Question

A parallel-plate capacitor is made of two conducting plates of area A separated by a distance d. The capacitor carries a charge Q and is initially connected to a battery that maintains a constant potential difference between the plates. The battery is then disconnected from the plates and the separation between the plates is doubled. The electric potential energy stored in the capacitor will:

Answer 1

After disconnecting the battery and doubling the distance between the plates of a charged parallel-plate capacitor, the electric potential energy stored in the capacitor will double.

Step-by-step explanation:

When a parallel-plate capacitor is initially charged, the electric potential energy (U) stored in it is given by the expression U = ½ QV, where Q is the charge and V is the potential difference between the plates. Once the battery is disconnected and the distance between the plates is doubled, the capacitance (C) of the capacitor changes because capacitance is inversely proportional to the distance (C = εA/d). As the capacitance decreases, the voltage across the plates increases since Q remains constant (V = Q/C). However, the increase in voltage is proportional to the increase in distance, meaning that V will double if d is doubled. Therefore, the electric potential energy stored in the capacitor after doubling the distance, using the new voltage, is U = ½ Q(2V), which equals 2U, or twice the initial energy.

Answer 2

Stored potential energy will increase.

Step-by-step explanation:

Given that

Area = A

Capacitance =C

Charge =Q

We know that stored potential energy given as

`U=(1)/(2)CV^2`

`U=(1)/(2C)Q^2`

`C=(\varepsilon _oA)/(d)`

So when distance (d) between plates become double then C will become C/2.

So new value of capacitance C'=C/2

`U'=(1)/(2C')Q^2` (Charge will remain same)

`U'=(1)/(2(C)/(2))Q^2`

`U'=2\frac{1}{2{C}}Q^2`

U' =2 U

It means that stored potential energy will increase.