Electric Charges and Field
Electric Charges and Field
Introduction
When a glass rod is rubbed with a piece of silk cloth, it repels another similar glass rod. Similarly, a plastic measuring scale,
when rubbed with a cloth attracts small bits of paper. Greek people discovered that some naturally occurring stones can
attract iron. From these simple discoveries, a branch of physics came into existence. This was named electromagnetism. This
new science deals with electric charges and their interactions.
Electrostatics is the study of charges while they are static i.e. at rest. In this part of physics we deal with interaction between
charges, not taking into account the motion of the charges. In the coming chapters, we will deal with moving charges i.e.
electric current and its effects.
ELECTRIC CHARGE
Charge is an intrinsic property of fundamental particles due to which they can attract or repel each other. It accompanies them
wherever they exist. In practical life, there are a large number of phenomena which involve charged particles. For Example,
lightning, attraction between a comb and dry hair etc.
All electric charge is not same. There are two kinds of charges. It was observed that a plastic ruler rubbed with a cloth repels
another similarly charged plastic ruler but it attracts the charged glass rod. Therefore, the charge on the glass must be
different from that on the plastic. Thus, we conclude that there are two type of electric charge. One kind of charge is called
positive and the other negative.
SI Unit: The SI unit of charge is coulomb (C). When a current of 1 ampere (A) flows across a section of conductor for 1
second, the charge flown is taken as 1 coulomb.
Dimensional Formula: [M0 L0 A T]
Electric charge in matter
The positively charged particles in ordinary matter are protons, and the negatively charged particles are electrons. The charge
of a proton is +e = + 1.6 × 10 – 19 C and the charge of an electron is – 1.6 × 10 – 19 C
Here are some important facts about atoms:
1.
Every atom consists of a positively charged nucleus surrounded by negatively charged electrons.
2.
The electrons of all atoms are identical. Each has the same quantity of negative charge and same mass.
3.
Protons and neutrons compose the nucleus. Protons are about 1800 times more massive than electrons, but they carry an
amount of positive charge equal to the negative charge of electrons. Neutrons have slightly more mass than protons and have
no net charge.
4.
Atoms usually have as many electrons as protons, so the atom has zero net charge.
Benjamin Franklin assigned (+) and (-) sign to the two kind of charges.
Check Yourself
How does the charge of an electron differ from the charge of a proton?
Ans. The charge of an electron is equal in magnitude, but opposite in nature to the charge of a proton.
Charging by Friction
Material objects are made of atoms, which mean they are composed of electrons and protons (and neutrons). Objects
ordinarily have equal number of electrons and protons and are, therefore, electrically neutral. An imbalance comes about
when electrons are added to, or removed from, an object. The amount of energy required to tear an electron away from an
atom varies from one substance to other. The electrons are held more firmly in rubber and plastic than in your hair. Thus,
when a comb is passed through your hair, electrons transfer from the hair to the comb. The comb thus has an excess of
electrons, it becomes negatively charged. Your hair, in turn, has a deficiency of electron and they become positively charged.
This is known as charging by friction.
Charging by Induction
When a charged object is brought near a conducting surface, it causes electrons to move in the conductor, even though there
is no physical contact. Consider the two insulated metal spheres A and B in figure.
Electric Charges and Field
- -A
B
(a)
-B --
++
+A
++
- --
--
++
+A
++
(b)
B -
--
(c)
- B - -
+
+A+
+ +
(d)
(a)
They touch each other, so in effect they form a single uncharged conductor.
(b)
When a negatively changed rod is brought near A, electrons in the metal, being free to move, are repelled as far as possible.
(c)
A and B are separated while the rod is still present.
(d)
Each of the two, A as well as B will be equal and oppositely charged.
The charging rod has never touched them, and the rod retains the same charge it had initially. This is called charging by Induction.
When a neutral body acquires a negative charge, its mass increases
•
Electrostatic induction can be demonstrated using small pieces of paper and a polythene rod. The rod is negatively charged
prior to the experiment by vigorously rubbing it with a cloth. It is then placed over the pieces of paper. The pieces of paper
jump up through the air to the polythene rod and "stick" to it. This is because when the negatively charged polythene is
brought close to the paper, the negative electrons in the paper are repelled downwards. The upper side of the paper becomes
positively charged and is therefore attracted towards the negative rod.
Check Yourself:
A negatively charged rod is brought close to same small pieces of neutral paper. The positive sides of molecules in the paper are
attracted to the rod and the negative sides of the molecules are repelled. Since the negative and positive sides are equal in
number, why don't the attractive and repulsive forces cancel out?
Ans. The force of attraction is more as the positive charges are closer to the rod.
Electroscope
An electroscope is a device that can be used for detecting charge. As shown inside of a case are two
movable metal leaves often made of gold. The leaves are connected by a conductor to a metal knob on
the outside of the case but are insulated from the case itself. If a positively charged object is brought
close to the knob a separation of charge is induced, electrons are attracted up into the knob leaving the
leaves positively charged. The two leaves repel each other as shown, because they are both positively
charged. If, instead, the knob is charged by conduction the whole apparatus acquires a net charge. In
either case the greater the amount of charge, the greater the separation of the leaves.
PROPERTIES OF ELECTRIC CHARGE
The important properties and characteristic of electric charge are given below.
1.
There are two kinds of charges. As a matter of convention, charge of a proton is taken as positive and that of an electron is
taken as negative.
2.
Like charges repel each other. Unlike charges attract each other.
3.
Charge is quantized: The charge on an object is always an integral multiple of electronic charge.
Mathematically q = ±ne
4.
5.
Here n = 1, 2, 3 and e = 1.6 × 10-19 coulomb.
Electric charge is conserved: The total charge of the universe is constant. The total charge of an isolated system is constant.
Unlike mass, charge is non-relativistic: The mass of an object is found to depend on speed, according to famous Einstein's
m0
formula, m =
No such variation in charge has been observed.
v2
1− 2
c
Electric Charges and Field
The total number of charged particles in the universe in not conserved, while the total charge is conserved. It has been observed
that an electron and a positron (anti-particle of electron) can annihilate each other i.e. mass is converted into energy. Here, charged
particles are lost, but total charge does not change.
•
There is strong evidence for the existence of fundamental particles called quarks that have charges of ±e/3 or ±2e/3. But these
particles are always found to be in combinations and the combinations always have charge in multiples of electronic charge.
Example 1:
How many electrons are there in 1C?
Solution:
Using q = ne, n = q/e
Example 2:
In a process known as β-decay, a neutron in an unstable atomic nucleus becomes a proton and in the
process ejects an electron and an anti-neutrino. Use conservation of charge to determine the charge of an
anti-neutrino.
Solution:
The decay of a neutron can be represented as n → p + e + v (anti-neutrino)
Or
n=
1
⇒
1.6 × 10 −19
n = 6.25 × 1018
Since the neutron is electrically neutral, total charge on left side is zero.
∴ Total charge on right side must also be zero.
⇒
q + (+e) + (−e) = 0
Here (+e) is charge of a proton, (−e) is charge of an electron. So, we get q = 0. So, the antineutrino is an electrically
neutral particle.
COULOMB'S LAW
Charles Coulomb experimentally established the fundamental law of electric force between two charged particles.
He observed that an electric force has the following properties:
1.
It is directed along a line joining the two particles and is inversely proportional to the square of the separation distance r,
between them.
2.
It is proportional to the product of the magnitudes of the charges, |q1| and |q2|, of the two particles.
3.
It is attractive if the charges are of opposite sign and repulsive if the charges have the same sign.
From these observations, Coulomb proposed the following mathematical form for the electric force between two charges.
The magnitude of the electric force F between charges q1 and q2 separated by a distance r is given by
F=
k
q1 q 2
…(i)
r2
Here k is a constant called the Coulomb constant. Its value is given below.
k = 8.9875 × 109 Nm2C-2, [k] = [ML3 T-4A-2]
The constant k is often written in terms of another constant, ε0, called the permittivity of free space. It is related to k by
k=
1
4πε 0
Here ε 0 =
∴
1
4π k
F=
1 Q1Q2
4πε 0 r 2
…(ii)
= 8.85 × 10−12 C 2 / N .m 2 . [ε 0 ] = M −1 L−3T 4 A2
Equations (i) and (ii) apply to objects whose size is much smaller than the distance between them. Ideally, it is precise for point
charges.
•
•
Coulomb's law is like Newton's law of gravity but, unlike gravity, electric forces can be attractive or repulsive:
When two charged objects are surrounded by a medium, the net force on each charged object becomes
the dielectric constant of the medium. K is also written as εr.
1
times, where K is
K
Electric Charges and Field
Check yourself
1.
The proton that is the nucleus of the hydrogen atom attracts the electron that orbits it. Relative to this force, does the electron
attracts the proton with less force, with more force, or with the same amount of force?
2.
If a proton at a particular distance from a charged particle is repelled with a given force, by how much will the force decrease
when the proton is three times farther away from the particle? When it is five times farther away?
Ans: 1. The same amount of force; 2. It decreases to 1/9 times of its original value; 1/25 times.
Properties of electrostatic force
(i)
The force between the two charge particles is a central force. Following two figures describe the difference between a central
and non-central force.
F
F
q2
q1
F
Non central force
Central force
(ii)
F
The force is attractive, when the charges have opposite nature, i.e., one of the charge is positive while the other is negative.
See the following figure. The positive charge is attracted to right and negative charge is attracted to left.
+q1
(iii)
q2
q1
F
- q2
F
The force is repulsive, when the charges have same nature, i.e., either both are positive or both are negative.
F
F
+
+
F
-
-
F
Example 3:
Two small objects each with a charge of 1 micro coulomb are located 3 cm apart in air. What is the electric
force between them?
Solution:
F= k
Example 4:
The electron and proton in hydrogen atom are separated by a distance of about 5.3 × 10-11 m. Find the
magnitudes of the electric force and the gravitational force that each particle exerts on the other. Determine
the ratio of the electric force Fe to the gravitation force Fg.
Solution:
Fe = ke
q1q2
r2
Fg = G
e
=
2
r2
=
me m p
r2
9 × 109 × 10 −6 × 10 −6
= 10 N
(3 × 10 −2 )2
9 × 109 × (1.6 × 10−19 )2
(5.3 × 10−11 ) 2
=
= 8.2 × 10−8 N
6.67 × 10−11 × 9.11 × 10−31 × 1.67 × 10−27
(5.3 × 10−11 )2
The ratio of the two forces is given as:
Example 5:
= 3.6 × 10 −47 N
Fe
= 2.27 × 1039
Fg
How many excess electrons must be added to each neutral sphere of mass 1 kg to balance the force of gravity
between them?
1.00 kg
1.00 kg
1.00 m
Solution:
G
6.67 × 10−11
m=
× 1.00 = 8.61 × 10−11
k
9 × 109
To find the number of excess electrons, take q = ne
kq 2
m2
=
G
⇒ kq 2 = Gm 2
r2
r2
So, n =
q 8.61 × 10 −11
=
e 1.6 × 10 −19
⇒
So, q =
n = 5.38 × 108
Electric Charges and Field
Example 6:
A charge Q is to be divided into two parts x and Q − x such that they experience maximum electrostatic
repulsion when placed at a certain distance apart. Find x.
Solution:
Force between them is given by F =
When the force is maximum,
⇒
∴
Example 7:
Solution:
1
(Q − 2 x )
4πε 0
2
r
=0
1
x(Q − x)
4πε 0
r2
dF
=0
dx
x=
⇒
d  1 xQ − x 2 
=0
dx  4πε 0
r2 
⇒
Q
2
The charge Q must be divided into two equal parts.
Two identically charged spheres are suspended by strings of equal length. The strings
make an angle of 300 with each other. When suspended in a liquid of density 0.8
gm/cc, the angle remains same. What is the dielectric constant of liquid? Density of
spheres = 1.6 gm/cc.
30 °
In air, we have T cos 15° = mg; T sin 15° = F, dividing, we get
tan 15° =
F
mg
…(1)
30 °
F
When set-up is immersed in medium, the electric force will be
and the
K

ρ 
effective gravitational force will becomes mg 1 − l  .
 ρs 
F/K
Then tan 15° =
 ρl 
m 1 − 
 ρs 
⇒
…(2)
T
15 °
F/k

