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Q.

Two electric charges 12 μC and-6 μCare placed 20 cm apart in air. There will be a point P on the line joining these charges and outside the region between them, at which the electric potential is zero. The distance of P from -6 μC charge is

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a

0.15 m

b

0.25 m

c

0.10 m

d

0.20 m

answer is C.

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Detailed Solution

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Point P will lie near the charge which is smaller in magnitude i.e., - 6 μC. Hence, potential at P

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$V = \frac{1}{4 {πε}_{0}} \frac{(- 6 \times 10^{- 6})}{x} + \frac{1}{4 {πε}_{0}} \frac{(12 \times 10^{- 6})}{(0.2 + x)} = 0 \Rightarrow x = 0.2 m$

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Two electric charges 12 μC and-6 μCare placed 20 cm apart in air. There will be a point P on the line joining these charges and outside the region between them, at which the electric potential is zero. The distance of P from -6 μC charge is