Question

A spherical conductor of radius 0.12 m has a charge of 1.6 × 10⁻⁷C distributed uniformly on its surface. What is the electric field (i) inside the sphere (ii) on the sphere (iii) at a point 0.18 m from the centre of the sphere?

Answer 1

Hey There!!

Here, we have a conducting sphere, with charges distributed over the surface.


We have the following data:

$Radius = R = 0.12 \, \, m \\ \\ Charge = Q = 1.6 \times 10^{-7} \, \, C$


Now, let us consider the three cases:

1) Inside The Sphere:


Electric Field inside the conducting sphere is zero. The inside region of conducting sphere is shielded from outside effects.


2) On The Surface Of Sphere


On the surface, the electric field is going to be:

$E=\frac{kQ}{R^2} \\ \\ \\ \implies E = \frac{9 \times 10^9 \times 1.6 \times 10^{-7}}{(0.12)^2} \\ \\ \\ \implies \boxed{E=1.0 \times 10^5 \, \, N / C}$



3) Outside The Sphere


Here, distance of point from center of sphere is :
$r = 0.18 \, \, m$

Electric Field is given by:

$E=\frac{kQ}{r^2} \\ \\ \\ \implies E=\frac{9\times 10^9 \times 1.6 \times 10^{-7}}{(0.18)^2} \\ \\ \\ \implies \boxed{E \approx 4.44 \times 10^4 \, \, N / C}$


Hope it helps
Purva
Brainly Community

Answer 2

We have the following data:



Now, let us consider the three cases:
1) Inside The Sphere:

Electric Field inside the conducting sphere is zero. The inside region of conducting sphere is shielded from outside effects. 

2) On The Surface Of Sphere

On the surface, the electric field is going to be:




3) Outside The Sphere

Here, distance of point from center of sphere is :

Electric Field is given by