring of radius R has uniformly distributed
positive charge Q. A small part of the ring, of length d, is
removed (d << R). The electric field at the centre of the ring
will now be
(1) Directed towards the gap, inversely proportional to R3
(2) Directed towards the gap, inversely proportional to
R2
(3) Directed away from the gap, inversely proportional to R3
Directed away from the gap, inversely proportional to R2
(4)
18. If charges are distributed uniformaly in the sphere and charge
r 

density is ρ = ρ0 1 +  where ‘r’ is the distance from the
R


centre of the sphere then
R
r
(2) Sphere is non conducting
(3) Information is incomplete
(4) None of these
34
(2) d 1 3 to the right of q
2
(3) d 1 5 to the left of q
2
(4) d 1 5 to the right of q
2
21. If the electric flux entering and leaving an enclosed surface
respectively are ϕ1 and ϕ2 the electric charge inside the
surface will be
(1)
φ2 − φ1
εo
(2)
φ2 + φ1
εo
(4) εo ( φ2 − φ1 )
(3) (ϕ1 – ϕ2)ε0
22. Electric charges q, q, –2q are placed at the corners of an
equilateral triangle ABC of side l. The magnitude of electric
dipole moment of the system is
(1)
(1) Sphere is conducting
(1) d 1 3 to the left of q
2
a
3q
(2)
2q
(3) ql
(4) 2ql
23. An electron is moving around the nucleus of a hydrogen
atom in a circular orbit of radius r. The Coulomb force F

between the two is  where, k =

1 

4πε0 
NEET (XII) Moduel-1 PW
(1) k
e2 ^
r
r3
(2) −k
e2 
r
r3
(1) E
(3) k
e2 ^
r
r
(4) −k
e2 ^
r
r
(3)
24. A uniformly charged and infinitely long line having a linear
charge density ‘λ’ is placed at a normal distance y from a
point O. Consider a sphere of radius R with O as centre and
R > y. Electric flux through the surface of the sphere is
(1) Zero
(2)
2
(3)
2λ R − y
ε0
2
2λR
ε0
(4)
λ R +y
ε0
2
(3)
25. An electron moving with the speed 5 × 106 m/s is shooted
parallel to the electric field of intensity 1 × 103 N/C. Field
is responsible for the retardation of motion of electron.
Now evaluate the distance travelled by the electron before
coming to rest for an instant (mass of e = 9 × 10–31 kg,
charge = 1.6 × 10–19 C)
(1) 7 m
(2) 0.7 m
(3) 7 cm
(4) 0.7 cm
(2) 8 × 105 ms–1
(3) 16 × 105 ms–1
(4) 4 2 × 105 ms −1
O
R
1/r
E 1/r
O
E
2
2
(3)
(4)
R
2
1/r
O
r
r
E
O
R
(2)
θ
sin  
4
(4)
Q
θ
sin  
4π ε 0 R
2
2
2
Q
4π2 ε 0 R 2
θ
sin  
8
30. 4 charges are placed each at a distance ‘a’ from origin. The
dipole moment of configuration is
y
3q
r
(1) 2qa j
(3) 2aq i + j


