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A positive charge of $q_{1}$ is located $3 m$ to the left of a negative charge $- q_{2}$ . Magnitude of charge $q_{1}$ greater than magnitude of negative charge $q_{2}$ . The net electric field intensity is zero at a point on the line passing through charges and $1 m$ to the right of $- q_{2}$ . On this line there are also points where the potential is zero. Location of these two points relative to the negative charge is/are

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at $0.2 m$ left of negative charge

at $0.2 m$ right of negative charge

at $\frac{3}{17} m$ left of negative charge

at $\frac{3}{17} m$ right of negative charge

answer is B, C.

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Detailed Solution

$\frac{k q_{2}}{1^{2}} = \frac{k q_{1}}{16} \Rightarrow q_{1} = 16 q_{2}$

Potential can be zero both inside and outside the if on the line joining then

For a point P lying outside the two charges

$V_{at P} = 0 \Rightarrow - \frac{k q_{2}}{x} + \frac{k q_{1}}{3 + x} = 0$

$\Rightarrow + \frac{q_{2}}{x} = \frac{16 q_{2}}{3 + x} \Rightarrow 3 + x = 16 x \Rightarrow 3 = 15 x \Rightarrow x = \frac{1}{5} = 0.2 m F o r a p o i n t P l y i n g b e t w e e n t h e c h a r g e s V_{at P} = 0$

$\Rightarrow \frac{k q_{1}}{3 - x} = \frac{k q_{2}}{x} \Rightarrow 16 x = 3 - x \Rightarrow x = \frac{3}{17} meter$

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