Question

A positive charge $Q$ is placed on a conducting spherical shell with inner radius $R_1$ and outer radius $R_2$ . A particle with charge $q$ is placed at the center of the spherical cavity. The magnitude of the electric filed at a point in the cavity, a distance $r$ from center is

• A. zero
• B. $\frac{Q}{4 \pi \varepsilon_0 R^2}$
• C. $\frac{q}{4 \pi \varepsilon_0 r^2}$
• D. $\frac{(q + Q)}{4 \pi \varepsilon_0 r^2}$

Answer 1

Answer: (B)

Total charge on shell $=+Q$
Here, total charge on shell is $+Q$. So, the $(-Q)$
is on the inner surface of shell.
Hence, electric field inside the conductor $=0$
According the Gauss's law,
$\rho E d s=\frac{Q}{\varepsilon_{0}}$
$ E \oint d s =\frac{Q}{\varepsilon_{0}} $
$E\left(4 \pi r^{2}\right) =\frac{Q}{\varepsilon_{0}} $
Electric field, $E=\frac{Q}{4 \pi r^{2} \varepsilon_{0}}$