Question

Four point charges A, B, C and D are placed at the four corners of a square of side a. The energy required to take the charges C and D to infinity (they are also infinitely separated from each other) is

• A. $\frac{q^2}{4\pi\epsilon_0 a}$
• B. $\frac{2q^2}{\pi\epsilon_0 a}$
• C. $\frac{q^2}{4\pi\epsilon_0 a}(\sqrt{2} + 1)$
• D. $\frac{q^2}{4\pi\epsilon_0 a}(\sqrt{2} - 1)$

Answer 1

Answer: (C)

$\frac{q^2}{4\pi\epsilon_0 a}(\sqrt{2} + 1)$

Answer 2

The initial potential energy of the system of four charges is the sum of the potential energies of all pairs of charges: $U_{initial} = \sum_{i<j} \frac{q_i q_j}{4\pi\epsilon_0 r_{ij}}$

The charges are $q_A = +q$, $q_B = +q$, $q_C = -q$, $q_D = -q$. The side length of the square is $a$.

The distances between the charges are: $r_{AB} = r_{BC} = r_{CD} = r_{DA} = a$ (adjacent sides) $r_{AC} = r_{BD} = a\sqrt{2}$ (diagonals)

$U_{initial} = \frac{1}{4\pi\epsilon_0} \left( \frac{q_A q_B}{r_{AB}} + \frac{q_A q_C}{r_{AC}} + \frac{q_A q_D}{r_{AD}} + \frac{q_B q_C}{r_{BC}} + \frac{q_B q_D}{r_{BD}} + \frac{q_C q_D}{r_{CD}} \right)$

Substituting the charges and distances: $U_{initial} = \frac{1}{4\pi\epsilon_0} \left( \frac{(+q)(+q)}{a} + \frac{(+q)(-q)}{a\sqrt{2}} + \frac{(+q)(-q)}{a} + \frac{(+q)(-q)}{a} + \frac{(+q)(-q)}{a\sqrt{2}} + \frac{(-q)(-q)}{a} \right)$ $U_{initial} = \frac{1}{4\pi\epsilon_0} \left( \frac{q^2}{a} - \frac{q^2}{a\sqrt{2}} - \frac{q^2}{a} - \frac{q^2}{a} - \frac{q^2}{a\sqrt{2}} + \frac{q^2}{a} \right)$ $U_{initial} = \frac{q^2}{4\pi\epsilon_0 a} \left( 1 - \frac{1}{\sqrt{2}} - 1 - 1 - \frac{1}{\sqrt{2}} + 1 \right)$ $U_{initial} = \frac{q^2}{4\pi\epsilon_0 a} \left( (1 - 1 - 1 + 1) - \left(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}\right) \right)$ $U_{initial} = \frac{q^2}{4\pi\epsilon_0 a} \left( 0 - \frac{2}{\sqrt{2}} \right) = \frac{q^2}{4\pi\epsilon_0 a} (-\sqrt{2}) = -\frac{\sqrt{2}q^2}{4\pi\epsilon_0 a}$

The final state is when charges C and D are at infinity. Charges A and B remain at their original positions. The potential energy of the system in the final state is the potential energy of the remaining charges A and B. $U_{final} = \frac{q_A q_B}{4\pi\epsilon_0 r_{AB}} = \frac{(+q)(+q)}{4\pi\epsilon_0 a} = \frac{q^2}{4\pi\epsilon_0 a}$

The energy required to take charges C and D to infinity is the work done by an external agent, which is equal to the change in potential energy of the system: $W_{ext} = U_{final} - U_{initial}$ $W_{ext} = \frac{q^2}{4\pi\epsilon_0 a} - \left( -\frac{\sqrt{2}q^2}{4\pi\epsilon_0 a} \right)$ $W_{ext} = \frac{q^2}{4\pi\epsilon_0 a} + \frac{\sqrt{2}q^2}{4\pi\epsilon_0 a}$ $W_{ext} = \frac{q^2}{4\pi\epsilon_0 a} (1 + \sqrt{2})$