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Figures $(a)$ and $(b)$ show the field lines of a positive and negative point charge respectively
$(a)$ Give the signs of the potential difference $V_{ P }-V_{ Q } ; V_{ B }-V_{ A }$
$(b)$ Give the sign of the potential energy difference of a small negative charge between the points $Q$ and $P ; A$ 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 $A?$
Solution
$(a)$ As $V \propto \frac{1}{r}, V_{P}>V_{Q}$. Thus, $\left(V_{P}-V_{Q}\right)$ is positive. Also $V_{B}$ is less negative than $V_{A} .$ Thus, $V_{B}>V_{A}$ or $\left(V_{B}-V_{A}\right)$ is positive.
$(b)$ A small negative charge will be attracted towards positive charge. The negative charge moves from higher potential energy to lower potential energy. Therefore the sign of potential energy difference of a small negative charge between $Q$ and $P$ is positive. Similarly, $(P.E.)_{ A } > (P.E.)_{ B }$ and hence sign of potential energy differences is positive.
$(c)$ In moving a small positive charge from $Q$ to $P$, work has to be done by an external agency against the electric field. Therefore, work done by the field is negative.
$(d)$ In moving a small negative charge from $B$ to $A$ work has to be done by the external agency. It is positive.
$(e)$ Due to force of repulsion on the negative charge, velocity decreases and hence the kinetic energy decreases in going from $B$ to $A.$
