Question

Figure shows the field lines due to a negative point charge. Give the sign of the potential energy difference of a small negative charge between the points A and B.

Answer 1

$U=\frac{1}{4\pi ϵ0}.\frac{q_1{q}_2}{r}$

Since $r_A<r_B$

$\therefore \frac{k{q}_1{q}_2}{r_A}>\frac{k{q}_1{q}_2}{r_B}$

$\therefore$ $U_A>U_B$

Therefore, $U_A-U_B$ is positive.