Question

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

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

Answer 1

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

Two charges $9e$ and $3e$ separation $=r$

The distance of the point where the electric field intensity will be $=0$

Let, distance at which electric field intensity $=0$ be ${}^{\prime }{a}^{\prime }$ from $9e$ charge. So, Putting formula, the electric field for $3e$ and $9e$ will be equal.

$\frac{1}{4\pi {ϵ}_0}\frac{9e\times 1}{a^2}=\frac{1}{4\pi {ϵ}_0}\times \frac{3e}{{(r-a)}^2}$

$\Rightarrow 3{(r-a)}^2=a^2$

Solving we get,

$\sqrt{3}(r-a)=a$

or, $\sqrt{3}r-\sqrt{3}a=a$

or, $\sqrt{3}r=a(1+\sqrt{3})$

or, $a=\frac{r\sqrt{3}}{1+\sqrt{3}}$

from $9e$ charge.