d
Now the charge on outer surfaces should be same
$q=Q-q \Rightarrow q=Q / 2$
So $(A)=-2 Q+Q / 2=-3 Q / 2$
Charge on $\mathrm{A}=\mathrm{Q} / 2$ and charge on $\mathrm{B}=3 \mathrm{Q} / 2$
So as $E \propto Q$ and $V \propto E \Rightarrow V \propto Q$
and hence $C$ is correct.
