Chapter 1: Electric Charges and Fields

 Soln. :

Electric field intensity (E) at a distance (d) from the centre of a sphere containing net charge q is given by the relation,

Where, q = Net charge = 1.5 × 10 ³ N/C

d = Distance from the centre = 20 cm = 0.2 m

ε ₀ = Permittivity of free space and

Therefore, the net charge on the sphere is 6.67 n C.

Soln. :

(a) The diameter of the sphere, d = 2.4 m

The radius of the sphere, r = 1.2 m

Surface charge density, σ = 80.0 μC /m ² = 80 × 10 ^{– 6} C/m ²

The total charge on the surface of the sphere can be calculated as follows,

Q = Charge density × Surface area

= σ × 4πr ²

= 80 × 10 ^{– 6} × 4 × 3.14 × ( 1.2 ) ²

= 1.447 × 10 ^{– 3} C

Therefore, the charge on the sphere is 1.447 × 10 ^{– 3} C.

(b) Total electric flux ( φ _Total  ) leaving out the surface of a sphere containing net charge Q is given by the relation,

Here,

Electric flux = 1.63 x 10 ⁸ N C ^{– 1} m ²

Therefore, the total electric flux leaving the surface of the sphere is 1.63 × 10 ⁸ N C ^{– 1} m ²

Soln. :

The electric field produced by the infinite line charges at a distance d having linear charge density λ is given by the relation,

Here,

d = 2 cm = 0.02 m

E = 9 x 10 ⁴ N / C

ε ₀ = Permittivity of free space and

Soln. :

The situation given in the question can be depicted in the following figure.

Parallel plates A and B are placed close to each other, as shown in the figure. The region between plates A and B is labelled as II, the outer region of plate A is labelled as I, and the outer region of plate B is labelled as III.

(a) The charge density of plate A can be calculated as,

σ = 17.0 × 10 ^{– 22} C/m ²

(b) Similarly, the charge density of plate B can be calculated as,

σ = – 17.0 × 10 ^{– 22} C/m ²

(c) In regions I and III, electric field E is zero. This is because the charge is not enclosed by the respective plates.

Electric field E in region II is given by the relation,

here,

ε ₀ = Permittivity of free space = 8.854 × 10 ^{– 12} N ^{– 1}C ² m ^{– 2}