Question

A charge $+q$ is kept inside the cavity present in a solid conducting sphere, as shown in figure.


If the sphere is given a charge $Q$, then total charge appearing on outer surface of sphere will be:

• A. $Q-q$
• B. $Q$
• C. $-(q+Q)$
• D. $Q+q$

Answer 1

The correct option is D $Q+q$

Let us consider a spherical Gaussian surface passing through region of conductor outside cavity.


Electric field inside metal is always zero i.e $E=0$
On applying Gauss law,
$\oint \overrightarrow{E}.d\overrightarrow{A}=\frac{q_{en}}{{ϵ}_0}$
Where, $q_{en}=$ charge enclosed by the Gaussian surface.
$dA=$ Elemental Gaussian surface area
Thus, $q_{en}=0$
This is only possible when a charge $-q$ is induced at the inner surface of the cavity.
The charge $Q$ given to sphere will remain at its outermost surface and as per law of conservation of charge, +q will also appear on the outer surface of the sphere. The $+q$ charge will quickly redistribute uniformly on the surface.


$\therefore$ Total charge appearing at outer surface $=Q+q$
Hence, option (d) is correct.

$\begin{array}{c}\text{Why this question}?\\ \text{Note: The electric field exists inside a cavity present in conductor}\\ \text{if a charge is placed there. It is in accordance of}\\ \text{Gauss's law}(q_{en}\ne 0,E\ne 0)\end{array}$