Question

Two point charges $Q$ and $-\frac{Q}{4}$ placed along the $\text{x-}$axis are separated by a distance $r$. Take $-\frac{Q}{4}$ as origin and it is placed to the right of $Q$. Then the potential is zero

• A. at $x=\frac{r}{4}$ only
• B. at $x=-\frac{r}{5}$ only
• C. both at $x=\frac{r}{3}$ and at $x=-\frac{r}{5}$
• D. there exist two points on the axis, where the electric field is zero.

Answer 1

The correct option is C both at $x=\frac{r}{3}$ and at $x=-\frac{r}{5}$

Let $\text{P}$ and $\text{R}$ are the points at distance $x_1$ and $x_2$ respectively from the origin where $V_{net}=0$.


Equating the net potential at point $\text{P}$ to zero, we get

$\frac{kQ}{r-x_1}-\frac{kQ}{4\left(x_1\right)}=0$

$\Rightarrow \frac{r-x_1}{x_1}=4$

$\Rightarrow r_1-x_1=4x_1$

$\Rightarrow x_1=\frac{r}{5}$

Since, $x_1$ is along negative $\text{x}-$axis

$\therefore x=-\frac{r}{5}$

Equating the net potential at point $\text{R}$ to zero, we get

$\frac{kQ}{r+x_2}-\frac{kQ}{4x_2}=0$

$\Rightarrow \frac{r+x_2}{x_2}=4$

$\Rightarrow x_2=\frac{r}{3}$

Since, $x_2$ lies on the positive $\text{x-}$ axis,

$\therefore x=\frac{r}{3}$

Hence, option (c) is correct answer.