Question

# Equipotential surfaces are shown in fig, then the electric field strength will be A
100 Vm-1 along X-axis
B
100 Vm-1 along Y-axis
C
200 Vm-1 at an angle 1200 wirh X-axis
D
50 Vm-1 at an angle 1800 wirh X-axis
Solution
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#### Using,dV=−→E⋅→drΔV=−E⋅ΔrcsθE=−ΔVΔrcosθE=−(−20−10)10×10−2cos1200=−1010×10−2(−sin300)=−102−1/2=200V/mdirection of E be perpendicular to the equipotential surface i.e., at 1200 with x− axis.Hence,option (C) is correct answer.

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