Page No 47: - Chapter 1 Electric Charges & Fields Exercise Solutions class 12 ncert solutions Physics - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Question 1.13:

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:

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

Initially, the charge on each sphere, q = 6.5 × 10⁻⁷ C

When sphere A is touched with an uncharged sphere C,  amount of charge from A will transfer to sphere C. Hence, charge on each of the spheres, A and C, is.

When sphere C with charge  is brought in contact with sphere B with charge q, total charges on the system will divide into two equal halves given as,

Each sphere will share each half. Hence, charge on each of the spheres, C and B, is.

Force of repulsion between sphere A having charge  and sphere B having charge  =

Therefore, the force of attraction between the two spheres is 5.703 × 10⁻³ N.

Question 1.14:

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.

Question 1.15:

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:

(a) Electric field intensity,  = 3 × 10³ î N/C

Magnitude of electric field intensity, = 3 × 10³ N/C

Side of the square, s = 10 cm = 0.1 m

Area of the square, A = s² = 0.01 m²

The plane of the square is parallel to the y-z plane. Hence, 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 = 

= 3 × 10³ × 0.01 × cos0°

= 30 N m²/C

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

Flux, Φ = 

= 3 × 10³ × 0.01 × cos60°

 = 15 N m²/C

Question 1.16:

What is the net flux of the uniform electric field of Exercise 1.15 through a cube of side 20 cm oriented so that its faces are parallel to the coordinate planes?

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.

Question 1.17:

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:

(a) Net outward flux through the surface of the box, Φ = 8.0 × 10³ N m²/C

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

∈₀ = Permittivity of free space

= 8.854 × 10⁻¹² N⁻¹C² m⁻²

q = ∈₀Φ

= 8.854 × 10⁻¹² × 8.0 × 10³

= 7.08 × 10⁻⁸

= 0.07 μC

Therefore, the net charge inside the box is 0.07 μC.

(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.

Question 1.18:

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:

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, 

Where,

∈₀ = Permittivity of free space

= 8.854 × 10⁻¹² N⁻¹C² m⁻²

q = 10 μC = 10 × 10⁻⁶ C

∴

= 1.88 × 10⁵ N m² C⁻¹

Therefore, electric flux through the square is 1.88 × 10⁵ N m² C⁻¹.