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. $4\times {10}^{-6}\text{N}$
• B. $2\times {10}^{-6}\text{N}$
• C. $6\times {10}^{-6}\text{N}$
• 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
$E=-\frac{\partial V}{\partial x}$

$\Rightarrow E=-\frac{\partial }{\partial x}[5+4x^2]$

$\Rightarrow E=-8x$

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

The force on charge $q$ is given by, $F=qE$

Substituting the data obtained we get,

$F=(-2\times {10}^{-6})\times (-4)$

$\Rightarrow F=8\times {10}^{-6}\text{N}$

Hence, option (d) is the correct answer