The charge on the square plates of a parallel-plate capacitor is Q.The potential across the plates is maintained with constant voltage by a battery as they are pulled apart to twice their original separation,which is small compared to the dimensions of the plates.The amount of charge on the plates is now equal to
A) 4Q.
B) 2Q.
C) Q.
D) Q/2.
E) Q/4.

Question

Quizplus · Accepted Answer

The capacitance of a parallel-plate capacitor is given by: 
C = εA/d 
where ε is the permittivity of the material between the plates, A is the area of each plate, and d is the distance between the plates.

When the plates are pulled apart to twice their original separation, the capacitance becomes: 
C' = εA/2d

Since the potential difference across the plates remains constant, the charge stored in the capacitor is given by: 
Q = CV 
where V is the potential difference.

Substituting the expressions for C and C', we get: 
Q' = C'V = (εA/2d)V = (1/2)(εA/d)V = (1/2)Q

Therefore, the amount of charge on the plates is now half of the original amount, i.e. Q/2.

The Charge on the Square Plates of a Parallel-Plate Capacitor