Question

Two equal positive charges are kept at points A and B. The electric potential at the points between A and B (excluding these points) is situated while moving from A to B. The potential
(a) continuously increases
(b) continuously decreases
(c) increases then decreases
(d) decreases then increases

Answer 1

(d) decreases then increases

Let the distance between the points A and B be r.

Let us take a point P at a distance x from A (x < r).

Electric potential V at point P due to two charges of equal magnitude q is given by
$$\begin{gathered} V=\frac{q}{4\pi {\in }_{0}x}+\frac{q}{4\pi {\in }_0(r-x)} \\ \Rightarrow V=\frac{qr}{4\pi {\in }_{0}x(r-x)} \end{gathered}$$
Now, differentiating V with respect to x, we get

$\frac{\mathrm{d}V}{\mathrm{d}x}=\frac{-qr(r-2x)}{4\pi {\in }_0{x}^2(r-x{)}^2}$
Therefore, x = r/2.
It can be observed that $\frac{\mathrm{d}V}{\mathrm{d}x}<0$ for x < r/2. Thus, the potential is decreasing first. At
x = r/2, the potential is minimum.
As $\frac{\mathrm{d}V}{\mathrm{d}x}>0$ for x > r/2, the potential is increasing after x = r/2.