ρ 
mg 1 − l 
 ρo 
k = 2.
Vector form of Coulomb’s Law
A charge q1 is placed at A whose position vector is r1
y - axis
Another charge q2 is placed at B whose position vector is r2 , such that
| AB | = r
The magnitude of force is given by, F =
q1 q2
1
4πε 0
r
A
1
q1q2
4πε 0 r
2
F1
O
AB
Here AB is a unit vector along line joining A and B, pointing from A to B.
⇒
 1  q1q2 (r2 − r1 )
F2 = 
 4πε 0  r 2 r2 − r1
Similarly force on q1 due to q2 is given by,
Note that F1 = − F2
F1 =
F2
B
2
Force on q2 due to q1, in vector form, is given by
F2 =
AB
1
4πε 0
q1q2 (r2 − r1 )
q1q2
1
=
(r2 − r1 )
2
r2 − r1
4πε 0 r 3
r
x - axis
Electric Charges and Field
Example 8:
A particle of mass m and carrying a charge – q1 is moving around a fixed charge + q2 along a circular path of radius
r. Prove that period of revolution T is given by T =
Solution:
16π 3ε 0 mr 3
q1q2
When q1 revolves around q2, the required centripetal force is provided by the electrostatic force between them.
∴
kq1q2
r2
=
mv 2
where v is the velocity of q1.
r
16π 3ε 0 mr 3
2π r
⇒T=
v
q1q2
Also T =
Principle of Superposition
The force of interaction between a pair of charges q1 and q2 is independent of presence of other charges. Thus if we have a
number of charges exerting force on a given charge, we need to write individual forces in vector form and add them.
Example 9:
Solution:
Three charged particles are arranged in a line. Calculate the net force on Q3.
Let F31 be the force on Q3 by Q1. We have
F31 = k
Q3Q1
r312
=
(9.0 × 109 )(4.0 × 10−6 )(8.0 × 10−6 )
(0.50)2
= 1.15N
Where r31 = 0.50m is the distance from Q3 to Q1.
Similarly F32 = k
Q3Q1
r322
=
(9.0 × 109 )(4.0 × 10−6 C )(8.0 × 10−6 )
(0.20)2
= 2.7 N
Since F31 is repulsive and F32 is attractive, the directions of the forces are as shown in figure. F31 points in the right
direction and F32 points in the left.
The net force on particle 3 is then F = − F32 + F31 = −2.7 N + 1.15 N = −1.55 N
The magnitude of the force is 1.55 N, and it points to the left.
The charge Q1 exerts force on charge Q3 as if Q2 just were not there. This is the principle of superposition. The charge in the
middle, Q2, in no way affects the force exerted by charge Q1 on Q3.
Example 10:
Given below are four arrangements of charges: Let F1, F2, F3 and F4 be the magnitudes of force experienced by the
charge q placed at O in the respective cases. Arrange these magnitudes of forces in ascending order.
q
q
L
O
2q
q
O
L
(i)
L
(ii)
q
q
q
-q
q
O
(iii)
Solution:
2q
O
L
L
L
(iv)
Let us draw the free body diagram of the charge q at O in each case. They are shown below:
In case (i),
q
L
F1 =
2kq 2
2
−
L
In case (ii),
F2 =
2
kq
L2
kq 2 kq 2
= 2
L2
L
kq(2q)
L2
O
kq × q
q
L2
kq × q
O
2
L
3q
Electric Charges and Field
In case (iii),
F3 =
kq 2
−
kq 2
L2
L2
In case (iv),
F4 =
=0
kq × q
Solution:
kq × q
2
L
3kq 2 kq 2 4kq 2
+ 2 = 2
L2
L
L
kq × 3q
L2
q
O
kq × q
L2
L2
The correct ascending order is F3 < F1 = F2 < F4
Example 11:
q
O
Two charges q and 4q are placed at a distance L from each other. Where should another particle of charge q, be
placed on x-axis, so that net force on charge q becomes zero?
Let the charges be placed at x = 0 and x = L. The charge 4q repels
the charge q towards – x-axis. So the charge q has
to be placed
F2
F1
q
q
4q
on the left of charge q. Let it be placed at x = - r.
Let F1 be the force on q (at x = 0) due to charge at x = -r, given as
x=0
x=-r
x=L
 1 q ⌢
F1 = 
i
 4πε 0  r 2
2
 1  q × 4q ⌢
i
Let F2 be the force on q (at x = 0) due to charge x = 4q, given as F2 = 
 4πε 0  L2
According to the condition given,
F1 + F2 = 0
∴ The charge q should be placed at r =
Example 12:
Solution:
⇒
q 2 4q 2
= 2
r2
L
Or r =
L
2
−L
2
Four equal charges Q are kept at the four corners of a regular pentagon. What is the force on a charge q kept at the
centre of the pentagon? The distance of the centre from the corner of the pentagon is ‘a’.
If you place equal charges on each corner of a regular polygon then the force on a charge kept at the centre will
be zero. In the given question, one corner of the regular pentagon doesn't have any charge. So we put equal positive
and negative +Q and -Q at its corner so that the net charge on that corner is zero.
Now the five +Q at the corners will produce zero resultant force at the centre, so we can remove all these five +Q
and will be left with only one -Q as shown.
So, now the force on q as per Coulomb's law is
Qq
4πε 0 a 2
towards that corner which didn’t contain any charge
originally because –Q will attract +q towards itself.
Example 13:
Solution:
A charge Q1 = + 1 C and Q2 = - 4C are fixed at a separation of 1m. Where should you place the third charge q, so
that the entire system is in equilibrium?
In general remember if Q1 and Q2 are of same nature (means both +ve or both –ve) then the third charge should be
put between Q1 and Q2 on their joining straight line. But if Q1 and Q2 are of opposite nature then the third charge
will be put outside and close to that charge which is smaller in magnitude.
Here since Q1 and Q2 are of opposite nature, the third charge q will be kept at a distance x to the left of Q1 and
outside (on the straight line joining Q1 and Q2 as shown below) because Q1 has less magnitude compared to Q2.
First we will calculate x by making the net force on q equal to zero.
q
Q1 = 1C
Q2 = -4C
Q1 .q
Q2 .q
=
1m
x
4πε 0 x 2 4πε 0 (1 + x ) 2
Electric Charges and Field
Or
1
x
2
=
4
(1 + x)
Taking the positive square root
2
1
2
=
x (1 + x)
⇒ x = 1 meter.
For calculating q, make net force on Q1 equal to zero, as the entire system in equilibrium means, the net force on all
the three charges due to each other is equal to zero. Here do you make out that q has to be –ve because Q2 is –ve!
Because only then Q1 will experience forces in opposite direction.
Q1 .q
4πε 0 x
2
=
Q1 .Q2
4πε 0 1
2
∴
q
x
2
=
Q2
12
Put x = 1 m and Q2 = 4 C (just take the magnitude here)
Or q = 4 C
But as we predicted in the beginning q has to be of –ve sign. So, Q = - 4C and x = 1m to the left of Q1.
Example 14:
Solution:
Four charges, Q, q, Q and q are kept at the four corners of a square as shown below.
What is the relation between Q and q so that the net force on a charge q is zero?
Let us make the force on upper right corner ‘q’ equal to zero. Here both the ‘q’ will
have same sign (either positive or negative), so they will repel each other, say by force
F1. To balance F1, Q must apply attractive force on q, say F2. That also means Q and q
will have opposite signs.
2F2 (By Pythagoras theorem) and it will
Resultant of both these forces will be
exactly be opposite to F1. Or F1 = 2 F2
From Coulomb’s law
F1 =
∵
q2
4πε 0 (d 2)2
F1 = 2 F2
Qq
and
F2 =
∴
q2
=
(d 2 )2
4πε 0 d 2
2Qq
d2
∴
Q=
q
2 2
But as we said Q and q have opposite sign, so q = −2 2Q.
EXERCISE - 01
1.
The charge of an object cannot be
(a)
2.
4.
5.
6.
7.
8.
9.
(b)
1.6 × 10 −18 C
(c)
1.92 × 10 −18 C
^
(d)
2.4 × 10 −16 C
What must be the charge on each of two 100 kg spherical masses in order for the electrical force to be equal to the gravitation
force?
8.6 nC
(c)
7.4 × 10 −17 C
(d)
100 C
How many electrons form 1µC?
(a)
105
(b)
107
The Dimensional formula for Coulomb’s constant is
(c)
8 x 1042
(d)
6.25 x 1012
(a)
3.
1.9 × 10 −18 C
8.6 × 10 −19 C
(b)
(a)
ML3T−4A−2
(b)
M−1L3T4A2
(c)
ML2T3A−2
(d)
ML3T−4A2
The unit of permittivity of free space is
(a)
C2
(b)
Nm
(c)
NC2
(d)
C2/Nm2
Two charges Q1 and Q2 are separated by a certain distance. If the distance is doubled, the force between them
(a) increases by 100 % (b) decreases by 100 %
(c) Remains the same
(d) decreases by 75 %
If the separation between two charges is increased by 1%, the change in force of interaction between them changes by
(a)
+1%
(b)
-1%
(c)
2%
(d)
– 2%
Two charges q and 16 q are kept at a distance 20 r from each other. The net force on a third charge would be zero if it is kept
at a distance of
(a)
16 r from q and a distance 4 r from 4 q
(b)
16 r from 4q and a distance 4 r from q
(c)
10 r from each of the charge
(d)
16 r from each of the charge
Three charged particles each having a charge q are placed at the vertices of an equilateral triangle of side length a. The net
electrostatic force on any one of the charged particles is
Electric Charges and Field
(a)
10.
11.
12.
13.
14.
(b)
2kq 2
(c)
kq 2 3
(d)
kq 2 2
a2
a2
a2
a2
State True or False: The force exerted by a charge particle q1 on another charged particle q2 does not depend on the medium
between them.
Does charge depend on frame of reference?
Can charge exist without mass? Can mass exist without charge?
Two charged particles q and 4q are kept at distance 3 m, the force on a charge Q kept at a distance 1 m from q and 2m from
4q is
(a)
5 KQq
(b)
KQq
(c)
2K q2
(d)
Zero
If seven identical charged particles of charge ‘q’ each are kept at vertices of regular cube, then force acting on a charge ‘-q’ at
centre of cube is (Here, r is distance of vertex from, centre of cube)
(a)
15.
kq 2
Kq 2
(b)
5Kq 2
(c)
r2
r2
When a body is charged by friction, is there a change in its mass?
6Kq 2
r2
(d)
Zero
ELECTRIC FIELD
There is a force between two charged objects even when the objects are not touching. A
helpful way to look at the situation uses the idea of the field, developed by the British scientist
Michael Faraday (1791-1867). According to Faraday, an electric field extends outward from
every charge and permeates all of space (Figure (a)]. If a second charge (call it Q2) is placed
near the first charge, it feels a force exerted by the electric field (say at point P in Figure (a).
The electric field at point P is considered to interact directly with change Q2 to produce the
force on Q2.
We can investigate the electric field by measuring the force on a small positive test charge. By
a test charge we mean a charge so small that the force it exerts does not significantly alter the
distribution of those other charges that create the field. If a tiny positive test charge q is place
at various locations in the vicinity of a single positive charge Q as shown in Figure (b) (points
a, b, c), the force exerted on q is as shown.
In each case, the force on q is directed radially away from Q. The electric field E, at any point
in space is defined as the force F exerted on a tiny positive test charge placed at the point
divided by the magnitude of the test charge q.
E=
F
q
Figure (a)
Fa
a
a
c
Q
Fc
…(i)
b
F
More precisely, E is defined as the limit of
as q is taken smaller and smaller, approaching
q
zero. That is, q is so tiny that it exerts essentially no force on the other charges which created
the field. The electric field at any point in space is a vector whose direction is the direction of
the force on a tiny positive test charge at that point and whose magnitude is the force per unit
charge.
S.I. units: Newton per coulomb (N/C).
Dimensional formula: [E] = ML T-3 A-1
Fb
Figure (b)
If we are given the electric field E at a given point is space, then we can calculate the force F on any charge q placed at that
point by writing F = qE
Example 15:
A particle of mass 9.0 x 10−16 kg carrying 20 excess electrons is held stationary under the action of a vertical electric
field E. Taking g = 9.8 m/s2, determine the electric field intensity.
Solution:
Here, qE = mg and q = ne = 20 × 1.6 × 10−19C
⇒ E=
mg 9.0 × 10−16 × 9.8
=
= 2.75 × 103 N/C
q
20 × 1.6 × 10−19
Electric Charges and Field
Electric Field due to a Point Charge
According to Coulomb’s law, the force exerted by q on the test charge is Fe = k
qq0
r2
rˆ
Here r̂ is a unit vector directed from q toward q0. This force is directed away from the source charge q. Since the electric
q
field P is defined by E = F/q, the electric field created by q is E = k 2 rˆ
r
When the source charge q is positive the source charge sets up an electric field at point P, directed away from q.
When q is negative, the force on the test charge is towards the source charge, so the electric field at P is directed towards the
source charge.
E
q0
q0
+
P
q
q
+
+
P
P
q
+
E
q
-
P
+
Superposition Principle
The net or resultant electric field due to more than one point charge can be determined with the aid of principle of
superposition. If we place a test charge q0 near n point charges, the net force acting on it is
F = F01 + F02 + ....... + F0 n (where F0i is force on test charge due to ith charge).
⇒ E=
F0 F01 F02
F
=
+
+ ..... + 0 n
q0
q0
q0
q0
E = E1 + E2 + .... + En
⇒
Here En is the electric field that would be present at the position of test charge if only the nth charge existed.
Example 16:
Solution:
Calculate the magnitude and direction of the electric field at a point P at a distance 30 cm to the right of a point
charge Q = −3.0 x 10−6C.
The magnitude of the electric field is:
E=k
Q
r
2
=
9 × 109 × 3 × 10−6
(0.3)
2
q
= 3 × 105 N/C
+
P
The direction of the electric field is toward the charge Q, to the left as shown in Figure.
Check yourself
A +3 mC charge is at a point P where an external electric field is directed to the right and has a magnitude of 4 × 106 N/C. If
the charge is replaced by another charge of −3 mC, the external electric field at P.
(a) Remains the same
(b) Changes its direction
(c) Becomes indeterminate
Ans: (a)
Example 17:
Solution:
Two charges q1 = − 25 × 10−6C and q2 = 50 × 10-6 C are placed 10 cm apart. What is the magnitude of electric field
at a point at a distance 2 cm from q1 and 8 cm from q2.
The situation is shown in figure.
q1
q2
The resultant electric field at the concerned point is E = E1 + E2
kq1
kq2
+ 2 =
r12
r2
8
= 6.328 × 10 N/C.
⇒ E =
Example 18:
9 × 10 × 25 × 10
9
(2 × 10−2 )2
−6
+
9 × 10 × 50 × 10
9
−6
E1 E1
(8 × 10−2 )2
A point charge q is placed at x = 0. The electric field vector at x = L is E . When another charge 4q is placed at x =
2L, what is the new electric field vector at x = L?
Electric Charges and Field
Solution:
The two situations are as shown
1 q ⌢
−1 4q ⌢
E=
i &
E1 =
i
2
4πε 0 L
4πε 0 L2
x=0
The new electric field at x = L is given E = E1 + E
=
Example 19:
−1 3q ⌢
i = −3E
4πε 0 L2
E
q
E
E1
q
So, new electric field is −3E
Three charges Q, 2Q and – Q are placed at the vertices A, B and C respectively of an isosceles right angled triangle,
Determine the direction of the resultant field produced by these charges
at the midpoint of the hypotenuse.
q
2E
The distance of mid-point of hypotenuse is same from all the three
E2q
vertices. This means that if magnitude of electric field at the given
2√2E
⇒
point due to +Q charge is E, then the magnitudes of electrical field due to
Eq
the three charges are:
Solution:
E(-q)
E2Q+ = 2E
EQ+ = E; EQ− = E;
The resultant of three field vectors will be directed parallel to BC
Example 20:
2E
-q
2q
An infinite number of charges, each equal to q, are placed along the x-axis at x = 1, x = 2, x = 4, x = 8 and so on.
Find the electric field at point x = 0 due to this set of charges.
The field due to each of the charges is along negative x-axis.
Solution:
 1 q
⌢
1 q
1 1
Net field E0 = − 
+
+
+ ..... i
2
2
2
4πε 0 2
4πε 0 4
 4πε 0 1

