The potential at a point x (measured in μ | vedclass

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Question

Here, $V(x)=\frac{20}{x^{2}-4}$ $volt$

We know that $E=-\frac{d V}{d x}=-\frac{d}{d x}\left(\frac{20}{x^{2}-4}\right)$

or, $E=+\frac{40 x}{\left

Vedclass · Accepted Answer

$\frac{{10}}{9}$ $\frac{V}{{\mu m}}$ and in the $+ve\ x$ direction

Vedclass · Answer

Here, $V(x)=\frac{20}{x^{2}-4}$ $volt$

We know that $E=-\frac{d V}{d x}=-\frac{d}{d x}\left(\frac{20}{x^{2}-4}ight)$

or, $E=+\frac{40 x}{\left(x^{2}-4ight)^{2}}$

At $x=4\, \mu \mathrm{m}$

$E=+\frac{40 	imes 4}{\left(4^{2}-4ight)^{2}}=+\frac{160}{144}=+\frac{10}{9}$ $volt / \mu \mathrm{m}$

Positive sign indicates that $\vec{E}$ is in $+ ve\, x-$ direction.

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

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