Ncert Exercises & Solutions

Q 1.22)

A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of 80.0 μC /m ²

(a) Find the charge on the sphere.

(b) What is the total electric flux leaving the surface of the sphere?

At a distance ( d ) from the center of a sphere, the electric field intensity:

E = \frac{1}{4 π ϵ_{0}} . \frac{q}{d^{2}}

So the net charge on the sphere:

$q = E (4 {πε}_{0}) d^{2} = \frac{1 .5 \times 10^{3}}{9 \times 10^{9}} = 6 .67 \times 10^{9} C = 6 .67 nC$

The net charge on the sphere is 6.67 nC.

(a)

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:

Q = Charge density × Surface area

= σ × 4πr ²

= 80 × 10^{– 6} × 4 × 3.14 × ( 1.2 )²

= 1.447 × 10^{– 3} C

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

(b) Total electric flux ( $ϕ$ _Total ) leaving out the surface:

ϕ_{Total} = \frac{q}{ϵ_{0}}

$ϕ_{Total} = \frac{1 .447 \times 10^{- 3}}{8 .854 \times 10^{- 12}} = 1 .63 \times 10^{8} {NC}^{- 1} m^{2}$

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