Question

An ideal parallel-plate capacitor consists of a set of two parallel plates of area A separated by a very small distance d. When this capacitor is connected to a battery that maintains a constant potential difference between the plates, the energy stored in the capacitor is U0. If the separation between the plates is doubled, how much energy is stored in the capacitor?a. Uo/2 b. Uo c. Uo/4 d. 4Uo e. 2Uo

Answer 1

Answer:

option A

Explanation:

given,

area of the plate = A

distance = d

$U = \dfrac{1}{2}CV^2$

where

C is the capacitance

V is the potential difference

The capacitance of the parallel plate capacitor, at the beginning, is given by

$C = \dfrac{\epsilon_0 A}{d}$

where ε₀ is the permittivity of free space,

$U_0 = \dfrac{1}{2}\dfrac{\epsilon_0 A}{d}V^2$

now, the distance is doubled

d' = 2 d    

while the potential difference is kept constant. Therefore, we can calculate the new potential energy:

$U = \dfrac{1}{2}\dfrac{\epsilon_0 A}{d'}V^2$

$U = \dfrac{1}{2}\dfrac{\epsilon_0 A}{2d}V^2$

$U = \dfrac{U_0}{2}$

Hence, the correct answer is option A