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 Q₀on 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 new capacitance and the potential difference between the plates are A) $\frac { 1 } { 2 }$ C₀, $\frac { 1 } { 2 }$ V₀ B) $\frac { 1 } { 2 }$ C₀,V₀ C) $\frac { 1 } { 2 }$ C₀,2V₀ D)C₀,2V₀ E)2C₀,2V₀

Question

Quizplus · Accepted Answer

When the separation is doubled, the capacitance becomes half, $\frac{1}{2} C_0$. Since the charge remains constant and capacitance decreases, the voltage doubles, becoming 2V₀.

A Parallel Plate Capacitor of Capacitance C0 Has Plates of Area

A Parallel Plate Capacitor of Capacitance C₀ Has Plates of Area