−q
4πε 0
E0 =
1
 1 1
⌢
1 + 4 + 16 + 64 + ..... i




−q  1  ⌢
E0 =

i
4πε 0  1 
1−
 4
E0 =
a 

If r < 1, then the sum of infinite terms in a geometric progression is 1− r 


−4q ⌢
−q ⌢
i =
i
12πε 0
3πε 0
The resultant field intensity at O in each case shown below is zero.
+Q
+Q
+Q
r
O
r
+Q
r O
r
+Q
O
(a)
(b)
+Q
r
+Q
+Q
+Q
+Q
+Q
(d)
4q
+Q
O
+Q
+Q
+Q
O
+Q
(e)
+Q
+Q
(c)
+Q
+Q
Electric Charges and Field
Graphical Variation of Electric Field
The electric field due to a charge is E =
1
q
4πε 0 r 2
Before plotting the graph, you must look out for the relation between the two variables E and r. From the above relation, it is
1
clear that E ∝ 2
r
This implies the following points:
(a)
As r increases, E decreases and the variation is not a linear one,
(b)
As r → 0, E → ∞
(c)
As r → ∞, E → 0
Note that the curve will not intersect the r-axis or E-axis.
Example 21:
Solution:
Two point charges q each are placed at points (a,0) and (−a,0). Draw a graph to show the variation of electric field
strength.
E
(i)
Along y-axis with distance y from origin
(ii)
Along x-axis with distance x from origin
(i) The electric field at point P is the resultant of E1 and E2, given as
E2
E1
θ
Both E1 and E2 can be resolved into components E1cosθ and E1 sinθ
 1 
q
E1 = E2 = 

2
2
4
πε

0  (a + y )
θ
The components containing sin θ cancel out so that net field E = 2 E1 cos θ
 1 
 1 
q
y
2qy
= 2×
× 2
=

2
2
2 1/2
 4πε 0  (a + y ) (a + y )
 4πε 0  (a 2 + y 2 )3/ 2
E varies with y as described below.
(a)
For small values of y (i.e. y < < a):
a2 + y2
y
a
a
q
q
 1  2qy
E=
 4πε 0  a3
(∵ a 2 + y 2 ≈ a 2 )
So, E ∝ y
Also, at y = 0, E =0
 1  2q
E=
×
 4πε 0  y 2
(∵ a 2 + y 2 ≈ y 2 )
(b)
For large values of y (i.e. y > > a):
(i)
This becomes the field due to a point charge
In general we conclude the following.
Electric field starts from zero and increases linearly with y for small values of y and
(ii)
Electric field varies inversely as y2, for large values of y and approaches zero as y approaches ∞.
(c)
As the electric field increases from 0 at origin to again become zero as y → ∞, the electric field must attain a
maximum value somewhere in the region 0 < y < ∞.
At this point,
dE
=0
dy
3
( a 2 + y 2 )3/ 2 × 1 − y × ( a 2 + y 2 )1/ 2 × 2 y


d 
y
d  1 
2qy
2
⇒
=0

=0 ⇒

=0 ⇒
dy  (a 2 + y 2 )3/ 2 
dy   4πε 0  (a 2 + y 2 )3/2 
( a 2 + y 2 )3
Solving we get y =
this.
a
2
. The electric field attains a maximum at y =
a
2
. So, the complete graph will look like
Electric Charges and Field
(iii)
The whole x-axis is divided into three regions by the two charges.
Region -1
−∞ < × ≤ −a
Region -1
−a ≤ × ≤ a
Region - 3
a≤×<∞
The graphs for region 1 and region 3 should look similar
except that there will be a difference in direction and
therefore sign. First we shall discuss its variation.
Consider a point (x, 0) in region 3
E = E1 + E2
=
1  q
q 
1 2( x 2 + a 2 )
+
=
4πε 0  ( x − a)2 ( x + a)2  4πε 0 ( x 2 − a 2 ) 2
Clearly, at x = a, E = ∞
Also, as x > > a, E =
1 2q
4πε 0 x 2
[∵ x 2 + a 2 ≈ x 2 and x 2 − a 2 ≈ x 2 ]
So the electric field expression becomes similar to the point charge.
The only difference in variation of electric field in region 3 and region 1 is that of
sign. Let us take the electric field directed along +x-axis is positive. The graph
for E in region 1 and 3 will then be like the one given below.
In region 2. First thing that is very clear is that the electric field will be zero at the
origin (which is equidistant from the two charges). This also implies that electric
field changes direction at origin. The direction of electric field in the part 0 < x < a
will be determined by the charge at (a,0) where in the part –a < x < 0, the direction
will be determined by the charge at (-a,0)
For 0 < x < a: Clearly electric field intensity E1 due to charge at
(a,0)
is
more than electric field intensity E2 due to charge at (-a,0). So, electric field is
directed towards –x-axis.
By similar argument, the electric field in –a < x < 0 will be
towards positive x-axis. At x = 0, E = 0 At x = ±, E → ∞
So the variation of electric field will be like the one shown .
E
EXERCISE - 02
1.
The SI unit of electric field is:
(1)
N
(2)
N
(3)
N/C
(4)
N/m2
2.
For the given figure, q1 < q2 , q1 is negative while q2 is positive. The point where resultant electric field could be zero is
3.
(1)
A
(2)
O
(3)
C
(4)
None
In the given figure, q1 = q2 > 0. The point where electric field is directed downward could be
x
Electric Charges and Field
4.
5.
6.
7.
8.
9.
10.
11.
12.
(1)
A
(2)
B
(3)
C
(4)
D
Charge q is placed at x = 0 & another charge q is placed at x = L. The coordinates of the point where net electric field is zero
is
L
3L
x = 2 L (3)
x = −L
(1)
(2)
(4)
x=
x=
2
2
A charge q is placed at x =0 and another charge 4q is placed at x = 3L. The coordinates of the point resultant electric field
would be zero is
−L
x=L
x = 2L
x = −L
(2)
(3)
(4)
x=
(1)
2
A charge −q is placed at x = 0 and another charge 4q is placed at x = L. The coordinates of the point where resultant electric
field would be zero is:
L
−L
x = −L (4)
x = 2L
(1)
(2)
(3)
x=
x=
2
2
Figure shows charge beads with charges as shown in figure. The charges are placed upon the corners of a square. Rank the
magnitude of electric field at the centre in decreasing order.
State principle of superposition of electric field.
(i) To measure the electric field at a point P in space due to a source charge Q, the test chare q placed at the point P should be
negligibly small. Why?
(ii) To measure the electric field at a point in space, the source charge should be kept fixed. Why?
True /False
(i)
A charge interacts with its own electric field.
(ii)
A charge interacts with the gravitational field.
(iii)
A mass interacts with the electric field.
(iv)
Electric field can be added to gravitation field.
Select the choices in which electric field at point P is non zero.
The electric field at a distance r from a point charge q is E. The electric field at a distance 0.99r will be more than E by
(1)
13.
1%
(2)
2%
(3)
−2%
(4)
−1%
Two charges -4q and q are kept at (−a,0) and (a,0) respectively. The graph showing the variation of electric field along x-axis
is:
E
E
(b)
(a)
x= -a
x=a
x= -a
x=a
Electric Charges and Field
x= -a
14.
E
E
(c)
(d)
x= -a
x=a
x=a
For a given pair of charges q each kept at (−d, 0) and (d,0) , the electric field on y-axis will be maximum at y =
d
2
(1)
d
3
(2)
(3)
d
2
(4)
d
3
ELECTRIC DIPOLE
A configuration of two charges of same magnitude q, but of opposite sign, separated by a small
distance (say 2a) is called an electric dipole.
Dipole moment
Dipole moment for an electric dipole is a vector quantity directed from the negative charge to the
positive charge and its magnitude is p = q × 2a (charge × separation).
Units: SI unit of dipole moment is C-m (coulomb-metre)
Dimensional formula: M 0 LTA
Example 22:
In a water molecule a charge +q appears on H-atom due to movement of electrons
towards oxygen. Taking the separation between H-atom and oxygen
atom
as
r,
calculate the dipole moment of a water molecule (Take cos(52.5 ) ≈ 0.6 )
Solution:
The arrangement is equivalent to two electric dipoles each of dipole moment q × r.
 105 
The resultant dipole moment is p = 2 × q × r × cos 

 2 
(
)
⇒ p = 2 × q × r × cos 52.5 = 1.2 q r units
Electric Field due to an Electric Dipole
(a) On axial line
Consider an electric dipole kept along x-axis such that its centre lies at
origin.
Net electric field P = E1 + E2 , where E1 and E2 are fields due to +q and
−q respectively
E1 =
q
4πε 0 (r − a)
⇒ E = E1 + E2 =
⇒E=
2
iˆ, E2 =
−q
4πε 0 (r + a) 2
iˆ
q  1
1 ˆ
−
i

2
4πε 0  (r − a) (r + a)2 
2 pr
4πε 0 (r 2 − a 2 ) 2
⇒E=
2p
4πε 0 r 3
2
∵ p = 2aqiˆ 
For a short dipole, a is very small in comparison
⇒E=
ˆ
4ariˆ
2.(2a.q)ir
=
2 2
2
4πε 0 (r − a )
4πε 0 (r + a 2 ) 2
q
i.e., E ∝
∴r 2 − a 2 ≈ r 2
1
r3
(b) On equatorial line
p = 2aqiˆ
Electric field vector due to +q is E1 =
q
[− cosθ iˆ + sin θ ˆj ]
4πε 0 (r 2 + a 2 )
p
H(+q)
H(+q)
105 °
O(+2q)
Electric Charges and Field
Electric field vector due to −q is E2 =
q
[− cosθ iˆ − sin θ ˆj ]
4πε 0 (r 2 + a 2 )
Resultant field at P is E = E1 + E2 ⇒ E =
⇒E=
a
−2qiˆ
.
4πε 0 ( r 2 + a 2 ) r 2 + a 2
⇒E=
−2aqiˆ
4πε 0 (r 2 + a 2 )3/2
⇒E=
−2q cosθ iˆ
4πε 0 (r 2 + a 2 )

∵ cosθ =

−p
4πε 0 (r 2 + a 2 )3/2


r +a 
a
2
2
∵ p = 2aqiˆ 
For a short dipole ∴ r 2 + a 2 ≈ r 2
⇒E=
−p
4πε 0 r 3
i.e., E ∝
1
r3
For a short dipole, electric field Eax , at a distance from centre on the axial line is related to Eeq , the electric field at same
distance on equatorial line as Eax = −2Eeq
Electric field at a general point due to an electric dipole
The result that is being derived here is applicable only for a short dipole of
dipole moment p. Consider a point P, at a distance r from a short dipole such
the position vector of point P w.r.t. the dipole, makes an angle θ with the dipole
moment vector.
The net electric field E at point P is made of two components E1 and E2 . E1 is
generated by the component of dipole moment vector along the line OP (i.e.,
pcosθ) given as
| E1 | =
1
4πε 0
E
E2
2 p cos θ
×
r3
E1
r
p cos θ
Axial line
α
θ +α
θ
p
p sin θ
E2 is generated by the component of dipole moment vector ⊥ to line OP (i.e., psinθ) given as
| E2 | =
1
4πε 0
×
p cos θ
r3
Now, E = E12 + E22 =
The direction is given by α = tan −1
p
4πε 0
1 + 3cos 3 θ
r3
E2
tan θ
= tan −1
E1
2
 tan θ
The field vector E makes an angle θ + tan −1 
 2

 with the dipole moment vector.

