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 capacitance of the capacitor and the magnitude of the charge on the plates are A) $\frac { 1 } { 2 }$ C₀, $\frac { 1 } { 2 }$ Q₀ B) $\frac { 1 } { 2 }$ C₀,Q₀ C)C₀,Q₀ D)2C₀,Q₀ E)2C₀,2Q₀

Question

Quizplus · Accepted Answer

The capacitance of a parallel plate capacitor is given by C = ε₀A/d, where ε₀ is the permittivity of free space, A is the area of the plates, and d is the separation between the plates. When the plates are pulled apart to a separation 2d, the capacitance becomes C/2. The charge on the plates is given by Q = CV, where V is the voltage across the capacitor. When the capacitance is reduced by a factor of 2, the charge on the plates is also reduced by a factor of 2. Therefore, the correct answer is A: $\frac { 1 } { 2 }$ C₀, $\frac { 1 } { 2 }$ Q₀.

A Parallel Plate Capacitor of Capacitance C0 Has Plates of Area

A Parallel Plate Capacitor of Capacitance C₀ Has Plates of Area