The potential at a point x (measured in mu m) | vedclass

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(A) Given the potential function $V(x) = \frac{20}{x^2 - 4} 	ext{ volt}$.The electric field $E$ is related to the potential $V$ by the relation $E =

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

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(A) Given the potential function $V(x) = \frac{20}{x^2 - 4} 	ext{ volt}$.The electric field $E$ is related to the potential $V$ by the relation $E = -\frac{dV}{dx}$.$E = -\frac{d}{dx} \left( 20(x^2 - 4)^{-1} ight) = -20 \cdot (-1)(x^2 - 4)^{-2} \cdot (2x) = \frac{40x}{(x^2 - 4)^2}$.Now,substituting $x = 4 \mu m$ into the expression for $E$:$E = \frac{40(4)}{(4^2 - 4)^2} = \frac{160}{(16 - 4)^2} = \frac{160}{12^2} = \frac{160}{144}$.Simplifying the fraction by dividing both numerator and denominator by $16$,we get $E = \frac{10}{9} 	ext{ V/}\mu m$.Since the result is positive,the electric field is in the $+ve \ x$ direction.

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

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