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Question: Two charges 9*e* and 3*e* are placed at a distance *r*. The distance of the point where the electric...

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

r(3+1)\frac{r}{\left( \sqrt{3} + 1 \right)}from 9e charge

B

r1+13\frac{r}{1 + \sqrt{\frac{1}{}3}} from 9e charge

C

r(1−3)\frac{r}{\left( 1 - \sqrt{3} \right)}from 3e charge

D

r1+13\frac{r}{1 + \sqrt{\frac{1}{}3}} from 3e charge

Answer

r(1−3)\frac{r}{\left( 1 - \sqrt{3} \right)}from 3e charge

Explanation

Solution

Let third charge be placed at a distance x1x_{1} from +4q charge as shown

Now x1=L1+q4q=2L3x_{1} = \frac{L}{1 + \sqrt{\frac{q}{4q}}} = \frac{2L}{3} ⇒\Rightarrow x2=L3x_{2} = \frac{L}{3}

For equilibrium of q, Q=+4q(L/3L)2=4q9⇒Q=−4q9Q = + 4q\left( \frac{L/3}{L} \right)^{2} = \frac{4q}{9} \Rightarrow Q = - \frac{4q}{9}.