Question

A parallel plate capacitor is charged and then isolated. The effect of increasing the plate separation on charge, potential, capacitance respectively are.

• A. Constant, decreases, decreases
• B. Increases, decreases, decreases
• C. Constant, decreases, increases
• D. Constant, increases, decreases.

Answer 1

Answer: (D)

Constant, increases, decreases.

Answer 2

: As the capacitor is isolated after charging. Charge on it remains constant. Plate separation d increases, capacitance decreases as $\mathrm { C } = \frac { \varepsilon _ { 0 } \mathrm {~A} } { \mathrm {~d} }$and hence, potential increases at