Figure shows, two equipotential surfaces V1 and V2. The component of electric field (in SI units) along x and y direction will be -
A
100^iand100^j
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B
−100^iand100^j
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C
100^iand−100^j
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D
−100^iand−100^j
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Solution
The correct option is B−100^iand100^j Electric field lines are perpendicular to the equipotential surface and are directed from high potential to low potential. ∴ Electric field will have one component in negative x direction and one component in +y direction. Ex=−ΔVΔx=−20.02=−100 Vm−1 Ey=−ΔVΔy=20.02=100 Vm−1