Question

Two charges 9e and 3e are placed at a distance r. The distance of the point where the electric field intensity will be zero, is:

• A. $\frac{r}{(1+\sqrt{3})}$ from 9e charge
• B. $\frac{r}{(1+\sqrt{\frac{1}{3}})}$ from 9e charge
• C. $\frac{r}{(1-\sqrt{3})}$ from 3e charge
• D. $\frac{r}{(1+\sqrt{\frac{1}{3}})}$ from 3e charge

Answer 1

The correct option is B $\frac{r}{(1+\sqrt{\frac{1}{3}})}$ from 9e charge

Let electric field at a point $x$ from the charge $9e$, is zero.

It can be written as,

$\frac{1}{4\pi {\epsilon }_0}\times \frac{9e}{x^2}=\frac{1}{4\pi {\epsilon }_0}\times \frac{3e}{{\left(r-x\right)}^2}$

$3{\left(r-x\right)}^2=x^2$

$\sqrt{3}\left(r-x\right)=x$

$x=\frac{\sqrt{3}}{\left(\sqrt{3}+1\right)}r$

The distance of the point from $9e$ charge where the electric field intensity will be zero is $\frac{\sqrt{3}}{\left(\sqrt{3}+1\right)}r$.