Two electric charges of 9µC and –3µC are placed 0.16 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 9µC charge is

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0.08 m

0.12 m

0.14 m

0.06 m

answer is B.

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Detailed Solution

For zero potential point P
$X = \frac{d}{1 + \frac{Q_{2}}{Q_{1}}} = \frac{0 .16}{1 + \frac{9 X 10^{- 6}}{3 \times 10^{- 6}}} = 0 .04 m from - 3 μC$
or 0.12cm from 9µC

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