Question

Two point charges Q1 = 44 C and Q2 = -5  C are separated by a distance of 17 cm. Find the electric potential energy of the system

Answer 1

Potential energy is given by: PE = $\frac{k * Q1 * Q2}{r}$

where Q1, Q2 and r are the charge Q1, charge Q2 and distance between the two charges respectively.

Given, Q1 = 44C

Q2 = -5 C

r = 17 cm = 0.17 meter

and k = 8.99 * $10^9 \frac{Nm^2}{c^2}$

Therefore, PE = $\frac{8.99*10^9*44*(-5)}{0.17}$

On solving above, we get,

PE = -1.163 * $10^{13} J$