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Q.

Two electric charges 12 μC and -6 μC are 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. Toe 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

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

 

                    V=14πε0-6×10-6x+14πε012×10-6(0.2+x)=0

⇒               x=0.20 m

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