Question

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

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

Answer 1

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

Given,

Electric potential, $V=5+4x^2$

We know that,

Electric field E is given by,

$\overrightarrow{E}=-(\frac{\partial V}{\partial x}\hat{i}+\frac{\partial V}{\partial y}\hat{j}+\frac{\partial V}{\partial x}\hat{k})$

$\overrightarrow{E}=-(\frac{\partial (5+4x^2)}{\partial x}\hat{i}+\frac{\partial (5+4x^2)}{\partial y}\hat{j}+\frac{\partial (5+4x^2)}{\partial x}\hat{k})$

$\overrightarrow{E}=-8x$

Now, force experienced by a charged particle in an electric field is given by,

$F=qE=(-2)\times {10}^{-6}\times (-8)\times 0.5=8\times {10}^{-6}N$.

(Since q = 2 $\mu$C and x = 0.5 m).