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Question

Physics Question on electrostatic potential and capacitance

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

A

constant, decreases, decreases

B

increases, decreases, decreases

C

constant, decreases, increases

D

constant, increases, decreases

Answer

constant, increases, decreases

Explanation

Solution

After separation charge is constant.
Capacity, C=ε0AdC=\frac{\varepsilon_{0}A}{d}
Capacitance decreases with increase in distance and V=QCV=\frac{Q}{C}
Potential increases with decrease in capacitance (C).