Question

The electric potential along the x-axis is $V=100e^{-2x}V$, where$x$ is in meters. What is $E_x$ at (a)role="math" localid="1649581712352" $x=1.0m$ and (b) role="math" localid="1649581719374" $x=2.0m$?

Short answer

$$\begin{gathered} \left(a\right)E_x=\mathbf{27}\mathbf{}\mathit{V}\mathbf{/}\mathit{m} \\ \left(b\right)E_x=\mathbf{3}\mathbf{.}\mathbf{7}\mathbf{}\mathit{V}\mathbf{/}\mathit{m} \end{gathered}$$

Step-by-step solution

Given information and theory used  

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

(a) $x=1.0m$and (b) $x=2.0m$

Theory used :

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

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

Calculating Ex

The negative derivative of the potential $V=100e^{-2x}V$is equivalent to the component of the electric field. As a result,

role="math" localid="1649581602113" $E\left(x\right)=-\frac{dV\left(x\right)}{dx}=-100(-2e^{-2x})V/m=\underline{200e^{-2\mathrm{x}}V/m}$

(a) We have $E_x=\mathbf{27}\mathbf{}\mathit{V}\mathbf{/}\mathit{m}\mathrm{at}a=1.0m$.

(a) We have $E_x=\mathbf{}\mathbf{3}\mathbf{.}\mathbf{7}\mathbf{}\mathit{V}\mathbf{/}\mathit{m}\mathrm{at}x=2.0m$.