NCERT Solutions for Electric Charges and Fields Class 12 Physics
Book Solutions1
What is
the force between two small charged spheres having charges of 2 × 10−7
C and 3 × 10−7 C placed 30 cm apart in air?
Answer
Repulsive
force of magnitude 6 × 10−3 N
Charge on
the first sphere, q1 = 2 × 10−7 C
Charge on
the second sphere, q2 = 3 × 10−7 C
Distance
between the spheres, r = 30 cm = 0.3 m
Electrostatic
force between the spheres is given by the relation,
Hence,
force between the two small charged spheres is 6 × 10−3 N. The
charges are of same nature. Hence, force between them will be repulsive.
2
The
electrostatic force on a small sphere of charge 0.4 μC due to another small
sphere of charge − 0.8 μC in air is 0.2 N. (a) What is the distance between the
two spheres? (b) What is the force on the second sphere due to the first?
Answer
(b) Both the spheres attract each other with the same force. Therefore, the force on the second sphere due to the first is 0.2 N
3
Check
that the ratio ke2/G memp is
dimensionless. Look up a Table of Physical Constants and determine the value of
this ratio. What does the ratio signify?
Answer
e = 1.6 ×
10−19 C
G = 6.67
× 10−11 N m2 kg-2
me= 9.1 ×
10−31 kg
mp = 1.66 ×
10−27 kg
Hence,
the numerical value of the given ratio is
The ratio signify that the Electrostatic force is much larger than gravitational force.
4
Answer
(i) Electric charge of a body is quantized means charge on a
body can only exist in the form of integral multiple of any particular charge
that is charge of electron. This because charge is acquired or lost by transfer
or gain of electrons only.
(ii) Considering large scale or macroscopic charges, the charge
which are used over there are comparatively too large to the magnitude of the
electric charge. Hence, on a macroscopic level the quantization of charge is of
no use Therefore, it is ignored and the electric charge is considered to be
continuous.
5
Answer
When two bodies are rub together then, charges are developed on
both of the bodies and the charge developed over the bodies are equal but
opposite in nature. And this phenomenon of getting a charge is called as charge
by friction. The net charge on both of the bodies. When we rubbed the glass rod
with the silk cloth, there will be charge with opposite magnitude is generated
over there. This phenomenon is in consistence with the law of conservation of
charge. A similar phenomenon is observed with many other pairs of bodies.
6
Four
point charges qA = 2 μC, qB = −5 μC, qC
= 2 μC, and qD = −5 μC are located at the corners of a square
ABCD of side 10 cm. What is the force on a charge of 1 μC placed at the centre
of the square?
Answer
The given
figure shows a square of side 10 cm with four charges placed at its corners. O
is the centre of the square.
Where,
(Sides) AB = BC = CD = AD = 10 cm
(Diagonals) AC = BD =10√2 cm
AO = OC = DO = OB = 5√2 cm = 5√2 cm
A charge of amount 1μC is placed at point O.
Force of repulsion between charges placed at corner A and centre O is equal in magnitude but opposite in direction relative to the force of repulsion between the charges placed at corner C and centre O. Hence, they will cancel each other. Similarly, force of attraction between charges placed at corner B and centre O is equal in magnitude but opposite in direction relative to the force of attraction between the charges placed at corner D and centre O. Hence, they will also cancel each other. Therefore, net force caused by the four charges placed at the corner of the square on 1 μC charge at centre O is zero.
7
(a) An
electrostatic field line is a continuous curve. That is, a field line cannot
have sudden breaks. Why not?
(b) Explain
why two field lines never cross each other at any point?
Answer
(a) An
electrostatic field line is a continuous curve because a charge experiences a
continuous force when traced in an electrostatic field. The field line cannot
have sudden breaks because the charge moves continuously and does not jump from
one point to the other.
(b) If two
field lines cross each other at a point, then electric field intensity will
show two directions at that point. This is not possible. Hence, two field lines
never cross each other.
8
We
have two point charges on is qA=3μC and other is qA=–3μC and the distances between these two
charges are 20 cm and these are placed inside a vacuum.
(i) Let us consider O as the mid – point of the line. Then,
measure the electric field at the point O of the line AB joining the two
charges.
(ii) If
we will place a negative charge having a magnitude of 1.5×10-9 at this point, then
calculate the force experienced by the test charge?
Answer
9
A system
has two charges qA = 2.5 × 10−7 C and qB
= −2.5 × 10−7 C located at points A: (0, 0, − 15 cm) and B: (0, 0, +
15 cm), respectively. What are the total charge and electric dipole moment of
the system?
Answer
The given system can be represented on coordinate plane as seen in the figure.
At A,
amount of charge, qA = 2.5 × 10−7C = 2.5 ×10 -7
C
At B,
amount of charge, qB = −2.5 × 10−7 C = -2.5×10-7
C
Total
charge of the system,
q=qA+qB
= 2.5 ×10
-7 C - 2.5 ×10 -7 C
= 0
Distance
between two charges at points A and B,
d = 15 +
15 = 30 cm = 0.3 m
Electric
dipole moment of the system is given by,
p = qA × d = qB× d
= 2.5 ×
10−7 × 0.3
= 7.5 ×
10−8 C m along positive z-axis
Therefore,
the electric dipole moment of the system is 7.5 × 10−8 C m along
positive z−axis.
