Question

In a region, electric potential varies with position $x$ as $V=4x^2$. The magnitude of electric field at $x=2\text{m}$ is

• A. $9\text{V/m}$
• B. $2\text{V/m}$
• C. $32\text{V/m}$
• D. $16\text{V/m}$

Answer 1

The correct option is D $16\text{V/m}$

Relation between electric field and potential is

${\overrightarrow{E}}_x=\frac{-\partial V}{\partial x}\hat{i}$

$\left|\overrightarrow{E}\right|=\left|\frac{\partial V}{\partial x}\right|=\frac{\partial }{\partial x}\left(4x^2\right)=8x$

at $x=2\text{m}$

$\left|\overrightarrow{E}\right|=8\times 2=16\text{V/m}$