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Two electric charges of $6 μ C a n d - 3 μ C$ are placed 0.15 m apart in air. There will be a point P at which electric potential is zero on the line joining the two charges and in between them. The distance of P from $6 μ C$ charge is

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0.14 m

0.1 m

0.08 m

0.06 m

answer is B.

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Detailed Solution

$x = \frac{d}{\frac{q_{2}}{q_{1}} + 1} = \frac{0.15}{\frac{3}{6} + 1} = 0.1 m$

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