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Solve it, 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

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

 
Answers (1)
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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}       ,           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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