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Question

Physics Question on electrostatic potential and capacitance

Two electric charge 12 μ\muC and -6 μ\muC 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, The distance of P from -6 μ\muC charge is

A

0.10 m

B

0.15 m

C

0.20 m

D

0.25 m

Answer

0.20 m

Explanation

Solution

Let PP be at distance xx from 6μC-6 \mu C charge.
We know, potential V=kqrV =\frac{ kq }{ r }
Given, potential at P:VP=0P : V _{ P }=0
k×12×106(0.2+x)+k×(6)×106x=0\Rightarrow \frac{ k \times 12 \times 10^{-6}}{(0.2+ x )}+\frac{ k \times(-6) \times 10^{-6}}{ x }=0
or, 120.2+x=6x\frac{12}{0.2+x}=\frac{6}{x}
or, 2x=0.2+x2 x =0.2+ x
or, x=0.2mx=0.2 \,m