Question

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

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

Answer 1

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

Given that,
Electric potential, $V=5+4x^2$
Charge, $q=-2\mu \text{C}=-2\times {10}^{-6}\text{C}$

The relation between electric field and potential is given by
$E=-\frac{dV}{dx}$
$\Rightarrow E=-\frac{d}{dx}[5+4x^2]$
$\Rightarrow E=-8x$

At $x=0.5\text{m}$
$E=-8\times 0.5=-4{\text{NC}}^{-1}$

The force on charge $q$ is
$\Rightarrow F=qE$
$\Rightarrow F=(-2\times {10}^{-6})\times (-4)$
$\Rightarrow F=8\times {10}^{-6}\text{N}$

Hence, option (d) is correct answer