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Q.

Some equipotential surfaces are shown in fig. The electric field strength is

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a

100V/m along -axis

b

4003v/m at an angle 120° with x-axis

c

100V/m along yaxis

d

400V/m at an angle 120° with x-axis

answer is C.

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Detailed Solution

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E=dvdx

E=-10-205×10-2× sin300V/m=400 V/m

as the direction of electric field is perpendicular to equipotential surface  and along the direction of decreasing potential so it makes an angle  1200 with the positive x axis

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