Question

The electric potential along the $x-axis\mathrm{is}V=100x^2$V, where $x$is in meters. What is $E_x$at

(a) $x=0m$and

(b) $x=1m$?

Short answer

The field is $\mathbf{0}\mathbf{}\mathrm{at}\mathbf{}\mathit{x}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}$ and $\mathbf{-}\mathbf{}\mathbf{200}\mathbf{}\mathit{V}\mathbf{/}\mathit{m}\mathbf{}\mathrm{at}\mathbf{}\mathit{a}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{}\mathit{m}$.

Step-by-step solution

Given information and theory used   

Given : The electric potential along the $x$-axis is : $V=100x^{2}V$

(a) $x=0m$and (b) $x=1m$

Theory used :

The electric field is expressed as a function of potential as :

$E\left(x\right)=-\frac{dV\left(x\right)}{dx}$

Calculating the required electric potential 

$E\left(x\right)=-\frac{dV\left(x\right)}{dx}=-100·2xV=\underline{-200xV}$ is the $x$component of the electric field that is equivalent to the derivative of the potential with respect to $x$

This gives $E_x=0$at $x=0m$and

$E=-200V/m$at $x=1m$.