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The potential at a point x (measured in μ m ) due to some charges situated on the x-axis is given by V(x)=20 / (x^2-4 ) ~Volts. The electric field E at x=4 ~μ m is given by

10 / 9 ~Volt / μ m and in the +ve x direction.

5 / 3 ~Volt/ μ m and in the -ve x direction

5 / 3 ~Volt/ μ m and in the +ve x direction.

10 / 9 ~Volt / μ m and in the -ve x direction

Electrostatics

JEE Main level, single correct, medium

227 of 311

Question 227 of 311

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

Show answer and solution

Answer: (D) $10 / 9 \mathrm{~Volt} / \mu \mathrm{m}$ 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}}

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