When θ = tan −1√2, the angle θ + α = 90 ° i.e. E ⊥ P.
Electric Dipole in a Uniform Electric Field
An electric dipole is placed in a uniform electric field E
The force experienced by the dipole is Fnet = qE − qE = 0
∴ In a uniform electric field, Fnet = 0
The two forces form a couple and it tries to turn the dipole. Torque due to couple is given
as
τ = either force × perpendicular distance between the forces
⇒ τ = qE × (2a sin θ ) = 2aqE sin θ
τ
2 a sin θ
Electric Charges and Field
⇒ τ = pE × sin θ
(∵ p = 2aq)
∴τ = p × E
(This is clockwise)
Let us find out τ for different values of θ.
When θ = 0, τ = 0
When θ = 180ο, τ = 0
When θ = 90ο, τ = pE (i.e., maximum)
The torque acts so as to decrease the angle between p & E
Potential energy Associated with Dipole
A dipole placed in electric field is associated with a potential energy. The change in potential energy is related to the work
done by electric field as
dU = −dWE
If the dipole is turned to increase θ by dθ,
(∵ τ is clockwise while dθ is anticlockwise.
dWE = τ .dθ = τ dθ
⇒ dU = pE sin θ dθ
θ2
U2
⇒
∫
U1
dU =
∫ pE sin θ dθ ⇒ U 2 − U1 = − pE [cos θ 2 − cos θ1 ]
θ1
Let θ = 90 is taken as reference position or zero of potential energy.
⇒ U 2 − 0 = − pE cos θ 2 − cos 90 
∴θ = 90 ⇒ U1 = 0
θ = 0 ,U min = − pE , τ = 0.
This is stable equilibrium position.
θ = 180 ,U max = + pE , τ = 0.
This is unstable equilibrium.
θ = 90 ,U = 0,τ = pE.
This is the maximum value of torque.
⇒ U = − pE cos θ
Example 23:
When an electric dipole is placed in a uniform electric field making angle θ with electric field, it experiences a
torque τ. Calculate the minimum work done in changing the orientation to 2θ.
Solution:
τ = pE sinθ
⇒ pE =
τ
sin θ
W = ∆U = − pE cos θ 2 + pE cos θ1 ⇒ W = pE [cos θ − cos 2θ ]
⇒W =
τ
sin θ
(∵θ1 = θ , θ2 = 2θ )
[cos θ − cos 2θ ]
Example 24:
A charge q is placed at (1, 2, 1) and another charge −q is placed (0,1,0), such that they form an electric dipole. There
exists a uniform electric field E = (2iˆ + 3 ˆj ). Calculate the dipole moment vector and torque experienced by the
dipole.
Solution:
(i) p = q × r
⇒ p = q × ( r2 − r1 )
Now, r1 = ˆj , r2 = iˆ + 2 ˆj + kˆ
∴ p = q (iˆ + ˆj + kˆ )
(ii) τ = p × E ⇒ τ = q (iˆ + ˆj + kˆ ) × (2iˆ + 3 ˆj )
Electric Charges and Field
iˆ + ˆj + kˆ
τ = q 1 2 3 = q ( −3iˆ + 2 ˆj + kˆ) N − m
2 3 0
When a dipole is placed in a non-uniform electric field, it may experience a force as well as torque.
Example 25:
A point chare Q is placed on the axis of an electric dipole (having dipole moment p) as shown.
p
r
Q
Assuming the dipole to be short, determine the torque and force experienced by the dipole.
Solution:
The electric field due to Q acts anti-parallel to dipole moment. So, τ = ( p × E ) is zero.
To calculate the force on dipole, we calculate the force on Q.
The force on dipole would be equal and opposite to it.
As F = QEax
⇒F=
∴ Force on dipole is
Q × 2p
4πε 0 r 3
2 pQ
4πε 0 r 3
F = Q E ax
(towards right)
, leftward
Example 26:
Two short dipoles (A and B) with their moments p1 and p2 respectively are placed as shown in figure. Find
(i)
Torque on B due to A
(ii)
Potential energy of B in the field of A
p1
p2
r
(iii)
Force on B due to A
Solution:
(i) τ = p2 × E1 = p2 × E1 sin180 = 0
(ii) Potential Energy = = − p 2 × E1 = − p2 E1 cos180 = p2 E1
= P2 ×
1
.
p1
4πε 0 r
3
=
1
4πε 0
(iii) For force F = −
.
p1 p2
r3
dU
d  1 p1 p2 
1
d −3
3 p1 p2
=− 
. 3 =−
. p1 p2
(r ) = +
.
dr
dr  4πε 0 r 
4πε 0
dr
4πε 0 r 4
Example 27:
A light mass-less rod of length l lies in x-y plane with its centre at origin and it makes an angle θ with x - axis. A
particle of mass m and charge −q is attached at its left end and another particle of
mass m and charge q at the other end. Now an electric field of constant
magnitude
E
and directed along x-axis is switched on. What is the angular speed of the rod at the
instant when it becomes parallel to x-axis?
Solution:
The dipole moment has a magnitude p = q × l
The electric field is given as E = Eiˆ
To calculate angular speed, we can apply mechanical energy conservation.
U i = − p.E = − pE cos θ
Kf =
Using,
&
U f = − p.E = − pE [rod becomes parallel to x-axis]
2
2
1 2 1 l 
l 
Iω = m   + m    ω 2
2 
2
2   2 

U i + Ki = U f + K f ,
we get − pE cosθ = − pE +
ml 2 2
ω
4
ω
4qE[1 − cos θ ]
ml
Electric Charges and Field
EXERCISE-03
1.
Find the dipole-moment of system
2.
A particle of charge q is to be released in the electric field produced by an electric dipole. Is it possible to find a point, where
the particle can remain stationary, when released.
3.
The permanent electric dipole moment of the water molecule (H2O) is 6.2 × 10−30 Cm. What is the maximum possible torque
on a water molecule in 5.0 × 105 N/C electric field?
The angle between dipole moment and electric field for the state of minimum potential energy is
4.
5.
(1) Zero
(2)
π/2
(3)
π
(4)
3π/2
The magnitude of the each of two opposite charges that form an electric dipole is increased by a factor of 5 while the
separation between the charges is tripled. What is the change in magnitude of the torque on the dipole in a uniform field?
6.
A dipole having charges +q and −q is fixed on a rough surface as shown. Another charge Q of
mass m is placed at distance d from this dipole. Find the minimum value of d for which charge
Q will not move. (Given a << d, co-efficient of static friction = µs= 0.5)
7.
A dipole is placed in constant electric field E as shown. If it is released from rest from this
position, find the kinetic energy gained when it becomes parallel to electric field vector. (Assume
dipole is small and placed in a gravity free space)
A small dipole is initially placed perpendicular to a constant electric field E and then released. If it
is allowed to rotate about its centre. Find its angular velocity (ω) in term of its angular
displacement (θ). Moment of inertia of dipole about its centre is I.
8.
9.
Two small and ideal dipoles are placed in space as shown. Dipole
to rotate about its hinged centre. Find the direction of rotation of
10.
p1 is fixed and p2 is allowed
p2 in the following cases:
Determine the force of interaction between the short dipoles arranged such that their dipole moments are collinear and the
separation between them is r.
OSCILLATIONS OF A DIPOLE
When an electric dipole is kept in a uniform electric field, it experiences a torque given by τ = pE sin θ. When θ = 0, dipole
will be in equilibrium. On disturbing the dipole from equilibrium position, the torque due to field makes it oscillate. Let a
dipole be disturbed by angle θ from its equilibrium position and let the dipole possess a moment of inertia I about O. It
experiences a torque τ = − pE sin θ , where negative sign indicates that this is the restoring torque which tries to restore dipole
to its mean position.
For small θ, sin θ ≈ θ.
⇒ τ = − pEθ ,
Further τ = I α = I
⇒
d 2θ
d 2θ
dt 2
 pE 
θ =0
+
 I 
dt
2
⇒I
d 2θ
dt 2
= − pEθ
Electric Charges and Field
Comparing the above result with standard differential equation of simple harmonic motion,
d 2θ
pE
+ ω 2θ = 0, we have ω =
2
I
dt
This equation shows that the dipole will execute angular SHM when a small angular displacement is given from its mean
position. The time period is given by
T=
Example 28:
Solution:
2π
ω
= 2π
I
pE
⇒ Frequency of oscillation = f =
1
1
=
T 2π
pE
I
A charged particle of charge q, mass m is released from rest in a uniform electric field. Obtain its velocity,
displacement and kinetic energy at time ‘t’.
Force on the particle is given by F = qE (in the direction of field since q is +ve)
⇒a=
F qE
=
= constant
m m
(a)
Velocity after time t is v = u + at = 0 +
(b)
Displacement s = ut +
(c)
Kinetic energy
K=
qE
.t
m
1 2
1 qE
at = 0(t ) + . .t 2
2
2 m
1 2 1
qE
mv = m
t
2
2
m
or
v=
qE
.t ⇒ v ∝ t
m
or
s=
qE 2
.t ⇒ s ∝ t 2
m
or
K=
2
q 2 E 2t 2
⇒ K ∝ t2
2m
Example 29:
A charged particle of charge q, mass m is projected in a uniform electric field E with initial velocity u at right
angles to the field. Show that it moves in a parabolic path.
Solution:
Let u = uiˆ and E = Ejˆ
⇒a=
F qE qE ˆ
=
=
j
m m
m
Applying equation of motion in vector form, r = OP = ut +
Comparing with r = xiˆ + yjˆ, x = ut,
 qE
y=
 2mu 2
Example 30:
Solution:
y=
…(i)
1 2
1 qEt 2
at = (ut )iˆ +
2
2
m
ˆj
1  qE  2

t
2 m 
 2
x . This is the equation of a parabola.

The distance between the two plates of a parallel plate capacitor is 1 cm and potential difference between them is
1200 volts. If an electron of energy 2000 eV enters at right angles to the field, what will be its deflection if the plates
are 1.5 cm long?
V
1200
=
= 1.2 × 105 Vm −1
d 1 × 10−2
Deflection can be obtained directly from equation of trajectory.
Electric field between the plates is: E =
 qE  2 qE 2
y0 ( x = L ) = 
L =
.L
 2mu 2 
4K
qEL2
4K
Where K = Kinetic energy of electron. Given: L = 1.5 × 10−2 m, K = 2000 × 1.6 × 10−19 J
⇒ Deflection = 3.375 mm
⇒Deflection =
Charge Distribution and Charge Density
(i)
Volume Charge Distribution: When charge is distributed over a volume in space, we define volume charge density as
dq
ρ=
where ‘dq’ is the charge that occupies the volume dV.
dV
Electric Charges and Field
The unit of ρ is C/m3 in SI units. The total amount of charge within the entire volume V is Q = ρdV
v
(ii)
The concept of charge density here is analogous to mass density ρm.
Surface Charge Distribution and Surface Charge Density: In a similar manner, the charge can be distributed over a surface
S of area A with a surface charge density σ. Here surface charge density can be written as
σ=
dq
The unit of σ is C/m2 in SI units.
dA .
The total charge on the entire surface is Q = σ dA
S
(iii)
Linear Charge Distribution and Linear Charge Density: If the charge is distributed over a line of length l, then the linear
dq
charge density λ is λ =
. The unit of λ is C/m in SI units. The total charge is now an integral over the entire length
dl
Q=
∫
λ dl
line
If charges are uniformly distributed throughout the region, the densities can be written as: ρ =
Electric field on the axis of a rod:
A non-conducting rod of length l with a uniform positive charge
density λ and a total charge Q is shown in figure. We will
calculate the electric field at a point P located along the axis of
the rod at a distance r from one end.
r
dE
Q
Q
Q
,σ = , λ = .
V
A
l
l
P
Q
.
dx
x
l
The amount of charge contained in a small segment of length dx
is dq=λdx. Since the source carries a positive charge Q, the field at P points is in the negative x direction. The electric field
due to dq is
The linear charge density is uniform and is given by λ =
dE =
1
dq
4πε 0 r
2
=
1
λ dx
4πε 0 x
2
=
1
Qdx
4πε 0 lx 2
.
The resultant field at P is
E = ∫ dE =
Q r +l dx
4πε 0 l ∫r x 2
1
⇒ E=
1 
Q 1
−

4πε 0 l  r r + l 
1
⇒ E=
1
Q
4πε 0 r (l + r )
When P is very far away from the rod, i.e., when r >> l the above expression reduces to E =
1 Q
4πε 0 r 2
At far distance away, the continuous charge distribution behaves similar to a point charge.
Example 31:
Solution:
Calculate the electric field on the axis of a very long uniformly charged, thin rod at a distance r from one end. The
charge per unit length of the rod is λ.
Field at point P due to any section of the rod will be directed leftward.
Consider a very small section of length dx at
P
r
a distance x from P. The charge on this
E
section is dq such that dq = λdx. The electric
λ
x
dx
field at P due to this section is
dE =
1 dq
(leftward)
4πε 0 x 2
Net field at P is E = ∫
∞
r
1
dq
4πε 0 x
2
=
λ ∞ dx
λ 1 1 
λ
2k λ
=
− =
=
∫

