The point charges +4 μC and –6 μC are separated by a distance of 20 cm in air. At what point of the line joining the two charges is the electric potential zero?

Solution

Let, the point charges be given by q_A and q_B.
Charge, q_A= $+ 4 \times 10^{- 6} C$
Charge, q_B= $- 6 \times 10^{- 6} C$
Distance betwen charges, d=20 cm
Let, the point where the potential is 0 be O.
Assume, the distance of O from point charge A be x.
then, distance of O from point charge B will be (.2-x)

Since the electric potential is zero at point O therefore,
Potential at point O due to q_A=Potential at point O due to q_B
$\Rightarrow V_{A} + V_{B} = 0 \frac{1}{4 {πε}_{\circ}} \frac{4 \times 10^{- 6}}{x} = - \frac{1}{4 {πε}_{\circ}} \frac{- 6 \times 10^{- 6}}{0.2 - x} \frac{4}{x} = \frac{6}{0.2 - x} 0.8 - 4 x = 6 x 0.8 = 10 x \Rightarrow x = 0.08 m$

Therefore, the point O is at a distance of 0.08 m from charge placed at A.

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Electrostatic Potential And Capacitance

The point charges +4 μC and –6 μC are separated by a distance of 20 cm in air. At what point of the line joining the two charges is the electric potential zero?

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