The equation of an equipotential line in an electric field is y=2x, then the electric field strength vector at (1,2) may be :
A
4^i+3^j
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B
4^i+8^j
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C
8^i+4^j
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D
−8^i+4^j
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Solution
The correct option is D−8^i+4^j Now equation of equipotential surface is y=2x Now electric field along the euipotential surface should be zero therefore angle made by equipotential surface with x-axis is tan−1(2) Now since net electric field should be perpendicular to the equipotential surface therefore for any electric field which makes an angle tan−1(−1/2) with x-axis can be the electric field at point (1,2) which is true only for option (D)
because for two perpendicular line, product of their slope should be equal to -1 i.e., m1×m2=−1