Physics - Electrostatics Question with Solution | TestHub

The charge on inner shell $=Q=+15$ mC, outer shell $=-Q=-15$ mC.

Potential on inner shell after earthing is zero:

$\frac{1}{4\pi {\epsilon }_0}\left[\frac{Q^{\prime }}{a}-\frac{Q^{\prime }}{2a}+\frac{q}{3a}\right]=0$

By charge conservation $-Q^{\prime }+q=-q$, i.e. $Q^{\prime }$ and $q$ redistribute:

$\frac{Q^{\prime }}{a}-\frac{Q^{\prime }}{2a}+\frac{q}{3a}-\frac{q}{3a}=0$

$Q^{\prime }\left(1-\frac{1}{2}+\frac{1}{3}\right)=\frac{Q}{3}$

$\frac{5Q^{\prime }}{6}=\frac{Q}{3}\Rightarrow Q^{\prime }=\frac{2Q}{5}=\frac{2\times 15}{5}=6\mathrm{mC}$

$q=(6-15)=-9\mathrm{mC}$