Question

Two equal positive charges are kept at points $\text{A}$ and $\text{B}$. The electric potential at the points between $\text{A}$ and $\text{B}$ (excluding the points $A\text{and}B$) is studied while moving from $\text{A}$ to $\text{B}$. Then the electric potential will

• A. continously increase
• B. continously decrease
• C. first increase and then decrease
• D. first decrease and then increase

Answer 1

The correct option is D first decrease and then increase

Suppose the charges are $+\text{q}$ and kept at $r_0$ distance apart.


Now, let us take a point at distance $x$ from $\text{A}$.
The net electric potential at point $\text{P}$,
$V_P=\frac{kq}{x}+\frac{kq}{r_0-x}$
$\Rightarrow V_P=\frac{kq{r}_0-kqx+kqx}{x(r_0-x)}$
$\text{Or,}{V}_P=\frac{kq{r}_0}{x(r_0-x)}....\left(i\right)$
At point $\text{C}$,
$x=\frac{r_0}{4}$
$C$ is taken as the midpoint of $AO$
$V_C=\frac{16kq}{3r_0}=5.3\times \frac{kq}{r_0}$

Similarly at point $\text{O}$,
$x=\frac{r_0}{2}$
$AO$ is taken as the mid-point of $AB$,
$\Rightarrow V_O=\frac{4kq}{r_0}$
(from Eq.$\left(i\right)$)
$\therefore V_O=4\times \frac{kq}{r_0}$

At point $\text{D}$,
$x=AD=\frac{3}{4}{r}_0$
$\Rightarrow V_D=\frac{16kq}{3r_0}=5.3\times \frac{kq}{r_0}$

From above observations of potential at points $C,O,D$ we can see that while moving from $\text{A}$ to $\text{B}$, the electric potential first decreases and then increases.
Therefore, option$\left(d\right)$ is correct choice.