First slide
Capacitance
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/x24volt.
The electric field E at x = 4 μm is given by:
 

Easy
Solution

E=dVdx=20ddx1x24=-40xx242=+40×4(164)2=+109

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