2
r
4πε 0
4πε 0  r ∞  4πε 0 r
r
x
Electric Charges and Field
Electric Field on the axis of a Ring
A ring of radius carries a uniformly distributed positive total charge Q. We shall calculate the electric field due to the ring at a
point P lying at a distance x from its centre along the central axis perpendicular
to the plane of the ring.
The magnitude of the electric field at P due to the segment of charge dq is
dq
dE = k 2
r
Figure (a): The field at P on the x-axis due to an element of charge dq.
Figure (b): The total electric field at P is along the x-axis. The perpendicular component
of the field at P due to segment 1 is cancelled by the perpendicular component
due to segment 2.
This field has an x component dEx = dE cosθ along x axis and a component
dE⊥ perpendicular to the x-axis. As we see in figure, however, the resultant
field at P must lie along the x-axis because the perpendicular components of
all the various charge segments sum to zero. That is, the perpendicular
component of the filed created by any charge element is cancelled by the
perpendicular component created by an element on the opposite side of the
ring.
r = ( x 2 + a 2 )1/2 and cosθ = x/r, we find that
Now,
dE x = dE cos θ = k
ke x
dq x
=
dq
r 2 r ( x 2 + a 2 )3/2
All the segments of the ring make the same contribution to the field at P because they are all equidistant from this point.
Thus, we can integrate to obtain the total field at P:
Ex = ∫
⇒
Ex =
kx
2
2 3/ 2
(x + a )
dq =
kx
2
( x + a 2 )3/2
∫ dq
kx
Q
( x + a 2 )3/2
2
This result show that the field is zero at x = 0.
Suppose a negative charge is placed at the center of the ring in above situation and displaced slightly by a distance x<<a
along the x-axis. When released, what type of motion does it exhibit? In the expression for the field due to a ring of charge,
we let x<<a, which results in
kQ
Ex = 3 x
a
 kQq 
Thus, the force on a charge –q placed near the center of the ring is Fx = −  3  x
 a 
Because this force has the form of Hooke’s law, the motion will be simple harmonic.
The expression for this field is similar to that obtained for a pair of identical charges. The field will attain a maximum value
a
at x =
. The graph is show below.
2
E
x= −
a
2
x=
a
2
Electric Charges and Field
Electric Field at the Centre of an Arc
Consider a small element of charge dq as shown. The electric field vector due to this charge
element is
dE = −dE sin ϕ iˆ − dE cos ϕ ˆj Where dE =
Net electric field at O is E = ∫ dE
⇒
E=
dq
4πε 0 R
2
λ dϕ
4πε 0 R
=
+
++
+ ++
 θ /2 sin ϕ dϕ iˆ + θ /2 cos ϕ dϕ
∫−θ /2
4πε 0 R  ∫−θ /2
+ ++ + ++
ϕ
++
++
dϕ
θ/2
E = − ∫ dE sin ϕ iˆ − ∫ dE cos ϕ ˆj
⇒
−λ
[∵ dq = λ dl , dl = Rd ϕ ]
y - axis
x - axis
ˆj  = − λ sin  θ  ˆj
 2 
 2πε R
0
dE ϕ
Check yourself
1.
What is E when (i) θ = 180° (ii) θ = 360°?
Electric Field at a General Point due to a Rod
Consider a uniformly charged rod oriented along y-axis. The charge per unit
length of the rod is λ. We shall evaluate the electric field at a point P such that
the ends A and B of the thread subtend angles α and β at point P.
We have to use the principle of superposition. The electric
field at P is the
resultant of electric field due to different
sections of the thread. Consider
one such section of length dy of the thread at a distance y from origin.
Due to the section dy of charge dq, field at P is shown below.
y – axis
dy
y
ˆ
ˆ
dE = dE cosθ (iˆ) − dE sin θ ( ˆj ) = dq cos2θ (i )2 − dq sin2θ ( j )2
4πε 0 ( y + a ) 4πε 0 ( y + a )
λ dy cos θ
.
4πε 0 ( y 2 + a 2 )
⇒ Ex = ∫ dEx = ∫
α
−β
Ex =
α
β
dE cos θ
\θ
dE
dE sin θ
θ
x – axis
⇒
α
λ
λ
cos θ dθ ⇒ Ex =
[sin α + sin β ]
∫
β
−
4πε 0 a
4πε 0 a
Similarly, dE y =
=
As y = a tan θ, dy = a sec 2 θ dθ
λ a sec2 θ dθ cos θ
4πε 0 (a 2 sec2 θ )
y2 + a2
+
+
+
+
+
+
+
ˆ
ˆ
= λ dy cos θ (i ) − λ dy sin θ ( j )
2
2
2
4πε 0 ( y + a ) 4πε 0 ( y + a 2 )
dE x =
+
+
+
+
+
+
− λ dy sin θ
4πε 0 ( x + a )
λ
[cos α − cos β ]
4πε 0 a
2
2
⇒
⇒
E=
E y = ∫ dE y = − ∫
α
− λ a sec2 θ dθ sin θ
−β
4πε 0 (a sec θ )
2
2
⇒
Ey = −
λ
λ
{(sin α + sin β )(iˆ) + (cos α − cos β )( ˆj )}
4πε 0 a
Case 1:
Electric field at any point P lying on the perpendicular bisector of a finite uniformly charged thread. Her α = β
∴ Ex =
λ
λ sin α
sin α and E y = 0 ⇒ E =
, along the bisector
2πε 0 a
2πε 0 a
Case 2:
Electric field at any point P due to a very long uniformly charged thread. Here, α = β →
∴ Ex =
λ
2πε 0 a
and E y = 0
⇒ E=
λ
2πε0 a
along the bisector
π
2
α
4πε 0 a ∫− β
sin θ dθ
Electric Charges and Field
Case-3:
Electric field at any point P due to semi-infinite uniformly charged thread. Here, α =
λ
∴ Ex =
4πε 0 a
and E y = −
λ
⇒ E=−
4πε 0 a
λ
4πε 0 a
π
2
,β = 0
(iˆ + ˆj )
Electric field due to a uniformly charge disc
A uniformly charge disc of radius R with a total charge Q lies in the XY-place. We
shall find the electric field at a point P, along the Z-axis that passes through the centre
of the disc perpendicular to its plane.
By treating the disc as a set of concentric uniformly charged rings, the problem could
be solved by using the result obtained in example for ring. Consider a ring of radius x
and thickness dx, as shown in figure.
By symmetry arguments, the electric field at P points in the + z direction. Since the
ring has a charge dq = σ (2π x dx), we see that the ring gives a contribution.
dE z =
dE
P
z
1
zdq
1 z (2πσ xdx )
=
4πε 0 ( x 2 + z 2 )3/2 4πε 0 ( x 2 + z 2 )3/2
R
x
Integrating from x = 0 to x = R. The total electric field at P becomes
Ez = ∫ dEz =
=−
1.
2.
σz
2ε 0
R
∫0
xdx
( x 2 + z 2 )3/2
=
σz
4ε 0
R2 + z 2
∫z
2
du
u 3/2

