Question

The potential at a point $ x $ (measured in pm) 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 \mu m $ is given by

• A. $ \left(\frac{10}{9}\right) volt/ \mu m $ in the $ +ve\,x- $ direction
• B. $ \left(\frac{5}{3}\right) volt/ \mu m $ in the $ -ve \,x- $ direction
• C. $ \left(\frac{5}{3}\right) volt/ \mu m $ in the $ +ve \,x- $ direction
• D. $ \left(\frac{20}{9}\right) volt/ \mu m $ in the $ -ve \,x- $ direction

Answer 1

Answer: (A)

Given, $V(x) = \frac{20}{(X^2 -4)}$ volt
We know that
$E = -\frac{dV}{dx}$
$ = - \frac{d}{dx}\left[\frac{20}{\left(X^{2} -4\right)}\right]$
$ = \frac{20\times (2x)}{(X^2 - 4)^2}$ volt
At $ x = 4\,\mu m$
$ v = \frac{20\times (2\times 4)}{[(4)^2 -4]} = \frac{10}{9} \,V\,\mu m$
Since, $V$ decreases as $x$ increases, it follows from
$E = -\frac{dV}{dx}$ that $E$ is along the $+$ ve $x$-direction.