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The potential at a point x (measured in μm) due to some charges situated on the X-axis is given by V(x)=20/x2−4 volt. The electric field E at x = 4 μm is given by

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a

10/9 volt/μm and in the - ve X direction

b

10i9 volt/μm and in the +ve X direction

c

5/3 volt/μm and in the - ve X direction

d

5/3 volt/μm and in the +ve X direction

answer is B.

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Detailed Solution

E=−dVdxGiven V(x)=20x2−4∴ E=20x2−42×2xAt x = 4,we haveE=20(4)2−42×2×4=109Hence at x = 4 μm, the electric field will be (10/9) along positive X-direction.

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