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The potential at a point $x$ (measured in $\mu \mathrm{m}$ ) due to some charges situated on the $\mathrm{x}$-axis is given by $\mathrm{V(x)=20 /\left(x^2-4\right)} \mathrm{~Volts}$. The electric field $\mathrm{E}$ at $\mathrm{x=4 ~\mu m}$ is given by
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The correct answer is:
$10 / 9 \mathrm{~Volt} / \mu \mathrm{m}$ and in the $+ve$ $x$ direction.
$10 / 9 \mathrm{~Volt} / \mu \mathrm{m}$ and in the $+ve$ $x$ direction.
$\mathrm{V_x=\frac{20}{x^2-4}}$
$\mathrm{E=-\frac{d V}{d x}=\frac{20}{\left(x^2-4\right)^2}(2 x-0)=\frac{160}{144}=\frac{10}{9}}$
$\mathrm{E=-\frac{d V}{d x}=\frac{20}{\left(x^2-4\right)^2}(2 x-0)=\frac{160}{144}=\frac{10}{9}}$
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