## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

(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. Hence, (a) The capacitor charge remains constant. (b) The electric field strength remains constant. (c) Potential difference doubles.