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.The plates are pulled apart to a separation 2d while the capacitor remains connected to the battery.After the plates are 2d apart,the magnitude of the charge on the plates and the potential difference between them are A) Q₀, V₀ B) Q₀,V₀ C)Q₀,V₀ D)2Q₀,V₀ E)2Q₀,2V₀

Question

Quizplus · Accepted Answer

The capacitance of a parallel plate capacitor is given by C=ε*A/d, where ε is the permittivity of the medium between the plates (assuming it is a vacuum or air, ε=ε0). Therefore, when the distance between the plates is doubled, the capacitance becomes half of its original value, i.e. C' = ε*A/(2d) = 1/2 * C. Since the battery is still connected, the potential difference between the plates remains the same, V=V₀. The charge stored on a capacitor is given by Q=C*V, so the charge on the plates after the separation is doubled becomes Q' = C'*V = 1/2 * Q. Therefore, the correct answer is B.

A Parallel Plate Capacitor of Capacitance C0 Has Plates of Area

A Parallel Plate Capacitor of Capacitance C₀ Has Plates of Area