CHAPTER 1 Electric Charges and Fields
EXERCISES PAGE:42
Question1 What is the force between two small charged spheres having charges of 2 × 10–7C and 3 × 10–7 C placed 30 cm apart in the air?
Solution :
The formula used to find the force, F is given as,
Where, q1 and q2 are the charges. r is the distance between the charges. is a constant and its value is is the permittivity of free space. Its value is
Since, both the charges are positive, thus, the nature of force will be repulsive.
F12 is the force on charge q1 caused by charge q2.
Now, Given:
Charge on the first sphere,
Charge on the second sphere,
Distance between the spheres, r = 30 cm = 0.3 m
Putting the values in equation (1), we get,
Hence, the force between the given charged particles will be 
Since the nature of the charges is the same i.e. they are both positive. Hence, the force will be repulsive.
Question2. 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?
Solution :
(a) Electrostatic force on the first sphere, F = 0.2 N
Charge on this sphere, q1 = 0.4 μC = 0.4 × 10−6 C
Charge on the second sphere, q2 = − 0.8 μC = − 0.8 × 10−6 C
Electrostatic force between the spheres is given by the relation,
Where, ∈0 = Permittivity of free space
The distance between the two spheres is 0.12 m.
(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
Question 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?
Solution :
The given ratio is 
.
Where,
G = Gravitational constant
Its unit is N m2 kg−2.
me and mp = Masses of electron and proton.
Their unit is kg.
e = Electric charge.
Its unit is C.
∈0 = Permittivity of free space Its unit is N m2 C−2.
Hence, the given ratio is dimensionless.
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
This is the ratio of electric force to the gravitational force between a proton and an electron, keeping distance between them constant.
Question4. (a) Explain the meaning of the statement ‘electric charge of a body is quantised’.
(b) Why can one ignore quantisation of electric charge when dealing with macroscopic i.e., large scale charges?
Solution : (a) Electric charge of a body is quantized. This means that only integral (1, 2, …., n) number of electrons can be transferred from one body to the other. Charges are not transferred in fraction. Hence, a body possesses total charge only in integral multiples of electric charge.
(b) In macroscopic or large scale charges, the charges used are huge as compared to the magnitude of electric charge. Hence, quantization of electric charge is of no use on macroscopic scale. Therefore, it is ignored and it is considered that electric charge is continuous.
Question5. When a glass rod is rubbed with a silk cloth, charges appear on both. A similar phenomenon is observed with many other pairs of bodies. Explain how this observation is consistent with the law of conservation of charge.
Solution :
Rubbing produces charges of equal magnitude but of opposite nature on the two bodies because charges are created in pairs. This phenomenon of charging is called charging by friction. The net charge on the system of two rubbed bodies is zero. This is because equal amount of opposite charges annihilate each other. When a glass rod is rubbed with a silk cloth, opposite natured charges appear on both the bodies. This phenomenon is in consistence with the law of conservation of energy. A similar phenomenon is observed with many other pairs of bodies.
Question6. 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?
Solution : 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 = 
cm
AO = OC = DO = OB = 
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.
Question7. (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?
Solution :
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.
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.
Question 8 Two point charges qA = 3 µC and qB= –3 µC are located 20 cm apart in a vacuum.
What is the electric field at the midpoint O of the line AB joining the two charges?
If a negative test charge of magnitude 1.5 × 10–9 C is placed at this point, what is the force experienced by the test charge?
Solution : (i) The situation is represented in the given figure. O is the mid-point of line AB.
Distance between the two charges, AB = 20 cm
AO = OB = 10 cm
Net electric field at point O = E
Electric field at point O caused by +3μC charge,
E1 =
along OB
Where,
= Permittivity of free space
Magnitude of electric field at point O caused by −3μC charge,
E2 = 
along OB
Question9 A system has two charges
Solution : The charges which are located at the given points are shown in the coordinate system as:
Question 10 An electric dipole with dipole moment
Question 11 A polythene piece rubbed with wool is found to have a negative charge of
Question 12 (i) Two insulated charged copper spheres, A and B, have their centres separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is
The radii of A and B are negligible compared to the distance of separation.
