Physics - Electrostatics Question with Solution | TestHub

Initially, the inner sphere has charge $Q$. This induces a charge $-Q$ on the inner surface of the outer shell. Let $Q_{outer}$ be the charge on the outer surface of the outer shell.

When the outer shell is connected to earth, its potential becomes zero. The potential at the outer surface of the outer shell ($r=2R$) is $V(2R)=\frac{k(Q-Q+Q_{outer})}{2R}=\frac{k{Q}_{outer}}{2R}$. Since $V(2R)=0$, it implies $Q_{outer}=0$. So, the charge on the outer surface of the outer shell is zero. (Option A is correct)

Since $Q_{outer}=0$, the total charge enclosed by a Gaussian surface outside the outer shell (for $r>2R$) is $Q+(-Q)+0=0$. Therefore, the electric field outside the outer shell is zero. (Option C is correct) For $R<r<2R$, the electric field is due to the charge $Q$ on the inner sphere, so $E(r)=\frac{kQ}{r^2}$. The potential at any point $r$ for $R<r<2R$ is $V(r)=V(2R)-{\int }_{2R}^{r}E(r^{\prime })d{r}^{\prime }=0-{\int }_{2R}^r\frac{kQ}{r^{\prime 2}}d{r}^{\prime }=-kQ[-\frac{1}{r^{\prime }}{]}_{2R}^r=kQ(\frac{1}{r}-\frac{1}{2R})$. So, $V(r)=\frac{kQ}{r}-\frac{kQ}{2R}$. (Option D is incorrect) The potential of the inner sphere is $V(R)=\frac{kQ}{R}-\frac{kQ}{2R}=\frac{kQ}{2R}$. (Option B is correct)