Question

The electric potential $V$ (in volt ) varies with distance $x$ (in meter) according to the relation $V=5+4x^2$. The force experienced by a negative charge of $2\mu \text{C}$ located at $x=0.5\text{m}$ is

• A. $4\times {10}^{-6}\text{N}$
• B. $2\times {10}^{-6}\text{N}$
• C. $8\times {10}^{-6}\text{N}$
• D. $6\times {10}^{-8}\text{N}$

Answer 1

The correct option is C $8\times {10}^{-6}\text{N}$

Given, $V=5+4x^2$

$E=-\frac{dV}{dx}=-\frac{d}{dx}[5+4x^2]$

$E=-(0+8x)=-8x$

At $x=0.5\text{m}$

$|E|=|-8\times 0.5|=4{\text{NC}}^{-1}$

Electric force
$\therefore F=Eq=\left(4\right)(2\times {10}^{-6})=8\times {10}^{-6}\text{N}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, (c) is the correct answer.