Question

The length of each side of the ABCD square is 2m. Charge at point A is + 6C and charge at point B is + 8C. How much work is required to take two electrons from point C to point D?

Answer 1

The work required to move from point C to point D is $A = q(\varphi_C - \varphi_D)$, where here $q = 2 e$ is the charge and $\varphi_C, \varphi_D$ are potentials due to electric field created by charges A, B at points C, D respectively. Let us denote the length of the side of the square $l = 2 m$.

Potential due to point charge is $\varphi = \frac{k q }{r}$, ($k = 9 \cdot 9^{10} \frac{N m^2}{C^2}$ is Coulomb constant), and it is additive, hence $\varphi_C = k \left( \frac{q_1}{l} + \frac{q_2}{\sqrt{2} l} \right)$, $\varphi_D = k \left( \frac{q_1}{\sqrt{2} l} + \frac{q_2}{l} \right)$.

Therefore, the work is $A = 2 e (\varphi_C - \varphi_D) = 2 e k\left[ q_1\left( \frac{1}{l} - \frac{1}{\sqrt{2} l}\right) + q_2\left( \frac{1}{\sqrt{2} l} - \frac{1}{l}\right)\right] \approx 8.43 \cdot 10^{-10} J$