σz 
1
1 σ 
z
− =

1 −

2ε 0  R 2 + z 2 z  2ε 0 
R 2 + z 2 
=
σ z u −1/2
4ε 0 (−1/ 2)
⇒ Ez =
R2 + z 2
dx

σ 
z
1 −

2ε 0 
R 2 + z 2 
EXERCISE-04
A wire is bent in the form of a regular pentagon of edge length a. A total of charge q is distributed over it. Find the electric
field at the centre of the pentagon.
A circular ring carries a charge Q, the variation of electric field with distance x measured from centre along axis for x < < R
can be given as [R → Radius of Ring]
1
1
(2) E ∝ x
(3) E ∝ 3
2
x
x
In the above situation, the maximum electric field on the axis will be
(1) E ∝
3.
(1)
4.
Q
4πε 0 R
2
×
2
3 3
(2)
Q
12 3πε 0 R
2
(3)
Q
4 3πε 0 R 2
(4) None of these
(4) None of these
The magnitude of linear charge density on the semi-circular ring, on both quadrants is same, the electric field at O is along
(1)
5.
Uniformly Charged Disc
z2
iˆ
(2)
− iˆ
(3)
ĵ
(4)
− ĵ
An infinite long thread carries a charge uniformly spread over its length. It is bent in the form as shown. If the charge per unit
length is λ, the electric field at point O is
(1)
λ
2πε 0 r
(2)
λ
2πε 0 R
(3)
λ
2 2πε 0 R
(4) Zero
Electric Charges and Field
6.
The electric field due to a uniformly charged one quarter (λ is the charge per unit length) of a circular arc with charge per unit
length ‘λ’ at centre O is
kλ
R
(1)
7.
2
(2)
kλ
R
kλ
R2
(3)
2
(4)
kλ
R2
If λ = 1 µ C/m, then electric field intensity at O is
O
(1)
9N/C
(2)
900 N/C
(3)
2m
9000 N/C
9×109 N/C
(4)
8.
Select the situation in which, the direction of electric field at point P is different from remaining three (Here, ‘λ’ and ‘− λ’
represent constant linear charge densities)
9.
A point charge q is kept at the centre of a fixed semicircular thread having a uniform charge per unit length. The radius of the
semicircular thread is R. The force required to keep the charge stationary is
(1)
10.
λq
2πε 0 r
(2)
+ + + + ++
++
++
+
++
+
+
+
O
+
++
+
+
+
++
++
+ + + ++
(2)
Uniform charge/length
(4) Zero
+ +++
++ λ = λ0 cos θ
++
+
+
+
+
+
+
+
+
O
+
+
+
+
+
+
+
++
+
+ + + ++
(3)
+ + + + ++ λ = λ0 sin θ
++
++
+
++
+
+
+
O
+
++
+
+
+
++
++
+ + + ++
(4)
--
Uniform charge/length
-
Variable charge/length
Variable charge/length
- +++
++
+
+
+
+
O
+
+
+
++
+ + -- ---
(2)
a
2
(3)
√2
(4)
a
2
Two fixed charges +4Q and −Q are located at A and B. Select the correct statements.
(1)
(2)
(3)
(4)
13.
λq
8πε 0 r
An infinitely large non-conducting plane sheet of charge density σ has a circular aperture of certain radius carved out of it.
σ
The electric field on the axis of the aperture, at a point which is at a distance ‘a’ from the centre of aperture is
. The
2 2ε 0
radius of aperture is
(1) a
12.
(3)
In which of the following figures, electric field at point O is non-zero?
(1)
11.
λq
4πε 0 r
A point (P) where a third charged particle will not experience a force lies outside AB
A point (P) where a third charged particle will not experience a force lies inside AB
A positive charge kept at the neutral point (P), can oscillate along horizontal.
If a negative charge kept at the neutral point (P), it can oscillate along horizontal.
Consider an annular disc of inner radius r and outer radius 2r, having uniform surface charge density σ. The electric field on
the axis of this disc at a distance r from the centre will be
Electric Charges and Field
(1)
σ
2 10ε 0
(2)
σ
2 10ε 0
( 5 − 2)
(3)
σ
2 10ε 0
( 3 − 2)
(4)
σ
1
1−
2ε 0
2
Electric Lines Of Force or Electric Field Lines
The electric field lines were invented by Faraday to visualize electric field. The electric lines of force are purely a geometrical
construction, which help us to visualize of the electric field in the region. They have the following characteristics:
(1)
The tangent to electric lines of force at any point gives the direction of electric field at that point.
(2)
In free space, they are continuous curves which emerge from positive charge and terminate at negative charge
(3)
They do not intersect each
ch other. If they do so, then it would mean two directions of electric field at the point of intersection,
which is not possible.
(4)
The density of field lines represents strength of electric field in the region.
Field Lines in Some Cases
(1) Positive point charge
(2) Negative point charge
(Field lines have spherical symmetry)
(3) Two similar charges of equal magnitude
(5) Two dissimilar charges of equal magnitudes
(4) Two similar charges of unequal magnitudes
(6) Uniform electric field
Check yourself
Rank the magnitudes of the electric field at points A, B and C in figure with the largest magnitude first.
Ans. EA > EB > EC
Electric Charges and Field
ELECTRIC FLUX
The electric flux is proportional to the number of field lines that pass through some area A.
Consider an electric field that is uniform in both magnitude and direction, as in figure. Let
the electric field lines penetrate a surface of area A, which is perpendicular to the field. The
number of field lines is proportional to the product of E and A. This is measure of electric
flux and represented by the symbol φ. In the above case φ = EA. If the surface under
consideration is not perpendicular to the field, as in following figure, the expression for the
electric flux is φ = EA cosθ.
A
θ
A
E
E
ϕ = EA.cos θ
ϕ = E. A = EA.cos 0 = EA
ϕ = EA cos90 = 0
It is to be noted that while calculating flux through an open surface A can be taken in any direction normal to the surface. But
for a closed surface (a surface that encloses some volume, like a cuboidal surface or a spherical surface), outward normal is
the positive direction of area vector (dA or A) . Thus while calculating flux through a closed surface, outward flux is taken as
positive and inwards flux is taken as negative.
General formula for flux through a surface:
Flux of electric field through an infinitesimally small surface is given by the relation
dϕ = E.dA = EdA cos θ
Total flux through a given surface area is
ϕ = ∫ E.dA = ∫ EdA cosθ
If E is uniform, ϕ = ∫ E.d A = EA cos θ
Units: SI unit of flux is N-m2/C or V-m (Volt-meter) Dimensional Formula: [φ] – ML3T-3A-1
Example 32:
Evaluate the flux of electric field through the surfaces bounded by the following curves, due to the field
(8iˆ + 9 ˆj + 5kˆ ) N / C
(i) x 2 + y 2 = 15
Solution:
(i)
x 2 + y 2 = 15 is a circle of radius √15 units lying in x-y plane.
Solution:
⇒
φ = E . A = (8iˆ + 9 ˆj + 5kˆ).(15π ) kˆ = 120π
A = π r 2 ˆj = 30π ˆj
⇒
φ = E. A = 200π
x 2 + z 2 = 70 is a circle of radius 12 units in z-x plane.
∴
Example 33:
A = π r 2 kˆ = 15π kˆ
y 2 + z 2 = 30 is a circle of radius √30 units in y-z plane.
∴
(iii)
(iii) x 2 + z 2 = 70
This is a case of uniform electric field. So you can straightway apply ϕ = E. A
∴
(ii)
(ii) y 2 + z 2 = 30
A = π r 2 ˆj = 70π ˆj
⇒
φ = E. A = 630π
A uniform electric field E = (3iˆ + 4 ˆj + 5kˆ ) N / c exists in a region. A cube of side
2m is placed with one vertex at origin and 3 edges coinciding with coordinates
axis. Calculate the flux of electric field through each face of the cube and total
flux through the cube.
Area of each face = 4m2
φABCD = 16 Nm2/C (outward),
φEFGO = −16 Nm2/C
(inward)
φABFE = 20 Nm2/C (outward),
φDCGO = −20 Nm2/C
(inward)
y - axis
4 N/C
D
C
A
B
3 N/C
G
O
E
z - axis
5 N/C
x - axis
F
Electric Charges and Field
φBCGF = 12 Nm /C (outward),
φADOE = −12 Nm /C
The total flux is the scalar addition of all the above values.
2
2
(inward)
φTotal = Σφ = 0
This is a point to remember that flux of electric field through any closed surface is zero in a uniform electric field.
Example 34:
Three objects are placed in a region of uniform electric field of magnitude E. The dimensions of the objects are
given in the figures. In each case, find the inward and outward flux through the object.
r
H
R
R
h
Solution:
(i)
ϕ base = Eπ R 2
[outward flux]
As ϕ curved + ϕ base = 0
[ ∵ϕtotal = 0 in uniform electric field]
⇒ ϕ curved = − Eπ R 2
2r
H
[inward flux]
As seen in case (i), in uniform electric field, φ depends on the area of crosssection which is normal to the electric field.
(ii)
ϕ inward = − E × ( H × 2r )
(see figure)
ϕ outward = + E × H × 2r
ϕ base = E × π R 2
(iii)
(outward)
ϕ curved = − E × π R 2
GAUSS’S LAW
“The flux of electric field through any closed surface is equal to the net charge enclosed by
the surface, Qinside divided by ε0”.
∫ E.dA = ϕ E =
E
ε0
Consider a point charge q surrounded by a spherical surface of radius r centered on the
charge, as in figure. The magnitude of the electric field everywhere on the surface on the
q
sphere is E = k 2 .
r
The electric field is perpendicular to the surface at all points on the surface. The flux of
this field through the surface is given by
ϕE =
q
∫ E.dA = ∫ EdA = EA = k r 2 (4π r
As k =
1
4πε 0
dA
Qinside
2
) = 4π kq
q
Gaussian Surface
(∵ E || dA at all points)
, the electric flux can also be expressed as ϕ =
q
ε0
This result says that the electric flux through a sphere that surrounds a charge q is equal to the charge divided by the constant
ε0. However, even if the surface surrounding q is irregular, the flux through that surface is also q/ε0.
A charge is placed at the centre of spherical Gaussian surface. The flux through the sphere will not change when
(a)
The sphere is replaced by a cube of the same volume.
(b)
The sphere is replaced by a cube of one tenth the volume.
Electric Charges and Field
(c)
(d)
The charge is moved off-centre in the original sphere, still remaining inside.
A second charge is placed near and outside, the original sphere.
Check yourself:
Find the electric flux through the surface in following figure.
-2 C
-
-3 C
+2 C
+
+4 C
+
-5C -
+1 C
+
Ans:
−5.310 × 10−11V − m
Example 35: Consider a point charge q placed at a corner of a cube. Determine the electric flux through bottom face of the cube.
Solution:
(a) A charge q is placed at the corner
of a cube
(b) The charge is surrounded by seven cubes such that
the charge is at the centre of a large cube.
To find the flux through each of the faces of our cube, we can use a technique that puts the single charge in the middle of a
large cube. It takes seven other similarly placed cubes to surrounds the points charge q completely in fig. (b). The charge is at
the dead centre of the new larger cube. So the flux through each of the six faces of the large cube will now have an electric
flux of one sixth of the total flux. The large face of the cube consist of four smaller square, one of which is in fact one of the
face of our given. So given that the total structure is completely symmetric, the flux through a face is one fourth of the flux
through the large side.
The total flux is
q
ε0
. So that the flux through each of the faces of the large cube is
through each of the far faces of the small cube. ∴ ϕ =
q
q
and one quarter of that
goes
6ε 0
24ε 0
q
24ε 0
Check yourself
Determine the total flux through cube?
Ans: q
8ε 0
1.
EXERCISE-05
A hemisphere surface of radius R is located in a uniform electric field E, as shown. The flux of electric field through the
hemispherical surface is
(1)
2.
Eπ R 2
(2)
E 2π R 2
(3)
E 4π R 2
Figure shows a cubical surface of face area A in the space in which E = E ( kˆ ).
(4)
Zero
Electric Charges and Field
Find the flux through
a.
The front face
b.
Rear Face
c.
Top Face
d.
The whole cube
3.
Which of the four particles contribute to the electric field at point P on the surface?
4.
(1)
q1
(2)
q2
(3)
q3
(4)
q4
Which of the four particles contribute to the net electric flux through the closed surface?
5.
(1)
q1
(2)
q2
Write dimensional formula of electric flux.
6.
9.
10.
11.
12.
13.
14.
15.
16.
E.π R 2
(2)
E.RH
(3)
E.2 RH
(4)
q4
E as shown in figure. The magnitude of flux
(4)
Zero
The electric field in a region of space is spherically symmetric and radially outward. The flux of electric field through a
sphere of radius r, centered at origin is φ = kr4. Which of the following is the correct relation between electric field strength
with radial distance r from origin?
1
(4)
E α r3
2
r
A charge q is placed just above the centre of a horizontal circle of radius r, and a hemisphere of this radius is erected about
the charge. Compute the electric flux through the closed surface that consists of the hemisphere and the planar circle.
(1)
8.
q3
A cylinder of height ‘H’ and radius of cross-section ‘R’ is kept in uniform field
that enters from left half of the cylinder is
(1)
7.
(3)
Eα r2
(2)
Eα r
(3)
Eα
The electric field due to an infinitely long, straight line of charge with uniform charge density λ points straight away from the
line and has magnitude E = λ/(2πε0r), where r is the distance from the wire. Calculate the flux of this electric field through a
right cylinder of height h and radius R, co-axial with the charged line. Repeat the calculation for a cylinder of radius 2R.
Consider a surface enclosing no electric flux. Discuss whether the electric field is also zero everywhere on the surface.
Use Gauss' law to show that electric field lines must be continuous in space and must originate from and terminate on charge.
A charge q is placed at the centre of a cube of side l. Find electric flux through each of six faces.
A charge q is placed at the corner of a regular cube of side l. Find the electric flux passing through the cube.
A charge q is placed at the corner of a regular cube of side l. Find the electric flux passing through each of the adjacent face.
The term q on the right side of Gauss law includes the sum of all type of charges enclosed by the surface, can the charges be
anywhere inside the surface?
A charge ‘q’ is placed at a point exactly above the centre of square (of side a) at a distance
through square?
a
. What is the flux passing
2
Electric Charges and Field
17.
18.
19.
20.
An electric field has the components Ex = 5x, Ey = −3y, and Ez = 4z. Calculate the electric flux through the sides of a Unit
cube, whose corners are (x, y, z) = (0, 0, 0), (1, 0, 0), (1, 1, 0), (0, 1, 0), (0, 0, 1), (1, 0, 1), (1, 1, 1) and (0, 1, 1). (All fields are
measured in N/C, all distances in m).
The electric field in a certain region of space points in the z-direction and has magnitude E = 5xz, where x and z are
measured from origin. Calculate the flux of that field through a square perpendicular to the z-axis, the corners of the square
are at (x, y, z) = (−1, −1,1), (−1, 2, 1), (2, 2,1) and (2,−1,1). (All fields are measured in N/C, all distance in m).
A test charge q0 is placed in the electric field of a source charge Q. The test charge will not following path of electric field
line if
(1)
It has zero velocity when released
(2)
It has non-zero velocity along outward radius, when released
(3)
It has non-zero velocity along inward radius when released
(4)
It has a velocity different from the above three
Which one of the following is correct regarding order of electric field intensity?
(1)
(3)
EA> EB < EC
EA= EB < EC
(2)
(4)
EA> EB < EC
EA> EB > EC
Application of Gauss Law to Calculate Electric Field
Gauss Law gives an alternate way to evaluate electric field. It can be applied effectively to evaluate electric field only to
systems that possess certain symmetry, namely, systems with cylindrical, planar and spherical symmetry. We shall learn to
use Gauss law by carefully examining the following cases:
Electric Field due to a Point Charge
To evaluate electric field at a point at a distance or using Gauss Law, we need to pass a closed Gaussian surface through the
point, enclosing the charge. Clearly the field is spherically symmetric. If we enclose the charge in a sphere of radius R, the
magnitude of electric field will be same at any point on the surface.
Now, φ =
∫ E.dA = ∫ EdA cosθ .
Here,
θ =0
for all points on the
dA
sphere.
⇒ ϕ ∫ EdA = E ∫ dA = E × 4π R 2
E
∴ ϕ = E × 4π R 2
By Gauss Law, ϕ =
⇒ E × 4π R 2 =
⇒E=
q
4πε 0 R 2
q
ε0
.
qenc
q
ε0
Gaussian Surface
[∵ qenc = q ]
This is the same result as given by Coulomb’s
law.
dE1
Electric Field due to a uniformly charged, long and thin rod
First, we shall look for the symmetry of electric field. Consider any point P, at
a distance r from the line charge. Let dE1 and dE2 be the electric field at the
dE2
point P due to elements shown. The resultant of dE1 and dE2 is along y-axis.
dx
dx
(i.e., in a direction normal to the line charge)
+++++++++++++++++++++++++++
x
If a long linear charge distribution is kept along x-axis, at any point, field is
+++++++|++ x
directed radially away from x-axis. The field has a cylindrical symmetry.
To find electric field, we enclose the distribution in a Gaussian cylinder of radius r and length l.
Electric Charges and Field
The flux linked with the cylinder
∫ E.dA = ∫
E.dA +
Curved
∫
Flat
E
E.dA = E.(2π rl ) cos 0 + 2 ∫ EdA cos 90
r
dA
++++++++++++++++++++++++++++++
+++++++++++++++
⇒ ϕ total = E × 2π rl
By Gauss law ϕ =
⇒ E × 2π rl =
∴E =
qenc
ε0
=
λ ×l
ε0
[where λ is linear charge density]
dA
Gaussian Surface
E
λ ×l
ε0
λ
2πε 0 r
Variation of E with r is also shown graphically here.
Electric Field due to a Hollow Cylinder of Charge
Consider a hollow uniformly charged cylinder with charge per unit length as
‘λ’.
+++++++++++++++
+++++++++++++++
We shall determine the field due to this shell in the region outside the shell
++
+++++++++++++++
as well as in the region inside the shell.
++
++
(a)
Inside the cylinder (radial distance < R):
When we draw a Gaussian cylinder of radius r, we find that the charge enclosed by it is zero.
⇒ ϕ total = 0
⇒
∫ E.dA = 0
This is possible if E = 0
(b)
Outside the cylinder (radial distance > R):
Consider the given shell being enclosed by a Gaussian cylindrical
surface.
Now,
∫ E.dA =
qenc
ε0
λ ×l
⇒ E × 2π rl =
ε0
⇒E=
Gaussian Surface
(∵ qenc = λ l )
λ
2πε 0 r
E
dA
++++++++++++++++++++
++++++++++++++++++++
++++++++++++++++++++
r
The Variation of E with r is also shown graphically
here.
Field due to a Long Uniformly Charged Solid Cylinder
Let ρ be the volumetric charge density and R be the radius of the cylinder. Here, the field is radially away from x-axis having
a cylindrical symmetry.
Case 1: r > R.
Consider a Gaussian cylinder of length l and radius r placed along x-axis. Electric field lines come out radially. Let E be the
magnitude of electric field, then φE = E × 2πr l
qenc = ρ × π R 2 × l. Now by Gauss law,
Electric Charges and Field
E
dA
E
dA
++++++++++++++
Infinte Cylinderical
charge
λ
ϕ=
qenc
ε0
⇒ E × 2π rl =
E=
= charge per unit length
ρ × π R2 × l
.
ε0
⇒E=
ρ × R2
2ε 0 r
λ
2π ∈0 r
1
∴ E ∝ ( r > R)
r
Case 2: r < R.
The electric field inside an infinite cylinder of uniform charge is radially outward (by symmetry), but a cylindridal Gaussian
surface would enclose less than the total charge Q. The charge inside a cylinder of radius r is given by qenc = ρ × π r 2 × l
The electric flux is given by
By Gauss Law
ϕ=
qenc
ϕ = E × 2π r l.
⇒ E × 2π rl ⇒
ε0
ρ × π R2 × l
ρr
⇒E=
ε0
2ε 0
∴ E ∝ r ( for r < R)
The graphical variation of E with r is also shown
Electric Field due to a uniformly charged Shell (Hollow Sphere)
Consider a shell having a charge Q uniformly distributed on its surface. The surface charge density is
Q
, where R is radius of shell.
4π R 2
Case 1: Field outside the shell (radial distance > R)
We enclose the shell in a Gaussian sphere
σ.
⇒σ =
ϕtotal =
∫ E.dA = E × 4π r
E
Q
2
r
R
By Gauss’ law
ϕ total =
qenc
ε0
=
Q
⇒ E × 4π r 2 =
ε0
Q
Or E =
ε0
Q
4πε 0 r 2
[The result is same as that of a point charge]
Case 2: Field inside the shell (radial distance < R)
As charge enclosed by the Gaussian sphere = 0
⇒ϕ = 0
⇒
∫ E.dA = 0
⇒E=0
The graphical variation is shown.
Electric Field due to Uniform Sphere of Charge
Consider a uniform spherical charge distribution in which a charge Q is uniformly distributed over
the volume of a sphere of radius R.
The volumetric charge density is given by, ρ =
Q
4
π R3
3
=
uniform)
Case 1: Field outside the sphere (radial distance > R).
The field has spherical symmetry
Q
3Q
4π R
E
3
(as the distribution of charge is
r
R
Electric Charges and Field
⇒
∫ E.dA = E × 4π r
⇒ E × 4π r 2 =
Q
ε0
By Gauss’s law, ϕ =
2
or E =
Q
ε0
Q
4πε 0 r 2
[Note that the result is same as that of a point charge]
Case 2: Field inside the sphere (radial distance < R).
Consider a Gaussian sphere inside the sphere of charge.
Q = total charge of sphere
Q′ = charge enclosed by the Gaussian sphere
4
Q
Q′ = ρ × π r3 = 3 r3
3
R
⇒E=
Q
4πε 0 r
2
=
Now, ϕ =
∫ E.dA = E × 4π r
2
=
Q′
ε0
ρr
Qr
=
3ε 0 4πε 0 R 3
The graphical variation is also shown.
Electric Field due to an infinite, thin non-conducting sheet
Consider an infinitely large non-conducting plane in the xy-plane with uniform surface charge density
determine the electric field everywhere in space.
We will use the following steps:
(1)
An infinitely large plate possesses a planar symmetry
(2)
Since the charge is uniformly distributed on the surface, the electric field must
point perpendicular away from the plane E = Ekˆ . The magnitude of the electric
σ . We shall
field is constant on planes parallel to the non-conducting sheet.
We choose our Gaussian surface to be a cylinder, which is often referred to as a
“pillbox”. The pillbox consists of three parts, two end-caps s1 and s2 and a
curved side s3.
(3)
(4)
Since the surface charge distribution is uniform the charge enclosed by the Gaussian “pillbox” is qenc = σ A. where A
= A1 = A2 is the area of the end-caps.
The total flux through the Gaussian pillbox is
∫ E.dA + ∫ E.dA + ∫ E.dA
s
s2
s3
= E1 A1 + E2 A2 + 0
= ( E1 + E2 ) A
Since the two ends are at the same distance from the plane, by symmetry, the magnitude of the electric field must
be the same: E1 = E2 = E. Hence, the total flux can be rewritten as ϕ E = 2 EA.
(5)
By applying Gauss’s law, we obtain
2 EA =
qenc
ε0
=
σA
ε0
E=
⇒
σ
2ε 0
Two parallel sheets are given surface charge densities σ 1 and σ 2 . Electric fields in different regions are as shown
E=
σ1 + σ 2
ε0
σ1
σ2
σ −σ
E= 1 2
ε0
E=
σ1 +σ2
ε0
Electric Charges and Field
Consider the following specific cases:
(i)
When σ 1 = σ ,σ 2 = σ , the situation will be like the one shown below:
E=
(ii)
σ
ε0
σ
σ
E=
σ
ε0
When σ 1 = σ ,σ 2 = −σ , the situation will be like the one shown below:
σ
−σ
σ
E=
ε0
EXERCISE - 06
1
1
rather than as 2 ?
r
r
1.
Would the Gauss law fail if the field of a point charge were to decrease as
2.
3.
4.
(i) Can Gaussian surface pass through a discrete charge
(ii) Can Gaussian surface pass through a continuous charge distribution? Give reason in support of your answer.
If the net outward flux through the surface of a black box were zero, will there be no charge inside the box? Why or why not?
What would Gauss' law look like for the gravitational field, which is defined by force/unit mass of a test body?
5.
In Gauss theorem
q
∫ E.dS = ε 0 .The surface integral is evaluated by choosing a closed surface, called the Gaussian surface.
Here the incorrect statement is
(1)
The closed surface can have any shape or size
(2)
'q' is the net charge enclosed by the Gaussian surface
(3)
(4)
6.
E must be the electric field due to all the charges inside the surface only
The exact location of the charges inside the surface does not affect the value of the integral
Consider Gauss law
q
∫ E.dS = ε 0
. Which of the following is not true?
(1)
(2)
(3)
E must be the electric field due to enclosed charge only
If net charge inside the Gaussian surface = 0, E must be zero everywhere over the Gaussian surface
If the only charge inside the Gaussian surface is an electric dipole, then the integral is zero
(4)
E is parallel to dA everywhere over the Gaussian surface
7.
Consider an electric field E that is zero at every point on a closed surface S. Does this mean that there are no charges within
this surface? Give an example for which there are charges inside a surface, while E = 0 on the surface.
8.
Two point charges (Q each) are placed at (0, y) and (0, −y). A point charge q of same polarity is constrained to move along xaxis. Select the correct alternatives
(1)
9.
The force on q is maximum at x = ±
y
2
(2)
The charge q is in equilibrium at origin
(3)
The charge q performs an oscillatory motion about origin
(4)
For any position of q other than origin, the force is directed away from origin
For a uniformly charged spherical shell of radius R1, the electric field, E, at a distance r from the centre is given as
Electric Charges and Field
(1)
10.
11.
(1)
E ∝ r , r < R,; E ∝
(3)
E∝
E = 0, r < R,; E ∝
1
r2
1
r
2
,r > R
1
r2
(2)
E = 0, r < R,; E ∝
(4)
E ∝r
1
r2
ρR
3ε 0 r 2
(2)
ρr
3ε 0
(3)
ρ R3
3ε 0 r 2
,r > R
,r > R
(4)
Charge is spread uniformly in whole x-z place with surface charge density
σ ˆ
j
2ε 0
(2)
−
σ ˆ
j
2ε 0
(3)
σ ˆ
i
2ε 0
ρ r2
3ε0 R
σ . The electric field at (0,2,0) is
(4)
−
σ ˆ
k
2ε 0
Two thin sheets of charge carrying surface charge density σ1 and σ 2 are placed parallel to each other. The strength of electric
field at a point in the space between the sheets has a magnitude.
σ1 + σ 2
2ε 0
(2)
σ1 − σ 2
2ε 0
(3)
σ1 + σ 2
ε0
(4)
σ1 − σ 2
ε0
Two thin sheets of charge carrying surface charge density σ1 and σ 2 are placed parallel to each other. The strength of
electric field at a point in the space outside the sheet has a magnitude.
(1)
15.
(2)
For a uniformly charged sphere with charge density ρ and radius R, the field inside the sphere, at a distance r from the centre
is
(1)
14.
,r > R
(4)
E ∝r
r2
For a uniformly charged sphere, the electric field E at a distance r from the centre is
E∝
(1)
13.
1
r2
1
(3)
(1)
12.
E ∝ r , r < R,; E ∝
σ1 + σ 2
2ε 0
(2)
σ1 − σ 2
2ε 0
(3)
σ1 + σ 2
ε0
(4)
σ1 − σ 2
ε0
Two concentric spheres of radii a and b (a<b) carry charges q and −q respectively. The charge distribution is uniform. The
electric field at a distance r from the common centre is
(1)
Problem 1:
(a)
0 for r < a
(2)
0 for r > b
(3)
q
4πε 0 r 2
for a < r < b
(4) Given by all the above choices
Miscellaneous Problems
A small plastic ball of mass m covered with a thin zinc coating, is suspended by means of an insulating thread from
a fixed point. There exists an electric field of magnitude E directed along horizontal. Now ultraviolet light is made to
incident on the ball due to which it acquires a positive charge. Acceleration due to gravity is g.
What is the origin of positive charge acquired by the ball?
(1) Excess of protons in plastic ball
(2) Deficiency of electrons in zinc coating
(3) Deficiency of electrons in plastic ball
(4) Some positively charged particles given by ultraviolet light
(b)
If the string makes an angle θ with the vertical when the ball comes to equilibrium, the number of electrons lost by
the ball is.
(c)
mg
mg sin θ
mg tan θ
mg cot θ
(2)
(3)
(4)
eE
eE
eE
eE
If the charge that suddenly appears on the ball is q, what is the maximum angle by which the string gets deflected?
 mg 
 2mg 
qE
qE
(1)
tan −1 
(2) tan −1 
(3)
2 tan −1
(4) tan −1


