Chapter 1: Electric Charges and Fields

Soln.:

(i) Since the wool is positively charged and the polythene is negatively charged, we can say that few amounts of electrons are transferred from wool to polythene.

Charge on the polythene, q =

Amount of charge on an electron,

Let the number of electrons transferred from wool to polythene be n

So, by using the given equation, we can calculate the value of n,

q = ne

Therefore, the number of electrons transferred from wool to polythene is

(ii) Yes,

Mass is also transferred as an electron is transferred from wool to polythene and an electron particle has some mass.

Mass of an electron,

Total mass transferred , m =

Here, the mass transferred is too low that it can be neglected.

Soln.:

(i) Charge on sphere A,

Charge on sphere B,

Distance between the spheres, r = 50 cm = 0.5 m

Force of repulsion between the two spheres,

Where

Therefore,

Therefore, the force between the two spheres is

(ii) After doubling the charge,

Charge on sphere A,

Charge on sphere B,

The distance between the spheres is halved.

Force of repulsion between the two spheres,

Therefore, the force between the two spheres is 0.243 N.

 Soln.:

Distance between the spheres, A and B, r = 0.5 m

Initially, the charge on each sphere

When sphere A is touched with an uncharged sphere C, then half of the charge will be transferred to sphere C. Hence the charge on both spheres A and C will be q/2.

After that, when sphere C with charge q/2 is brought in touch with sphere B with charge q, then the charge on each of the spheres will be divided into two equal parts, is:

Hence, the charge on each of the spheres, C and B, is

The Force of repulsion between spheres A and B is:

Therefore, the force of attraction between the two spheres is

Soln.:

We can see here that particles 1 and 2 are moving in the direction of the positive charge, and we know that the opposite charge attracts each other and the same charge repels each other. So, here we can say that the charged particles 1 and 2, which are going towards the charged particle, are negatively charged. And particle 3 is being attracted towards the negative charge. So, particle 3 will be the positively charged particle.

The EMF or charge-to-mass ratio is directly proportional to the amount of depletion or depletion at a given velocity. Here, we can see that particle 3 is depleting more as compared to the other two. Therefore, it will have a higher charge-to-mass ratio.

Soln. :

(a) Electric field intensity, E = 3 × 10 ³ î  N / C

Magnitude of electric field intensity, | E | = 3 × 10 ³ N / C

Side of the square,  s = 10 cm = 0.1 m

Area of the square, A = s ² = 0.01 m ²

The plane of the square is parallel to the y – z plane. Hence, the angle between the unit vector normal to the plane and electric field, θ = 0 °

Flux ( φ ) through the plane is given by the relation,

φ = | E |A cos θ

φ  = 3 × 10 ³ × 0.01 × cos 0 °

φ = 30 N m ² /C

(b) Plane makes an angle of 60 ° with the x-axis.

Hence, θ = 60°

Flux, φ = | E |A cos θ

Flux, φ  = 3 × 10 ³ × 0.01 × cos 60°

Flux, φ  = 30 x 0.5

Flux, φ  = 15 N m ² /C

Soln.:

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, the net flux through the cube is zero.

Soln.:

(a) Net outward flux through the surface of the box, φ = 8.0 × 10 ³ N m ² /C

For a body containing net charge q, flux is given by the relation,

= 8.854 x 10 ^{– 12} x 8.0 x 10 ³ C = 7.08 x 10 ^{– 8} C = 0.07

Thus, the total charge inside the box is 0.07

(b) No

Net flux piercing out through a body depends on the net charge contained in the body. If the net flux is zero, then it can be inferred that the net charge inside the body is zero. The body may have an equal amount of positive and negative charges.

 Soln.:

The square can be considered as one face of a cube of edge 10 cm with a centre where charge q is placed. According to Gauss’s theorem for a cube, total electric flux is through all its six faces.

Hence, electric flux through one face of the cube, i.e. through the square, is

Here,

Therefore,

Therefore, electric flux through the square is 1.88 × 10 ⁵ N m ² C ^{– 1}

Soln.:

Net electric flux ( φ _Net ) through the cubic surface is given by

Here,

Therefore,

The total electric flux through the surface of the cube given is,

Soln. :

(a) Electric flux, Φ = – 1.0 × 10 ³ N m ² /C

The radius of the Gaussian surface, r = 10.0 cm

Electric flux penetrating through a surface depends on the net charge enclosed inside the 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., – 10 ³ N m ² /C.

(b) Electric flux is given by the relation

Here,

q = total charge contained enclosed by the spherical surface = φ ε ₀

q = – 1.0 x 10 ³ x 8.854 x 10 ^{– 12}

q = – 8.854 x 10 ^{– 9} C

q = – 8.854 n C

Therefore, the value of the point charge is – 8.854 n C.