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Electrostatic Potential And Capacitance

Question
CBSEENPH12037294

Two charges 2μC and –2μC are placed at points A and B, 6 cm apart.
(a) Identify an equipotential surface of the system.
(b) What is the direction of electric field at every point on this surface?

Solution

Given two charges placed at points A and B.
 q1 = 2μC = 2 × 10-6Cq2 = -2μc = -2×10-6Cr =6 cm= 0.06 m



(a) Potential will be zero due to both charges at equipotential surface

   14πε0q1x+q2(0.06-x) =0

                q1x = -q2(0.06-x)

         2×10-6x = --2×10-6(0.06)-x

                     x = 0.06 - x
                             x=0.062 = 0.03 m

i.e., the plane normal to AB and passing through its mid-point has zero potential everywhere.

(b) The direction of electric field is normal to the plane in the AB direction.

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