mg
mg
 qE 
 qE 
(1)
Solution:
(a) Correct answer is (2). Ultraviolet light ejects electrons from zinc. This is known as photoelectric effect.
Let n be the number of electrons lost, so that the ball acquires a positive charge given by q = ne. So, ball experience
a force qE along horizontal direction.
Electric Charges and Field
The situation is shown in following figure.
For equilibrium:
T cos θ = mg
…(i)
T sin θ = qE
…(ii)
From (i) and (ii) we can infer that
qE
= tan θ
mg
or q =
mg tan θ
E
mg tan θ
mg tan θ
⇒ n=
eE
E
(b) As soon as a charge (q) appears on the ball it starts experiencing a horizontal force (qE) due to presence of
electric field. Due to the constraint of string, it will start moving on a circular path in a vertical plane. The state of
maximum deflection of the string will be characterized by zero velocity of the ball. (Recall that at maximum height
attained by a body projected up, velocity reduces to zero). The situati
situation
on is shown in the figure.
Between the two positions A and B, total change in KE is zero.
Now, q = ne. So, we have ne =
∴ Total work done by all the forces is zero. (Note that net force is not zero).
⇒ WTension +Wmg + WE=0
Now, WTension is zero, as T acts along radial direct
direction
ion while the ball moves along tangential direction.
Wmg = mg × (l – l cos θ) × cos 180 = − mg l (1 – cos θ)
l
WE = q E × l sin θ × cos 00
l cos θ
So, we have the condition 0 − m g l (1 − cos θ ) + q E l sinθ = 0

2  θ 
∵1 − cos θ = 2sin  2  
 

θ 
⇒ mgl × 2sin 2   = qEl sin θ
2

θ
θ 
θ 
2 sin   cos 
⇒ mg × 2 sin 2   = qE  2sin
2
2
2

θ 
θ 
⇒ mg sin   = qE cos  
2
2
Problem 2:
Solution:
v=0
B
l - l cos θ
θ
θ

∵ sin θ = 2sin 2 cos 2 


A v=0
 θ  qE
−1  qE 
⇒ θ = 2 tan 
⇒ tan   =

 2  mg
 mg 
Two positive charges q each are fixed at (−a,
a, 0) and (a,0) respectively. A particle free to move having a mass m and
charge q0 is kept at the origin so that it does not experience a force. Show that if q0 is positive, its equilibrium is
stable with respect to displacement along xx-axis and unstable for displacements with respects to y or along a-axis.
a
Further, for small displacements along xx-axis,
axis, show that motion would be simple harmonic and thus determine its
time period.
The situations are shown in figure a, b and c. FA represents the force on q0 due to charge at A.
FB represents the force on q0 due to charge at B.
q
FB
A(-a,0)
(0,0) F A
O(0,0)
FA = FB =
q
q
B(a,0)
A(-a,0)
kqq0
a2
. So, net force is zero.
FA q
FB
O(0,0)
a+x
B(a,0)
x
a-x
Now, q0 is displaced to the right i.e. along xx-axis
axis by a distance x. The forces change and are given as
FA =
kqq0
;
(a + x)2
FB =
kqq0
(a − x) 2
Clearly FA < FB, so q0 experiences a net force along negative xx-direction,
direction, which tries to restore the position of
equilibrium.
Same discussion holds good for displacement in negative xx-direction.
direction. Thus the equilibrium is stable with respect to
displacement along x-axis.
Electric Charges and Field
For displacement along y-axis,
axis, consider q0 to be displaced up along y-axis by a
distance y0 as shown.
FA
kqq0
. The resultant of FA & FB would be up, along y-axis. So it
(a + y 2 )
will cause q0 to move up and does not restore the equilibrium. Thus equilibrium is
unstable
stable to displacement along yy-axis. Same discussion holds good for z-axis also.
SHM: For the displacement in xx-direction, the net force will be
FA = FB =
2
F = FA − FB =
⇒ F=
FB
kqq0
(a + x)
2
−
kqq0
(a + x) 2
q
A(-a,0)
q
O(0,0)
B(a,0)
.
kqq0
 (a + x ) 2 − ( a − x ) 2 

