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

Options

A. increases , decreases , decreases
B. constant , increases , decreases
C. constant , decreases , decreases
D. constant , decreases , increases

Answer

B

Explanation

As the capacitor is isolated after charging charge Q on it remains constant . Plate seperation d increases , capacitance decreases as `C = (epsilon_(0)A)/(d)`
and hence , potential increases as `V = Q/C `