Question

The potential at a point x (measured in 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 $x=4m$ is given by :

• A. $5/3$ Volt/m and in the $-ve$ x direction
• B. $5/3$Volt/m and in the $+ve$ x direction
• C. $10/9$ Volt/m and in the $-ve$ x direction
• D. $10/9$ Volt/m and in the $+ve$ x direction

Answer 1

Answer: (D)

$10/9$ Volt/m and in the $+ve$ x direction

The relation between Electric field and electrostatic potential is given by

$\overrightarrow{E}=-\nabla V$

In one dimension, $\overrightarrow{E}=-\frac{dV}{dx}$

Given $V=\frac{20}{x^2-4}$

Thus, $E=-\frac{dV}{dx}=\frac{40x}{(x^2-4{)}^2}$

Thus Field at $x=4\text{m}$ is given by $E=\frac{40\times 4}{(4^2-4{)}^2}=+\frac{10}{9}\text{V/m}$

The field is in the positive $x$ direction