Figures $(a)$ and $(b)$ show the field lines of a positive and negative point charge respectively.
$(a)$ Give the signs of the potential difference $V_{P}-V_{Q}$ and $V_{B}-V_{A}$.
$(b)$ Give the sign of the potential energy difference of a small negative charge between the points $Q$ and $P$,and $A$ and $B$.
$(c)$ Give the sign of the work done by the field in moving a small positive charge from $Q$ to $P$.
$(d)$ Give the sign of the work done by the external agency in moving a small negative charge from $B$ to $A$.
$(e)$ Does the kinetic energy of a small negative charge increase or decrease in going from $B$ to $A$?

(A-D) As $V \propto \frac{1}{r}$,the potential is higher closer to the positive charge. Thus,$V_{P} > V_{Q}$,so $(V_{P}-V_{Q})$ is positive. For the negative charge,the potential is more negative closer to the charge. Thus,$V_{B} > V_{A}$ (since $V_{A}$ is more negative),so $(V_{B}-V_{A})$ is positive.
$(b)$ Potential energy $U = qV$. For a negative charge $q < 0$,if $V_{P} > V_{Q}$,then $U_{P} < U_{Q}$. The difference $(U_{P}-U_{Q})$ is negative. Similarly,for the negative charge,since $V_{B} > V_{A}$,$U_{B} > U_{A}$,so $(U_{A}-U_{B})$ is negative.
$(c)$ $A$ positive charge experiences a repulsive force from the positive source charge. Moving it from $Q$ to $P$ (closer) requires work against the field. Thus,the work done by the field is negative.
$(d)$ $A$ negative charge is attracted to the positive source charge but repelled by the negative source charge. Moving a negative charge from $B$ to $A$ (closer to the negative source) requires work against the repulsive force. Thus,the work done by the external agency is positive.
$(e)$ As the negative charge moves from $B$ to $A$,it moves closer to the negative source charge,experiencing a stronger repulsive force. This force does negative work,causing the kinetic energy to decrease.