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Q.

The potential at a point x ( measured in µ m) due to some charges situated on the x-axis is given by V(x) = 20/(x2–4) volt, then intensity at x=4µm is given by

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a

5/3 V/µm and the in the +ve x direction

b

10/9 V/µm and in the +ve x direction

c

5/3 V/µm and in the –ve x direction

d

10/9 V/µm and in the –ve x direction

answer is D.

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Detailed Solution

E=−dVdx;E=−x2−40−20(2x−0)x2−42E=20(2x)x2−42=40xx2−42
E=160144=109N/m along the + x-axis.

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