(ii) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?
Solution:
a) Charge on sphere A, qA = Charge on sphere B, qB = 6.5 × 10−7 C
Distance between the spheres, r = 50 cm = 0.5 m
Force of repulsion between the two spheres,
Where,
∈0 = Free space permittivity
(b) After doubling the charge, charge on sphere A, qA = Charge on sphere B, qB =
The distance between the spheres is halved.
∴
Force of repulsion between the two spheres,
= 16 × 1.52 × 10−2
= 0.243 N
Therefore, the force between the two spheres is 0.243 N.
Question13. Figure 1.30 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?
Solution : 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.
Question14. Consider a uniform electric field E = 3 × 103îN/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?
Solution : Given:
Electric field E = 3× 103 N/C
Side of square, s = 10 cm
a) Flux of field through square whose plane is parallel to yz plane.
We understand that the normal to the plane is parallel to the direction of field.
So, θ = 0°
Ø = E . A
Ø = E × A× cos(θ) …(1)
Where, E = Electric field
A = Area through which we have to calculate flux θ = Angle between normal to surface and the Electric field
A = s2
A = .01 m2
Plugging values, of E, A and θ in equation (1)
Ø = 3× 103 NC-1× 0.01m2× cos0°
Ø = 30 Nm2C-1
b) If normal to its (square’s) plane makes 60° with the X axis.
θ = 60°
Ø = E × A× cos(θ)
Ø = 3× 103 NC-1× 0.01m2 × cos60°
Ø = 15 Nm2C-1
Question15. What is the net flux of the uniform electric field of Exercise 1.14 through a cube of side 20 cm oriented so that its faces are parallel to the coordinate planes?
Solution : 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.
Question 16)
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 × 10 3 N m 2 /C.
What is the net charge inside the box?
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?
Solution :Given:
a) Ø = 8.0× 103 Nm2C-1
Let net charge inside the box = q
We know that,
Flux, Ø = q/ε0 ..(1)
Where, q = net charged enclosed ε0 = permittivity of free space
ε0 = 8.85× 10-12 N-1 m-2C2
Plugging values of Ø and ε0 in equation (1) we get, q = Ø × ε0
⇒ q = 8.0 × 103 Nm2C-1× 8.85× 10-12N-1C2m-2
⇒ q = 7.08 × 10-8 C
Hence, the net charge inside the box is 0.07 µC.
(b) No, we cannot conclude that the body doesn’t have any charge. The flux is due to the Net charge of the body. There may still be equal amount of positive and negative charges. So, it is not necessary that if flux is zero then there will be no charges.
Question17. 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. 1.31. 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.)
Solution :Given:
q = + 10 s = 10 cm
Assume the charge to be enclosed by a cube, where the square is one of its sides.
Now, let us find the total flux through the imaginary cube.
We know that,
Flux, Ø = q/ε …(1)
Where, q = net charged enclosed ε0 = permittivity of free space ε0 = 8.85× 10-12 N-1 m-2C2
Now plugging the values of q and ε0 in equation (2)
⇒ Ø = 11.28 × 105 Nm-2C-1
We understand that flux through all the faces of cube will be equal;
Let flux through the square = Øa
Hence,
Øa = Ø/6
Explanation: The net flux will be distributed equally among all 6 faces of the cube. Hence, the square will have one sixth of the total flux.
Øa = 1.88 Nm-2C-1
Question18. 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?
Solution :Given:
Total charge inside the cube, q = 2.0 µC
Edge length of Cube, a = 9.0 cm
Question19. A point charge causes an electric flux of – 1.0 × 10 3 N m 2 /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?
Solution :Given:
Ø = -1.0× 103 Nm2C-1 r-1 = 10.0 cm.
