Question

The potential at a point at distance $x$ (measured in $\text{m}$) due to some charge situated on $x$ - axis is given by $V\left(x\right)=\frac{20}{x^2-4}\text{V}$. The electric field at $x=4\text{m}$ is given by

• A. $\frac{5}{3}\text{V/m}$ in $+x$ direction.
• B. $\frac{10}{9}\text{V/m}$ in $-x$ direction.
• C. $\frac{5}{3}\text{V/m}$ in $-x$ direction.
• D. $\frac{10}{9}\text{V/m}$ in $+x$ direction.

Answer 1

The correct option is D $\frac{10}{9}\text{V/m}$ in $+x$ direction.

The potential at distance $x$ from the charge is given by
$V\left(x\right)=\frac{20}{x^2-4}=20{(x^2-4)}^{-1}$

The relation between electric field $\left(E\right)$ and electric potential is
$E=-\frac{dV}{dx}$
$\Rightarrow E=-20\times \frac{d{(x^2-4)}^{-1}}{dx}$

$\Rightarrow E=-20(-1)\times {(x^2-4)}^{-2}\times 2x$

$\Rightarrow E=\frac{40x}{{(x^2-4)}^2}$

So at $x=4\text{m}$

$\Rightarrow E=\frac{40\times 4}{{(4^2-4)}^2}$

$\Rightarrow E=\frac{160}{{(16-4)}^2}=\frac{160}{{12}^2}=\frac{160}{144}$

$\Rightarrow E=\frac{10}{9}\text{V/m}$

As this value is positive, $\overrightarrow{E}$ will be along the positive $x$ - direction.
Hence, option (d) is correct.