Four positive charges are fixed at the corners of a square as shown - AQA - A-Level Physics - Question 22 - 2018 - Paper 2

Let the charge of the fourth charge be $Q_4$. The total potential at the center due to all four charges is: $V_{final} = \frac{3Q}{4\pi\epsilon_0 d} + \frac{kQ_4}{d}$ Since we know from the question that the total potential is also: $V_{final} = \frac{5Q}{4\pi\epsilon_0 d}$ Equating these two expressions gives: $\frac{3Q}{4\pi\epsilon_0 d} + \frac{kQ_4}{d} = \frac{5Q}{4\pi\epsilon_0 d}$ Rearranging this leads to: $\frac{kQ_4}{d} = \frac{5Q}{4\pi\epsilon_0 d} - \frac{3Q}{4\pi\epsilon_0 d}$ Which simplifies to: $\frac{kQ_4}{d} = \frac{2Q}{4\pi\epsilon_0 d}$ Thus: $Q_4 = \frac{2Q}{4\pi\epsilon_0} \cdot \frac{d}{k} = \frac{2Q}{4\pi\epsilon_0} \cdot \frac{d}{\frac{1}{4\pi\epsilon_0}} = 2Q$