(a 2 − x 2 ) 2 
⇒ F=
−4qq0 ax
(a 2 − x 2 ) 2
⇒ F=−
When x < < a, a 2 − x 2 ≈ a 2
4kqq0 x
a3
Thus, F ∝ ( − x ) . This is the case of simple harmonic motion.
Time period: When F = −kx
kx the time period is T= 2π
m
k
.
∴ T = 2π
ma3
4kqq0
Problem 3:
A thin wire is bent in the form of a complete ring of radius R and charge Q. It is placed in y-z
y plane with centre at
origin. A long thread with charge per unit length ''λ' is placed along + x-axis
axis with one end at origin. Calculate the
force of interaction between them.
Solution:
The electric field vectors due to the thread at points P and Q on the ring
are
EP =
λ
4πε 0 R
[ −iˆ + ˆj ]
EQ =
λ
4πε 0 R
[ −iˆ − ˆj ]
The force experienced
xperienced by small charge elements dq at P and Q is
d FP =
λ dq
[ −iˆ + ˆj ],
4πε 0 R
d FQ =
λ dq
[ −iˆ − ˆj ]
4πε 0 R
Clearly, the components of these forces add up along -x-axis
and cancel out along y-axis. Thus, total force experienced by the ring is along -x-axis.
λ dq
− λQ
− λ Qiˆ
=
or F =
⇒ Net force, Fx = ∫ dFx ⇒ Fx = − ∫
4πε 0 R 4πε 0 R
4πε 0 R
Problem 4:
Solution :
A point charge q is placed at the centre of a shell carrying a charge Q uniformly distributed over its surface.
Calculate electric field at a point (i) inside (ii) outside the shell at radial distance r from the centre. (iii) Plot the
variation of electric field.
The entire space can be divided into two regions.
Region (i) Electric field at point A ((inside)
There is no electric field due to charge on the shell
⇒E=
q
4πε 0 r 2
directed radially outward
Region (ii) Electric field at B (outside)
E=
Q
4πε 0 r
⇒E=
2
(due to shell) +
Q+q
4πε 0 r 2
q
4πε 0 r 2
(due to charge at centre)
directed radially outward
The graphical variation is given as shown
Electric Charges and Field
Problem 5:
Solution:
Two concentric shells of radii a and b carry charge q1 and q2 (uniformly distributed) respectively (a < b). Calculate
electric field at radial distance r from the centre for (i) r < a (ii) b > r > a (iii) r > b. Represent this field graphically.
(i) r < a.
This point lies inside inner shell. E = 0 due to both the shells. Thus, net field is zero.
(ii) b > r > a.
This point lies in between the shells i.e., outside inner shell, but inside outer shell ⇒ E =
(iii)
directed radially outward (due t o inner shell only)
This point lies outside outer shell.
r > b.
⇒E=
⇒E=
q
4πε 0 r
2
q1 + q2
4πε 0 r 2
(due to inner shell) +
q
4πε 0 r 2
q
4πε 0 r 2
, (due to outer shell)
; directed radially outward
Graph:
The field remains zero for 0 < r < a. It suddenly becomes non-zero as soon
1
as exceeds a. Then it decreases as E ∝ 2 .
r
It suddenly increases (assuming q1 and q2 to be of same sign) as soon as r
exceeds b.
Again, for r > b, E ∝
Problem 6 :
Solution:
1
r2
. Here the following graph is obtained.
Figure shows a long thread along the axis of a long hollow cylinder. The charge per unit length of thread is λ, while
that of the cylinder is −λ. The radius of cylinder is R. What is the electric field at radial distance r from the axis for
(i) r < R (ii) r > R?
(i) r < R. This point lies inside the cylinder. There is no field at this point due to cylinder
∴E =
λ
2πε 0 r
(due to thread only)
(ii) r > R. This point lies outside the cylinder
∴E =
λ
2πε 0 r
(due to thread) +
−λ
(due to cylinder) = 0
2πε 0 r
∴ The field is limited to the region inside the cylinder.
Problem 7:
Three large non-conducting plane sheets of charge with charge densities σ, 2σ and 3σ are arranged as shown.
Described the electric field produced by the arrangement, in different regions marked.
Solution:
Region I : Field is in left direction due to all the sheets. So we get
⇒
σ + 2σ + 3σ 3σ
(left)
=
2ε 0
ε0
Region II: Field due to sheet-1 is towards right, while due to other
towards left.
⇒
2σ + 3σ − σ
2σ
(left) ⇒
2ε 0
ε0
two is
Electric Charges and Field
Region III: Field due to sheet 1 and sheet-2 is towards right, while due to third sheet is towards left.
∴E =
σ + 2σ − 3σ
=0
2ε 0
Region IV: Field due to all sheets is rightward.
∴E =
3σ
ε0
(right)
Problem 8:
A ball of radius R carries a positive charge throughout its volume, such that the volume density of charge depends
r

on distance r from the ball’s centre as ρ = ρ 0 1 −  , where ρ0 is a constant. Assuming the permittivity of ball to
 R
be one, find magnitude of electric field as a function of distance r, both inside and outside the ball.
Solution:
The field has a spherical symmetry. For a point outside the ball (r > R)
ϕ=
∫ E.dA = E × 4π r
By Gauss law, ϕ =
2
Q
ε0
⇒ Eout =
Q
4πε 0 r 2
, where Q is total charge
For a point inside the ball (r < R)
By Gauss law, ϕ =
Ein
qenc
ε0
qenc
4πε 0 r 2
Where qenc = charge in a sphere of radius (r < R)
To find the enclosed charge, consider a spherical shell of radius r and thickness dr
⇒ Charge dq = ρdV
Its volume dV = 4πr2dr
r

⇒ dq = ρ 0 1 −  4π r 2 dr
 R
3
 r 2

 r3 r 4 
r r
r
r

dr  = ρ0 4π  −
⇒ q = ∫ ρ 0 1 −  4π r 2 dr = ρ0 4π  ∫ r dr − ∫

0
0 R
0
 R


 3 4R 
(i)
Let us now find values of electric field inside and outside the sphere.
Charge Q enclosed by the outside Gaussian sphere can be found by replacing r by R in the above expression
 R3 R 4  ρ0 4π R3
⇒ Q = ρ 0 4π  −
=
12
 3 4R 
(ii)
⇒ Eout =
ρ0 R3
12ε 0 r 2
Also, charge q enclosed inside a sphere of radius r(<R) is
 r3 r4 
q = ρ 0 4π  −

 3 4R 
⇒ Ein =
ρ0
ε0r 3
 r 3 r 4  ρ 0 (4rR − 3r 2 )
 −
=
12ε 0 R
 3 4R 
Problem 9:
In a uniform sphere of charge (charge density ρ), a small cavity is created. The center of cavity is at a distance a
from the centre of sphere. Taking center of sphere as origin and as a unit vector along the line joining the centre of
ρa
cavity and origin, show that the electric field at any point inside the cavity is =
aˆ.
3ε 0
Solution:
Consider any point P inside cavity.
P
due
The position vector of P r . The field at P can be assumed to be a superposition of field
to a uniform sphere of charge ( E1 ) and field due to a negative charged sphere of the
size of cavity ( E2 ).
C
O
a = aaɵ
Electric Charges and Field
⇒ Net field , E = E1 + E2 ⇒
ρ
3ε 0
(r − r ) =
ρa ρa
=
aˆ
3ε 0 3ε 0
The field at point P does not depend on the position of point P in the cavity, So, the field inside cavity is uniform.
EXERCISE - 07
1.
Paragraph-1
The “Gaussian Surface” for an infinitely large charged plate is pillbox of end cap area A and length 2a
center about the origin. The plate has positive surface charge density σ.
How much charge is contained in the pillbox?
(1) σ A
2.
3.
4.
5.
(3) 2σ A
(2) σ A / 2
(4) Zero
The electric field points in the +x direction on the right side of the plate and −x on the other side because
(1)
The field due to a point charge in radial
(2)
Field is always normal to a charge distribution
(3)
Of the symmetry of charge distribution
(4)
None of these
How much electric flux passes through the curved surface of the pillbox
(1)
Same as that through flat surfaces
(2)
More than that through flat surfaces
(3)
Less than that through flat surfaces but not zero
(4)
Zero
The pillbox has a thickness a on either side of the plate. How do the magnitudes of the electric field compare at x = −a and x
= + a?
(1)
Equal on both sides
(2)
Lesser on right side
(3)
More on right side
(4)
It is not possible to predict this
Use Gauss’s Law to determine the electric field at x = a.
(1)
σ / 2ε 0
(2) σ / ε 0
2σ / ε 0 (4)
(3)
σ / 4ε 0
Paragraph-2
6.
7.
8.
A very long thick plate has a uniform positive volume charge density given by ρ(x) = ρ0 for
d
d
− ≤ x ≤ . where ρ 0 is a positive constant, d is the thickness of the plate and x = 0 is the
2
2
centre of the plate.
What is the symmetry of the electric field for this object?
(1)
Planar
(2)
Spherical
(3) Cylindrical (4) None of these
What is the electric field at x = 0?
(1)
Non-zero towards right
(2)
Non zero towards left
(3)
Zero
(4)
Cannot be determined
(2)
2ρ0 Ax
ρ0 x
ε0
(2)
(3)
ρ0 x / 2ε 0
(4)
4 ρ0 x / ε 0
(3)
ρ0 x
2ε 0
(4)
Zero
d
?
2
2 ρ0 x
ε0
How much charge is contained in a “Gaussian pillbox” of surface area A and thickness 2x where x >
(1)
11.
ρ0 Ax
What is the electric field at a distance x for x <
(1)
10.
x = −d/2
How much charge is contained in a “Gaussian pillbox” of cross-sectional area A and thickness 2x, where x <
(1)
9.
x - axis
ρ0 Ad
(2)
2ρ0 Ad
What is the electric field at a distance x for x >
(1)
ρ0 d
2ε 0
(2)
ρ0 d
ε0
(3)
ρ0 Ax
(4)
2ρ0 Ax
(3)
ρ0 d
4ε 0
(4)
Zero
d
?
2
d
?
2
d
?
2
x = d/2
Electric Charges and Field
12.
13.
14.
15.
Paragraph-3
A spherical shell of inner radius R and outer radius 2R, has a uniform charge distribution and total charge Q. A Gaussian
sphere of radius r having its center at O is drawn.
Calculate the charge inside the “Gaussian sphere” for the three regions
(i)
0<r<R
(ii)
R < r < 2R
(iii)
r > 2R
Use Gauss’s law to determine the electric field at the surface of the “Gaussian surface” when
(i)
0<r<R
(ii)
R < r < 2R
(iii)
r > 2R
Paragraph - 4
An early, but incorrect, model of the hydrogen atom proposed by J.J.Thomson, proposed that the atom can be considered as a
positive cloud of charge +e uniformly distributed throughout the volume of a sphere of radius R, with the electron (an equal
magnitude negatively charged particle) at the centre. Using Gauss law or otherwise, we can show that the electron will be in
equilibrium at the centre.
If the electron is displaced from the centre by a radial distance r, it would experience a force
 1  e2 r
(1) 
, radially inwards
 4πε 0  R3
 1  e2 r
(2) 
, radially outwards
 4πε 0  R3
 1  e2
(3) 
, radially inwards
 4πε 0  r 2
 1  e2
(4) 
, radially outwards
 4πε 0  r 2
Suppose the electron is not stationary at the centre, but oscillating about the centre with an amplitude less than R. The time
period of its oscillatory motion will be
(1) 2π
16.
4πε 0 mR3
(2) 2π
e2
4πε 0 R 2
(3) 2π
e2
e2
4πε 0 mR 3
(4) 2π
e2
4πε 0 mR 2
Suppose the electron is orbiting inside in a circle of radius r, the orbital speed of the electron will be
(1)
e2 r 2
4πε 0 mR 3
(2)
e2 R 3
4πε 0 mr
(3)
e2 r 3
2πε 0 mR 3
(4)
e 2 R3
8πε 0 mr
Paragraph - 5
A light insulating rod AB (with two small identical metal balls connected to its ends) is placed in a uniform electric field,
with its length parallel to the electric field. Charges on the balls are −q and + q as shown, mass of each ball is m and length of
the rod is l. Assume that the space is gravity free.
-q, m
17.
18.
19.
+q, m
Net force acting on the rod is
(1) 2qE leftwards
(2) 2qE rightwards
(3) Zero
(4) 2qE upwards
0
If the rod is rotated in anticlockwise direction by angle 90 , from the position shown, torque acting on the rod will be
(1) qEl, clockwise
(2) 2qEl, anticlockwise
(3) qEl, anticlockwise
(4) 2qEl, clockwise
If the rod is slightly disturbed from the position shown, it oscillates with time period.
(1) 2π
20.
l
ml
2qE
(2) 2π
ml
qE
(3) 2π
2ml
qE
(4) 2π
ml
4qE
A uniform electric field E = E0 ( − ˆj ) exists in space. An electron is given velocity v = v0 iˆ from origin. If equation of path of
 eE
electron is y =  0
 2m
(1) a = 2
 xa
. . Then
 v0b
(2) b = −2
(3) a = 1
Paragraph - 6
(4) b = 2
Electric Charges and Field
21.
There is a non-conducting rod of length L and negligible mass with two small balls each of mass m and electric charge Q
attached to its ends. The rod can rotate in the horizontal plane about a fixed vertical axis crossing it a distance L/4 from one
of its ends. A uniform horizontal electric field E is established.
At first the rod is in unstable equilibrium. If it is disturbed slightly from this position, the maximum velocity attained by the
ball which is closer to the axis in subsequent motion, is
(1)
2QEL
m
(2)
2QEL
5m
QEL
m
(3)
(4)
4QEL
m
22.
In what position, the rod must be set so that if displaced a little from that position, it begins simple harmonic oscillations
about the axis?
23.
What would be the angular frequency of SHM in the above questions?
(1)
24.
25.
(2)
5ml
QE
(3)
3QE
2mL
(4)
QE
mL
Paragraph - 7
Two balls of charges q1 and q2 initially have exactly same velocity. Both the balls are subjected to same uniform electric field
for same time. As a result, the velocity of the first ball is reduced to half of initial value and its direction changes by 60o. The
direction of velocity of second ball is found to change by 90o.
The electric field and initial velocity of the charged particles are inclined at angle
(2) 30o
(3) 90o
(4) 150o
(1) 60o
The new velocity of second charged particle has a magnitude ‘x’ times the initial velocity, where x is
(1)
26.
4QE
5mL
3
(2)
1
3
(3)
1
2
(4) 2
If the specific charge (charge to mass ratio) of first ball is K, the specific charge of second ball is
(1) 2K
(2) 4K
(3)
4
K
3
(4)
3
K
4