Due to some charges on the x-axis,the potenti | vedclass

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(D) The relation between electric field $E$ and potential $V$ is given by $E = -\frac{dV}{dx}$.Given $V(x) = 20(x^2 - 4)^{-1}$.Differentiating with re

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$10/9 \ V/\mu m$ and in $+ve \ x$-direction

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(D) The relation between electric field $E$ and potential $V$ is given by $E = -\frac{dV}{dx}$.Given $V(x) = 20(x^2 - 4)^{-1}$.Differentiating with respect to $x$:$\frac{dV}{dx} = 20 \cdot (-1) \cdot (x^2 - 4)^{-2} \cdot (2x) = -\frac{40x}{(x^2 - 4)^2}$.Therefore,$E = -\left(-\frac{40x}{(x^2 - 4)^2}\right) = \frac{40x}{(x^2 - 4)^2}$.At $x = 4 \ \mu m$:$E = \frac{40(4)}{(4^2 - 4)^2} = \frac{160}{(16 - 4)^2} = \frac{160}{12^2} = \frac{160}{144}$.Simplifying the fraction by dividing by $16$:$E = \frac{10}{9} \ V/\mu m$.Since the value is positive,the electric field is in the $+ve \ x$-direction.

Due to some charges on the $x$-axis,the potential at a point on the $x$-axis is given by $V(x) = 20/(x^2 - 4) \ V$. The electric field at $x = 4 \ \mu m$ is given by:

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