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Two charges $3 \times 10^{- 8} C$ and $- 2 \times 10^{- 8} C$ are separated by 15 cm in air. The distance of point where electric potential is zero is ( electric potential at infinity is zero)

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9 cm from positive charge in between the charges

9 cm from both the charges on perpendicular bisector of the line joining the in between the charges

30 cm from positive charge outside the charges

9 cm from negative charge in between the charges

answer is B.

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Detailed Solution

$∴ \frac{3}{x} = \frac{2}{(15 - x)} \Rightarrow 2 x = 45 - 3 x$

$x = 9 c m f r o m 3 \times 10^{- 8} charge$

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