(4) None of these
+++++
+
+
+
2
1/r
R
–2q
q
r
r
28. The electric field due to an electric dipole at a distance r
from its centre in axial position is E. If the dipole is rotated
through an angle of 90° about its perpendicular axis, the
electric field at the same point will be
P Electric Charges and Fields
W
Q
4π2 ε 0 R 2
sin(θ)
31. If a positively charged pendulum is oscillating in a uniform
electric field as shown in diagram. Its frequency compared
to that when it was uncharged
2
(2)
4π ε 0 R
2
(2) 3qa j
27. Which of the following represents the variation of the electric
field with distance from the centre of a uniformly charged
non-conduction sphere of radius R?
1/r
Q
2
x
(1) 2 3 × 105 ms −1
E
(4) 2E
–2q
26. An α-particle of mass 6.4 × 10–27 kg and charge 3.2 × 10–19
C is situated in a uniform electric field of 1.6 × 105 Vm–1.
The velocity of the particle at the end of 2 × 10–2 m path
when it starts from rest is
(1)
E
2
E
4
29. A circular wire of radius R carries a total charge Q distributed
uniformly over its circumference. A small length of the wire
subtending angle θ at the centre is cut off. Find the electric
field at the centre due to the remaining portion.
(1)
2
(2)
+++++
+
E
+
+
– – – – – – – – – –
(1) Will increase
(2) Will decrease
(3) Will not change
(4) Will first increase and then decrease
35
Exercise-3 (Multiconcept)
MATCH THE COLUMN MCQs
1. The Column-I gives the two point charge system separated
by 2a and the column-II gives the variation of magnitude of
electric field intensity along x-axis. Match the situation in
Column-I with the results in Column-II and indicate your
answer by darkening appropriate bubbles in the 4 × 4 matrix
given in the OMR.
Column-I
A.
–
p. Increases as x
increases in the
interval
0≤x<a
–
q. Decreases as x
increases in the
interval
0≤x<a
B.
C.
Column-II
r.
Zero at x = 0
A.
B.
C.
D.
Column-I
Uniform electric field, θ = 0
Electric field due to a point
charge, θ = 0
Electric field between the
two oppositely charged large
plates, θ = 90°
Dipole moment parallel to
uniformly charged long wire
(3) A-(p, q); B-(t); C-(t, q); D-(s)
(4) A-(r, s); B-(s); C-(r, p); D-(q)
3. Column-I gives certain situations involving two thin
conducting shells connected by a conducting wire via a
key K. In all situations one sphere has net charge +q and
other sphere has no net charge. After the key K is pressed,
Column‑II gives some resulting effect. Match the figures
in Column‑I with the statements in Column-II and indicate
your answer by darkening appropriate bubbles in the 4 × 4
matrix given in the ORS.
B.
C.
(1)
(2)
(3)
(4)
A-(p, q); B-(q); C-(p, q); D-(r)
A-(p, r, s); B-(p, s); C-(r, s); D-(q, s)
A-(p, q); B-(t); C-(t, q); D-(s)
A-(r, s); B-(s); C-(r, p); D-(q)
2. An electric dipole is placed in an electric field. The Column-I
gives the description of electric field and the angle between
the dipole moment and the electric field intensity and the
Column-II gives the effect of the electric field on the dipole.
Match the description in Column-I with the statements
in Column-II and indicate your answer by darkening
appropriate bubbles in the 4 × 4 matrix given in the ORS.
36
s. Force ≠ 0
(2) A-(p, r, s); B-(p, s); C-(r, s); D-(q, s)
A.
s. Decreases as x
increases in the
interval
a<x<∞
r. Torque ≠ 0
(1) A-(p, q); B-(q, s); C-(p, r); D-(p, r)
Column-I
D.
Column-II
p. Force = 0
q. Torque = 0
D.
Column-II
p. Charge flows
through
connecting
wire
q. Potential
energy of
system of
spheres
decreases.
r. No heat is
produced.
s.
The sphere
I has no
charge after
equilibrium is
reached.
NEET (XII) Moduel-1 PW
(1)
(2)
(3)
(4)
A-(p, q); B-(q, s); C-(p, r); D-(p, r)
A-(p, q); B-(p, q); C-(p, q, s); D-(r, s)
A-(p, q); B-(t); C-(t, q); D-(s)
A-(r, s); B-(s); C-(r, p); D-(q)
STATEMENT BASED MCQs
4. Column I gives a situation in which point charge(s) are
placed at different position with respect to a uncharged
thick conducting spherical shell. Column II gives resulting
effect. Match the figures in Column-I with the statements
in Column‑II and indicate your answer by darkening
appropriate bubbles in the 4 × 4 matrix given in the OMR.
Column-I
A.
Column-II
p.
Charge is induced
on inner surface
of shell and
is distributed
uniformly
q.
Charge is induced
on inner surface
of shell and is
distributed nonuniformly
r.
Charge is induced
on outer surface
of shell and
is distributed
uniformly
Positive point charge
q is placed at centre of
shell
B.
Positive point charge q
is placed inside gap of
shell, but not at centre
C.
