Question

Two particles with charges +5e and -5e are initially very far apart (effectively an infinite distance apart). They are then fixed at positions that are 7.71 x 10^-10 m apart. What is EPE_final - EPE_initial, which is the change in the electric potential energy?

Answer 1

Since the two charges are far apart

the electric potential energy is given by

$EPE = \frac{kq_1q_2}{r}$

so initially the distance r = infinite

$EPE_i = \frac{kq_1q_2}{r} = 0$

Now final position

$EPE_f = \frac{9*10^9*5e*(-5e)}{7.7*10^{-10}}$

Now we have