Flux if the radius of the Gaussian surface is doubled.
If the radius is doubled then the flux would remain same i.e. -1.0× 103 Nm2C-1.
The geometry of the Gaussian surface doesn’t affect the total flux through it. The net charge enclosed by Gaussian surface determines the net flux.
Let value of point charge enclosed by Gaussian surface = q
We know that,
Flux, Ø = q/ε0 …(1)
Where, q = net charged enclosed ε0 = permittivity of free space ε0 = 8.85× 10-12 N-1 m-2C2
Now plugging, the values Ø and ε0 in equation (1). q = Ø× ε0
⇒ q = -1.0× 103Nm2C-1 × 8.85× 10-12N-1m-2C2
⇒ q = -8.85 × 10-9 C
The charge enclosed by the surface is -8.85 × 10-9 C.
Question 20
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 × 10 3 N/C and points radially inward, what is the net charge on the sphere?
Solution :Given:
Radius of charged sphere, r = 10 cm
Electric field, 20 cm away from centre of sphere, E = 1.5× 103 NC-1
We know that, electric field intensity at a point P, located at a distance R, due to net charge q is given by,
…(1)
Now plugging the values of q and R in equation (1) q = 4× π× ε0× R2× E
⇒ q = 4× 3.14× 8.85× 10-12N-1m-2C-2× (0.2m)2 × 1.5× 103C
⇒ q = 6.67× 10-9C
The net charge on the sphere is -6.67× 10-9C. Since the electric field points radially inwards, we can infer that charges on sphere are negative.
Question21. 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?
Solution :Given: radius of sphere, r = 1.2 m
Surface charge density, σ = 80.0 μC/m2
a) Let charge on sphere = q
We understand that,
Total charge, Q = Surface charge density× surface area.
Surface area of sphere, S = 4πr2
S = 18.08 m2
Q = 80× 10-6Cm-2× 18.08m-2
⇒ Q = 1.447 × 10-3C
b) Let total electric flux leaving the surface of sphere = We know that,
Flux, Ø = q/ε0 …(1)
Where, Q = net charged.
ε0 = permittivity of free space ε0 = 8.85× 10-12 N-1 m-2C2
By, plugging the values of q and ε0 in equation (1), we get,(b)
Question22 An infinite line charge produces a field of 9 × 10 4 N/C at a distance of 2 cm. Calculate the linear charge density.
Solution :Given: E = 9× 104 NC-1 d = 2 cm
Let the linear charge density = λ Coulomb/metre
Where, λ = linear charge density. ε0 = permittivity of free space ε0 = 8.85× 10-12 N-1 m-2C2 d = distance
From equation (1) we have,
⇒ λ = E× 2× π× ε0× d
⇒ λ = 9× 104 NC-1× 2× π×8.85× 10-12 N-1 m-2C2× 0.02m
⇒ λ = 10 µ Cm-1
The linear charge density is 10 µ Cm-1.
Question23 Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of
opposite signs and magnitude 17.0 × 10 – 22 C /m 2
What is E:
In the outer region of the first plate?
In the outer region of the second plate and (c) between the plates?
Solution :Given:
Surface charge density on plate A, σA = -17.0 × 10-22 Cm-2
Surface charge density on plate B, σB = 17.0 × 10-22 Cm-2 The arrangement of plates are as shown:
Let electric field in region 1 = E1
Region 2 = E2
Region 3 = E3
The electric field in region I and region III is zero because no charge is present in these regions. E1 = 0
E2 = 0
We know that,
E3 = σ/ε0 …(1)
Where, σ Surface charge density
ε0 = permittivity of free space ε0 = 8.85× 10-12 N-1 m-2C2
Now, plugging the Values in equation (1).
⇒ E3 = 1.92 × 10-10 NC-1
The electric field in the region enclosed by the plates is found to be 1.92 × 10-10 NC-1.
The electric field in region III is 1.92 × 10-10 NC-1.