Question

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

• A. at $x=\frac{r}{3}$ only.
• B. at $x=-\frac{r}{5}$ only.
• C. both at $x=\frac{r}{3}$ and at $x=-\frac{r}{5}$.
• D. there exists 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 $x=-\frac{r}{5}$.

From the data given in the question, we draw schematic diagram as shown below.


Let the potential be zero at $A$.
$\therefore \frac{Q}{4\pi {\epsilon }_0(r-x_2)}-\frac{Q}{16\pi {\epsilon }_0{x}_2}=0$

$\Rightarrow x_2=\frac{r}{5}$ (towards left of origin)

Let the potential be zero at $B$

$\therefore \frac{Q}{4\pi {\epsilon }_0(r+x_1)}-\frac{Q}{16\pi {\epsilon }_0{x}_1}=0$

$\Rightarrow x_1=\frac{r}{3}$ (towards right of origin)

From the figure, it is clear that at point $A$, electric field due to $+Q$ is towards $A$ and electric field due to $-\frac{Q}{4}$ is away from it.

For electric field to be zero at point $A$

$\frac{Q}{4\pi {\epsilon }_0(r-x_2{)}^2}=\frac{Q}{16\pi {\epsilon }_0{x}_2^2}$

$\Rightarrow 4x_2^2=r^2+x_2^2-2r{x}_2$

$\Rightarrow 3x_2^2+2r{x}_2-r^2$

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

Similarly, for electric field to be zero at point $B$.

$\frac{Q}{4\pi {\epsilon }_0(r+x_1{)}^2}=-\frac{Q}{16\pi {\epsilon }_0{x}_1^2}$

$\Rightarrow 5x_1^2+2r{x}_1+r^2$

roots of the above quadratic equation are imaginary.

$\therefore$ we can conclude that, there exists only one point on the axis where the electric field is zero.

Hence, option (c) is the correct answer.