10
An
electric dipole with dipole moment 4 × 10−9 C m is aligned at 30°
with the direction of a uniform electric field of magnitude 5 × 104
N C−1. Calculate the magnitude of the torque acting on the dipole.
Answer
Therefore,
the magnitude of the torque acting on the dipole is 10−4 N m.
11
A
polythene piece rubbed with wool is found to have a negative charge of 3 × 10−7
C.
(a) Estimate
the number of electrons transferred (from which to which?)
(b) Is there
a transfer of mass from wool to polythene?
Answer
and polythene becomes negatively charged.
Amount of charge on the polythene piece, q = -3×10-7 C
Amount of charge on an electron, e = -1.6×10-19 C
Number of electrons transferred from wool to polythene = n
n can be calculated using the relation,
Therefore, the number of electrons transferred from wool to polythene is 1.87 × 1012.
(b) Yes.
There is
a transfer of mass taking place. This is because an electron has mass,
me = 9.1 × 10−3 kg
Total
mass transferred to polythene from wool,
m = me
× n
= 9.1 ×
10−31 × 1.85 × 1012
= 1.706 ×
10−18 kg
Hence, a negligible amount of mass is transferred from wool to polythene.
12
(a) Two
insulated charged copper spheres A and B have their centers separated by a
distance of 50 cm. What is the mutual force of electrostatic repulsion if the
charge on each is 6.5 × 10−7 C? The radii of A and B are negligible
compared to the distance of separation.
(b) What is
the force of repulsion if each sphere is charged double the above amount, and
the distance between them is halved?
Answer
13
Suppose the spheres A and B in Exercise 1.12 have identical
sizes. A third sphere of the same size but uncharged is brought in contact with
the first, then brought in contact with the second, and finally removed from
both. What is the new force of repulsion between A and B?
Answer
14
Figure 1.33 shows tracks of three charged particles in a
uniform electrostatic field. Give the signs of the three charges. Which
particle has the highest charge to mass ratio?
Answer
Opposite charges attract each other and same charges repel
each other. It can be observed that particles 1 and 2 both move towards the
positively charged plate and repel away from the negatively charged plate.
Hence, these two particles are negatively charged. It can also be observed that
particle 3 moves towards the negatively charged plate and repels away from the
positively charged plate. Hence, particle 3 is positively charged.
The charge to mass ratio (emf) is directly proportional to
the displacement or amount of deflection for a given velocity. Since the
deflection of particle 3 is the maximum, it has the highest charge to mass
ratio.
15
Consider a uniform electric field E = 3 × 103
îN/C. (a) What is the flux of this field through a square of 10 cm on a side
whose plane is parallel to the yz plane? (b) What is the flux through the
same square if the normal to its plane makes a 60° angle with the x-axis?
Answer
16
Answer
All the faces of a cube are parallel to the coordinate
axes. Therefore, the number of field lines entering the cube is equal to the
number of field lines piercing out of the cube. As a result, net flux through
the cube is zero.
17
Careful measurement of the electric field at the surface of
a black box indicates that the net outward flux through the surface of the box
is 8.0 × 103 N m2/C. (a) What is the net charge inside
the box? (b) If the net outward flux through the surface of the box were zero,
could you conclude that there were no charges inside the box? Why or Why not?
Answer
(b) No
Net flux piercing out through a body depends on the net
charge contained in the body. If net flux is zero, then it can be inferred that
net charge inside the body is zero. The body may have equal amount of positive
and negative charges.
18
A point charge +10 μC is a distance 5 cm directly above the
centre of a square of side 10 cm, as shown in Fig. 1.34. What is the magnitude
of the electric flux through the square? (Hint: Think of the square as
one face of a cube with edge 10 cm.)
Answer
19
A point charge of 2.0 μC is at the centre of a cubic
Gaussian surface 9.0 cm on edge. What is the net electric flux through the
surface?
Answer
20
A point charge causes an electric flux of −1.0 × 103
Nm2/C to pass through a spherical Gaussian surface of 10.0 cm radius
centered on the charge. (a) If the radius of the Gaussian surface were doubled,
how much flux would pass through the surface? (b) What is the value of the
point charge?
Answer
(a) Electric flux, Φ
= −1.0 × 103 N m2/C
Radius of the Gaussian surface,
r = 10.0 cm
Electric flux piercing out through a surface depends on the
net charge enclosed inside a body. It does not depend on the size of the body.
If the radius of the Gaussian surface is doubled, then the flux passing through
the surface remains the same i.e., −103 N m2/C.
21
A conducting sphere of radius 10 cm has an unknown charge.
If the electric field 20 cm from the centre of the sphere is 1.5 × 103 N/C
and points radially inward, what is the net charge on the sphere?
Answer
Therefore, the net charge on the sphere is 6.67 nC.
22
A uniformly charged conducting sphere of 2.4 m diameter has
a surface charge density of 80.0 μC/m2. (a) Find the charge on the
sphere. (b) What is the total electric flux leaving the surface of the sphere?
Answer
23
An infinite line charge produces a field of 9 × 104
N/C at a distance of 2 cm. Calculate the linear charge density.
Answer
Electric field produced by the infinite line charges at a distance d having linear charge density λ is given by the relation.
Therefore, the linear charge density is 10 μC/m.
24
Answer
