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The potential at a point x (measured in $μ$ m) due to some charges situated on the x-axis is given by $V (x) = 20 / (x^{2} - 4)$ volt.
The electric field E at x = 4 $μ$ m is given by:

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5/3 volt/μm and in the +ve x direction.

10/9 volt/μm and in the -ve x direction.

10/9 volt/μm and in the +ve x direction.

5/3 volt/μm and in the -ve x direction.

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detailed solution

Correct option is C

E=−dVdx=−20ddx1x2−4=-−40xx2−42=+40×4(16−4)2=+109

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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/x2−4volt.The electric field E at x = 4 μm is given by:

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Ready to Test Your Skills?

free study material

The potential at a point x (measured in $μ$ m) due to some charges situated on the x-axis is given by $V (x) = 20 / (x^{2} - 4)$ volt.
The electric field E at x = 4 $μ$ m is given by: