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, decrease, decrease
B. increase, decreases, decreases
C. constant, decreases, increases
D. constant, increases, decreases.

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Correct Answer - D
As the capacitor is isolated after charging charge on it reamins constant. Plate separation d increase, capacitance decrease as `C = (epis_(0)A)/(d)` and hence, potential increases as `V = (q)/(C)`

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