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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

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a

0.10 m

b

0.15 m

c

0.20 m

d

0.25 m

answer is C.

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Detailed Solution

Point P will lie near the charge which is smaller in magnitude i.e., - 6 μC. Hence, potential at PV=14πε0-6×10-6x+14πε012×10-6(0.2+x)=0⇒ x=0.2 m

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