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Question

Equipotential surfaces are shown in fig, then the electric field strength will be
1224828_1fdfa6cbc47c432ab0320203eec5d923.png
  1. 100 Vm-1 along X-axis
  2. 100 Vm-1 along Y-axis
  3. 200 Vm-1 at an angle 1200 wirh X-axis
  4. 50 Vm-1 at an angle 1800 wirh X-axis

A
200 Vm-1 at an angle 1200 wirh X-axis
B
100 Vm-1 along X-axis
C
100 Vm-1 along Y-axis
D
50 Vm-1 at an angle 1800 wirh X-axis
Solution
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Using, dV=Edr
ΔV=EΔrcsθ
E=ΔVΔrcosθ
E=(2010)10×102cos1200=1010×102(sin300)=1021/2=200V/m
direction 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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