Positive point charge
q is placed outside the
shell
Charge is induced
on outer surface
of shell and is
distributed nonuniformly
Positive point charge
q, is placed inside the
shell but not at centre.
Positive point charge
q2, is placed outside the
shell
(1)
(2)
(3)
(4)
A-(p, r); B-(q, r); C-(s); D-(q, s)
A-(p, r, s); B-(p, s); C-(r, s); D-(q, s)
A-(p, r); B-(p, q, r, s); C-(p, q, r); D-(p, s)
A-(p); B-(r, s); C-(p, q); D-(r, s)
P Electric Charges and Fields
W
Both Statement I and Statement II are correct
Both statement I and Statement II are incorrect.
Statement I is correct and statement II incorrect.
Statement I is incorrect and statement II is correct.
5. Statement-I: An electric dipole placed in the electric field
of a point charge can never experience zero resultant force.
Statement-II: Electric field of a point charge is non uniform.
6. Statement-I: When a charge is placed at rest in an electric
field its path will always be along electric line of force.
Statement-II: The force on the charge is along the tangent
drawn on electric field line.
7. Statement-I: Electric flux is a scalar quantity.
Statement-II: The solid angle subtended at the centre of a
sphere is 4π.
8. Statement-I: Electric lines are always straight and
continuous.
Statement-II: Crowding of electric field lines at a place
represents stronger electric field at that place.
9. Statement-I: Electric lines of force can never be closed
loops, as a line can never start and end on the same charge.
Statement-II: The number of electric lines of force
originating or terminating on a charge is proportional to the
magnitude of charge.
10. Statement-I: If a point charge q is placed in front of an
infinite grounded conducting plane surface, the point charge
will experience a force.
Statement-II: This force is due to the induced charge on the
conducting surface which is at zero potential.
11. Statement-I: When two charged spheres are touched, then
the total charge is always shared equally.
Statement-II: Induced charge is always equal to the inducing
charge.
ASSERTION & REASON MCQs
s.
D.
(1)
(2)
(3)
(4)
(1) Assertion (A) is True, Reason (R) is True; Reason (R) is a
correct explanation for Assertion (A)
(2) Assertion (A) is True, Reason (R) is True; Reason (R) is not
a correct explanation for Assertion (A)
(3) Assertion (A) is True, Reason (R) is False.
(4) Assertion (A) is False, Reason (R) is True.
12. Assertion (A): Electrons move away from a region of lower
potential to a region of higher potential.
Reason (R): Since an e– has negative charge.
13. Assertion (A): Work done in moving a charge between any
two points in an electric field is independent of the path
followed by the charge, between these points.
Reason (R): Electrostatic forces are non conservative.
14. Assertion (A): Force between two charges decreases when
air separating the charges is replaced by water.
Reason (R): Medium intervening the charges has no effect
on force.
37
15. Assertion (A): The no. of lines of force emanating from
1µC charge in vacuum is 1.13 × 105.
Reason (R): This follow from Gauss’s theorem in
electrostatics.
16. Assertion (A): Charge is invariant with speed.
Reason (R): Charge does not depend on speed or frame of
reference.
17. Assertion (A): Mass of ion is slightly different from its
element form.
Reason (R): Ion is formed, when some electrons are removed
or added so mass changes.
18. Assertion (A): Total number of positive ions in nature is
constant.
Reason (R): In an isolated system charge remains conserved.
19. Assertion (A): A point charge Q is rotated in a circle of
radius r around a charge q at the center. The work done will
be zero.
Reason (R): For this motion the force is along the radius
and direction of motion is perpendicular to the direction of
force.
Exercise-4 (Past 10 Years Questions)
1. Two point charges –q and +q are placed at a distance of L,
as shown in the figure
–q
+q
L
The magnitude of electric field intensity at a distance R
(R >> L) varies as
(2022)
1
1
(1)
(2)
6
R2
R
1
1
(3)
(4)
3
R4
R
2. A dipole is placed in an electric field as shown. In which
direction will it move?
(2021)
→
+q
–q
E
(1) Towards the right as its potential energy will decrease.
(2) Towards the left as its potential energy will decrease.
(3) Towards the right as its potential energy will increase.
(4) Towards the left as its potential energy will increase.
3. The electric field at a point on the equatorial plane at a
distance
r from the centre of a dipole having dipole moment

