First slide
NA
Question

The potential at a point x (measured in μm) due to some charges situated on the X-axis is given by V(x)=20/x24 volt. The electric field E at x = 4 μm is given by

Easy
Solution

E=dVdx
Given V(x)=20x24
 E=20x242×2x
At x = 4,we have
E=20(4)242×2×4=109
Hence at x = 4 μm, the electric field will be (10/9) along positive X-direction.

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