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Question
Two charges 2 μC and −2 µC are placed at points A and B 6 cm apart.
- Identify an equipotential surface of the system.
- What is the direction of the electric field at every point on this surface?
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Solution
Given: q1 = 2 μC = 2 × 10−6 C
q2 = −2 µC = 2 × 10−6 C
r = 6 cm = 0.06 m
(a) Potential will be zero due to both charges at the equipotential surface.
`1/(4piε_0)[q_1/x + q_2/((0.06 - x))] = 0`
`q_1/x = -q_2/((0.06 - x))`
`(2 xx 10^-6)/x = -((-2 xx 10^-6))/([(0.06) - x])`
x = 0.06 − x
x + x = 0.06
2x = 0.06
`x = 0.06/2`
x = 0.03 m
i.e., the plane normal to AB and passing through its mid-point has zero potential everywhere.
(b) The direction of the electric field at every point on this surface is normal to the plane in the direction of AB.
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