Question

A charge Q is placed at the centre of an uncharged, hollow metallic sphere of radius a. (a) Find the surface. (b) If a charge q is put on the sphere, what would be the surface charge densities on the inner and outer surfaces? (c) Find the electric field inside the sphere at a distance x from the centre in the situations (a) and (b).

Answer 1

Given:
Amount of charge present at the centre of the hollow sphere = Q

We know that charge given to a hollow sphere will move to its surface.

Due to induction, the charge induced at the inner surface = −Q
Thus, the charge induced on the outer surface = +Q

(a)

Surface charge density is the charge per unit area, i.e.
σ$=\frac{\mathrm{charge}}{\mathrm{total}\mathrm{surface}\mathrm{area}}$
Surface charge density of the inner surface, ${\sigma }_{\mathrm{in}}=\frac{-Q}{4\mathrm{\pi }{a}^2}$
Surface charge density of the outer surface, ${\sigma }_{\mathrm{out}}=\frac{Q}{4\mathrm{\pi }{a}^2}$

(b)
Now if another charge q is added to the outer surface, all the charge on the metal surface will move to the outer surface. Thus, it will not affect the charge induced on the inner surface. Hence the inner surface charge density,

${\sigma }_{\mathrm{in}}=-\frac{q}{4\pi a^2}$
As the charge has been added to the outer surface, the total charge on the outer surface will become (Q+q).

So the outer surface charge density, ${\sigma }_{\mathrm{out}}=\frac{q+Q}{4\pi a^2}$

(c)

To find the electric field inside the sphere at a distance x from the centre in both the situations,let us assume an imaginary sphere inside the hollow sphere at a distance x from the centre.
Applying Gauss's Law on the surface of this imaginary sphere,we get:

$$\begin{gathered} \oint E.ds=\frac{Q}{{\mathit{\in }}_0} \\ E\oint ds=\frac{Q}{{\mathit{\in }}_0} \\ E\left(4\mathrm{\pi }{x}^2\right)=\frac{Q}{{\mathit{\in }}_0} \\ \mathit{}E=\frac{Q}{{\mathit{\in }}_0}\times \frac{1}{4\mathrm{\pi }{x}^2}=\frac{Q}{4\mathrm{\pi }{\mathit{\in }}_0{x}^2} \end{gathered}$$
Here, Q is the charge enclosed by the sphere.

For situation (b):
As the point is inside the sphere, there is no effect of the charge q given to the shell.
Thus, the electric field at the distance x:
$\mathit{}E=\frac{Q}{4\mathrm{\pi }{\mathit{\in }}_0{x}^2}$