The potential at a point x (measured in) due to some charges situated on the x -axis is given by V(x) = 20/(x² - 4) volt The electric field E at x = 4 is given by (a) (10/9) volt / and in the +ve x direction (b) (5/3) volt/ and in the -ve x direction (c) (5/3) volt/ and in the +ve x direction (d) (10/9) volt/ and in the -ve x direction

Answer: (a). E = -dV/dx = -d/dx[(20)/(x² - 4)] = (40x)/((x² - 4)²) At x = 4: E = (160)/(144) = 10/9 V/, positive. So the answer is (a) 10/9 V/ μ m along +x.

The potential at a point x (measured in) due to some charges situated… | JEE Main 2007 (AIEEE)

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Electric Potential & Equipotentials

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

The electric field $E$ at $x = 4\ \mu\text{m}$ is given by

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Answer: (a)

$E = -\frac{dV}{dx} = -\frac{d}{dx}\left[\frac{20}{x^2 - 4}\right] = \frac{40x}{(x^2 - 4)^2}$
At $x = 4$ : $E = \frac{160}{144} = \frac{10}{9}\,\text{V}/\mu\text{m}$ , positive.
So the answer is (a) $\frac{10}{9}$ V/ $\mu$ m along $+x$ .

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