The electric potential V at any point (x,y,z) in space is given by V=4 x2 volt. The electric field at (1,0,2) m in Vm−1 is

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8, along neg. X-axis

8, along pos. X-axis

16, along neg. X-axis

16, along pos. Z-axis.

answer is A.

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Detailed Solution

E=−dVdr=−ddx(4 x2)=−8 x =−8(1)=−8 V m−1Negative sign indicates E→ is along negative direction of X-axis.

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