Answer to Question #159629 in Electric Circuits for Nike

Question

You charge a parallel-plate capacitor, remove it from the battery, and prevent the wires connected to the plates from touching each other. When you increase the plate separation, do the following quantities (a) increase , (b) decrease or (c) stay the same?
(i) Capacitance
(ii) Charge
(iii) Electric field between the plates
(iv) Voltage

Answer 1

We have

$C=\epsilon\frac{A}{d}$ Where $\epsilon$ is permitivity of dielectric

A is the area of the capacitor plates and d is the distance between them.

(i) d is inversely proportional to C implies Capacitance decreases when distance between parallel plate is increased.

(ii)Charge remains constant. As it is charged once and connection removed

(iii) Electric field remains constant. As

Electic field $E=V/d$ . As both V and d increase we have E constant.

(iv) Charge $Q=CV$

Q is constantly and C decreases hence by above equation Potential V increases