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Question

Some equipotential surfaces are shown in figure given below. The magnitude of electric field will be


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Solution

Since the equipotential planes are equally spaced, the electric field will be uniform.
The direction of E will be perpendicular to the equipotential surfaces, and it will flow from high potential to low potential as shown in the figure.


From relation,

ΔV=Ed

Where,
d = Perpendicular distance between two equipotential surfaces

60=E(10sin37×102)

E=6010sin37×102=60×10010×(35)

E=1000 V/m

Therefore, option (c) is the correct answer.

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