When a parallel-plate capacitor is connected across a battery, the energy stored in the capacitor is $W$ - AQA - A-Level Physics - Question 20 - 2020 - Paper 2

The energy stored in a capacitor is given by the formula:

U = rac{1}{2} CV^2

where:

• $U$ is the energy stored,
• $C$ is the capacitance,
• $V$ is the voltage.

When the distance between the plates of the capacitor is halved, the capacitance $C$ is doubled (since capacitance $C$ is inversely proportional to the distance between the plates). Therefore, we can express the new capacitance as:

$C' = 2C$

With the battery still connected, the voltage $V$ remains constant. Thus, the new energy stored in the capacitor can be calculated:

U' = rac{1}{2} C' V^2 = rac{1}{2} (2C) V^2 = 2 imes rac{1}{2} CV^2 = 2W

Hence, the energy now stored in the capacitor is $2W$. Therefore, the answer is:

C 2W.