A capacitor of capacitance C has a charge of Q stored on the plates - AQA - A-Level Physics - Question 19 - 2017 - Paper 2

The potential difference $V$ of a capacitor is related to charge and capacitance by:

$Q = C imes V$

When the potential difference is doubled, $V$ becomes $2V$, which means the new charge $Q'$ is:

$Q' = C imes (2V) = 2Q$.

The new energy stored in the capacitor, with the doubled charge, can be calculated using:

U' = rac{(Q')^2}{2C} = rac{(2Q)^2}{2C} = rac{4Q^2}{2C} = rac{2Q^2}{C}.

The change in energy can be determined by subtracting the initial energy from the new energy:

ext{Change in energy} = U' - U = rac{2Q^2}{C} - rac{Q^2}{2C} = rac{2Q^2}{C} - rac{Q^2}{2C} = rac{3Q^2}{2C}.

Hence, the energy stored in the capacitor increases, leading to a new stored change of: U' = rac{3Q^2}{2C}.