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(C) For a positive charge $+q$,the electric potential at a distance $r$ is given by $V = \frac{kq}{r}$. Since $V \propto \frac{1}{r}$,the potential is

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$+Ve, +Ve$

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(C) For a positive charge $+q$,the electric potential at a distance $r$ is given by $V = \frac{kq}{r}$. Since $V \propto \frac{1}{r}$,the potential is higher closer to the charge.For the positive charge,point $Q$ is closer to the charge than point $P$,so $r_Q  V_P$,therefore $V_Q - V_P > 0$ (positive).For a negative charge $-q$,the electric potential is given by $V = \frac{k(-q)}{r} = -\frac{kq}{r}$. Here,the potential is more negative (lower) closer to the charge.For the negative charge,point $A$ is closer to the charge than point $B$,so $r_A  0$ (positive).Thus,both potential differences are positive.

The figure shows the electric field lines of a positive and a negative charge,respectively. Determine the sign of the potential difference $V_Q - V_P$ and $V_B - V_A$.

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