Question

A solid spherical conductor of radius $R$ has a spherical cavity of radius $a(a<R)$ at its centre. A charge $+Q$ is kept at the centre. The charge at the inner surface,outer and at a position $r(a<r<R)$ are respectively

• A. $+Q,-Q,0$
• B. $-Q,+Q,0$
• C. $0,-Q,0$
• D. $+Q,0,0$

Answer 1

The correct option is B $-Q,+Q,0$

Consider a Gaussian sphere of radius $a$ just including the inner surface of the sphere.

Since the surface is inside the conducting region, $\overrightarrow{E}=\overrightarrow{0}$

Thus, flux through the gaussian surface is zero.

Using Gauss' Law, $\oint E.ds=\frac{q_{enclosed}}{{ϵ}_o}$

We get the charge enclosed within the Gaussian sphere is zero as $E=0$.

But, the cavity contains a charge of $+Q$.

Hence, the inner surface of the conductor contains a charge of $-Q$.

Now, consider a Gaussian sphere of radius $r(a<r<R)$ as shown in the figure.

The surface of the sphere lies within the conducting spherical shell.

Hence $\overrightarrow{E}=\overrightarrow{0}$

Thus, flux through the Gaussian surface is zero.

By Gauss' law, the total charge enclosed in the Gaussian surface is zero.

But, the the charge inside the cavity is $+Q$ and the inner surface of the conductor has a charge of $-Q$.

Hence, charge at the location $r$ is zero.

Initially, the charge on the conductor is zero.

Hence by principle of conservation of charge, total charge on outer surface and inner surface and the interior is zero.

Inner surface has a charge of $-Q$ and the interior has zero charge.

Hence, the outer surface has a charge of $+Q$.