is
given
by,
P
(r >> separation of two charges forming the dipole,
∈0 – permitivity of free space)
(2020 Covid)

2P
4π ∈0 r 3


P
(3) E = −
4π ∈0 r 3

(1) E =
38

P
4π ∈0 r 2


P
(4) E =
4π ∈0 r 3

(2) E = −
4. A spherical conductor of radius 10 cm has a charge of
3.2 × 10–7 C distributed uniformly. What is the magnitude of
electric field at a point 15 cm from the centre of the sphere?
1
9
2
2
 4π ∈ = 9 × 10 N m /C  0



(2020)
(1) 1.28 × 105 N/C
(2) 1.28 × 106 N/C
(3) 1.28 × 107 N/C
(4) 1.28 × 104 N/C
5. A hollow metal sphere of radius R is uniformly charged.
The electric field due to the sphere at a distance r from the
centre (2019)
(1) Increases as r increases for r < R and for r > R
(2) Zero as r increases for r < R, decreases as r increases
for r > R
(3) Zero as r increases for r < R, increases as r increases for
r>R
(4) Decreases as r increases for r < R and for r > R
6. Two parallel infinite line charges with linear charge densities
+λ C/m and –λ C/m are placed at a distance of 2R in free
space. What is the electric field mid-way between the two
line charges? (2019)
(1) Zero
λ
(3) πε R N / C
0
(2)
2λ
N/C
πε0 R
(4)
λ
N/C
2πε0 R
7. Two point charges A and B, having charges +Q and –Q
respectively, are placed at certain distance apart and force
acting between them is F. If 25% charge of A is transferred
to B, then force between the charges becomes (2019)
(1) F
(3)
16F
9
(2)
9F
16
(4)
4F
3
NEET (XII) Moduel-1 PW
8. Suppose the charge of a proton and an electron differ slightly.
One of them is –e, the other is (e + ∆e) If the net of electrostatic
force and gravitational force between two hydrogen atoms
placed at a distance d (much greater than atomic size) apart
is zero, then ∆e is of the order of [Given mass of hydrogen
mh = 1.67 × 10–27 kg]
(2017-Delhi)
–23
–37
(1) 10 C
(2) 10 C
(3) 10–47 C
(4) 10–20 C
9. Two identical charged spheres suspended from a common
point by two massless strings of lengths l, are initially at a
distance d (d << l) apart because of their mutual repulsion.
The charges begin to leak from both the spheres at a constant
rate. As a result, the spheres approach each other with a
velocity V. Then V varies as a function of the distance x
between the spheres, as
(2016-I)
(1) V ∝ x
(3) V ∝ x
1
2
sphere of radius ‘a’ centered at the origin of the field, will be
given by
(2015)
2
3
(1) Aε0a
(2) 4πε0Aa
(3) ε0Aa3
(4) 4πε0Aa2
11. Two pith balls carrying equal charges are suspended from
a common point by strings of equal length, the equilibrium
separation between them is r. Now the strings are rigidly
clamped at half the height. The equilibrium separation
between the balls now become
(2013)
(2) V ∝ x
1
−
2
(1)  2r 
(2)  1 


 2
 r 
(3)  3 
 2
 2r 
(4) 

 3
 3
(4) V ∝ x-1
10. The electric field in a certain region is acting radially
outward and is given by E = Ar. A charge contained in a
2
ANSWER KEY
CONCEPT APPLICATION
1. (3)
11. (3)
21. (3)
2. (4)
12. (2)
3. (4)
13. (1)
EXERCISE-1 (TOPICWISE)
1.
11.
21.
31.
41.
51.
61.
(2)
(3)
(2)
(4)
(2)
(3)
(1)
2.
12.
22.
32.
42.
52.
62.
(2)
(4)
(2)
(2)
(4)
(2)
(4)
3.
13.
23.
33.
43.
53.
63.
(4)
(4)
(2)
(3)
(4)
(1)
(3)
4. (4)
14. (1)
5. (4)
15. (3)
6. (4)
16. (1)
7. (4)
17. (2)
8. (2)
18. (3)
9. (3)
19. (3)
10. (2)
20. (3)
4.
14.
24.
34.
44.
54.
5.
15.
25.
35.
45.
55.
6.
16.
26.
36.
46.
56.
7.
17.
27.
37.
47.
57.
8.
18.
28.
38.
48.
58.
9.
19.
29.
39.
49.
59.
10.
20.
30.
40.
50.
60.
(1)
(4)
(3)
(4)
(1)
(4)
EXERCISE-2 (LEARNING PLUS)
1.
11.
21.
31.
(4)
(2)
(4)
(1)
2. (3)
12. (1)
22. (1)
3. (2)
13. (3)
23. (2)
4. (1)
14. (4)
24. (3)
EXERCISE-3 (MULTICONCEPT)
1. (2)
11. (4)
2. (1)
12. (1)
3. (2)
13. (3)
4. (1)
14. (3)
(1)
(4)
(4)
(4)
(3)
(2)
(3)
(4)
(1)
(1)
(1)
(3)
(1)
(4)
(2)
(2)
(4)
(4)
(2)
(1)
(1)
(1)
(3)
(3)
(3)
(3)
(3)
(2)
(2)
(3)
(1)
(1)
(4)
(2)
(4)
(4)
5. (1)
15. (2)
25. (3)
6. (1)
16. (1)
26. (4)
7. (3)
17. (1)
27. (4)
8. (1)
18. (2)
28. (3)
9. (2)
19. (1)
29. (2)
10. (3)
20. (1)
30. (1)
5. (1)
15. (1)
6. (4)
16. (1)
7. (1)
17. (1)
8. (4)
18. (4)
9. (1)
19. (1)
10. (1)
6. (3)
7. (2)
8. (2)
9. (3)
10. (2)
EXERCISE-4 (PAST 10 YEARS QUESTIONS)
1. (3)
11. (3)
2. (1)
3. (3)
P Electric Charges and Fields
W
4. (1)
5. (2)
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