Answer
(a) The capacitor charge remains constant.
(b) The electric field strength remains constant.
(c) Ptential difference doubles.
Work Step by Step
(a) No charge flows off the capacitor plate ( conservation of charge).
Therefore, the capacitor charge remains constant.
(b) Since, $E=\frac{Q/A}{\epsilon_{0}}$;$Q,A,\epsilon_{0}$ all are constants.
Therefore, the electric field strength remains constant.
(c) $|V_{c}|=\frac{Qd}{A\epsilon_{0}}$, hence, as $d$ becomes $2d$ , potential difference doubles.
