A parallel plate capacitor of capacitance C₀ has plates of area A with separation d between them.When it is connected to a battery of voltage V₀,it has charge of magnitude Q₀ on its plates.It is then disconnected from the battery and the plates are pulled apart to a separation 2d without discharging them.After the plates are 2d apart,the magnitude of the charge on the plates and the potential difference between them are A) $\frac { 1 } { 2 }$ Q₀, $\frac { 1 } { 2 }$ V₀ B)Q₀, $\frac { 1 } { 2 }$ V₀ C)Q₀,V₀ D)Q₀,2V₀ E)2Q₀,2V₀

Question

Quizplus · Accepted Answer

When the capacitor is disconnected from the battery, the charge Q₀ remains constant. Increasing the separation to 2d doubles the potential difference to 2V₀ because V = Q/C and C is halved when the distance is doubled.

A Parallel Plate Capacitor of Capacitance C0 Has Plates of Area

A Parallel Plate Capacitor of Capacitance C₀ Has Plates of Area