- #1
gracy
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Homework Statement
The potential at a point x(measured in µm)due to some charges situated on the x-axis is given by ##V(x)##=##\frac{20}{x^2-4}##V.The electric field ##E## at x=4µm is ?
Homework Equations
##E##=-##\frac{∂V}{∂r}##
The Attempt at a Solution
Actually I have solution but still I am unable to understand.So ,instead of my attempt at a solution .I 'll post solution itself.
##Ex##=-##\frac{∂V}{∂x}## (1)
= -##\frac{d}{dx}(\frac{20}{x^2-4})## (2)
=##\frac{40x}{(x^2-4)^2}## (3)
##Ex## at x=4µm =##\frac{10}{9}##V/µm
Here I am not able to understand how it proceeded from (2) to (3)