# 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 Option 1) $\frac{r}{\sqrt3 + 1}$ from 9e charge Option 2) $\frac{r}{1 + \sqrt{\frac{1}{3}}}$ from 9e charge Option 3) $\frac{r}{1 - \sqrt3}$ from 3e charge Option 4) $\frac{r}{1 + \sqrt{\frac{1}{3}}}$  from 3e charge

V Vakul

As we have learnt,

Neutral Point /Zero Electric Field -

Due to a system of two like Point Ooint charge.

$\dpi{100} x_{1}= \frac{x}{\sqrt{\frac{Q_{2}}{Q_{1}}}+1}$       ,           $\dpi{100} x_{2}= \frac{x}{\sqrt{\frac{Q_{1}}{Q_{2}}}+1}$

- wherein

Where, x = distance between Q1 and Q2

Suppose neutral point is obtained at a distance $x_1$ from charge 9e and $x_2$ from charge 3e

By using

$x_1 = \frac{x}{1 +\sqrt{\frac{Q_2}{Q_1}}} = \frac{r}{1 +\sqrt{\frac{3e}{9e}}} = \frac{r}{1 +\frac{1}{\sqrt3}}$

Option 1)

$\frac{r}{\sqrt3 + 1}$ from 9e charge

Option 2)

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

Option 3)

$\frac{r}{1 - \sqrt3}$ from 3e charge

Option 4)

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

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