Question

The magnitude $\mathbf{E}$ of an electric field depends on the radial distance r according to$\mathbf{E}\mathbf{}\mathbf{=}\mathbf{}\mathbf{A}\mathbf{/}{\mathbf{r}}^{\mathbf{4}}$, where is a constant with the unit volt–cubic meter. As a multiple of $\mathbf{A}$, what is the magnitude of the electric potential difference between$\mathbf{r}\mathbf{=}\mathbf{}\mathbf{2.00}\mathbf{\text{m}}$and$\mathit{r}\mathbf{}\mathbf{=}\mathbf{}\mathbf{3.00}\mathbf{\text{m}}$?

Short answer

The magnitude of the electric potential difference is${\mathrm{V}}_{\left(\mathrm{r}\right)}=+2.93\times {10}^{-2}\mathrm{A}$ .

Step-by-step solution

Given data:

The intensity of the electric field$\mathrm{E}=\frac{\mathrm{A}}{{\mathrm{r}}^4}$,

The value of distance are ${\mathrm{r}}_1=2.0\text{m}$and ${\mathrm{r}}_2=3.0\text{m}$

Understanding the concept

Use the relation between the electric field $\mathbf{E}$ and potential $\mathbf{V}$as given below.

role="math" localid="1662730170007" $\mathbf{}\mathbf{E}\mathbf{=}\frac{\mathbf{-}\mathbf{dV}}{\mathbf{dr}}$

Here,$\mathbf{r}$ is the distance.

Calculate the magnitude of the electric potential difference between r= 2.00 m and r= 3.00 m:

Since, the electric field is,

$\begin{array}{c}\mathrm{E}=\frac{-\mathrm{dV}}{\mathrm{dr}}\\ \underset{\mathrm{\infty }}{\overset{\mathrm{r}}{\int }}\mathrm{dV}=-\underset{\mathrm{\eta }}{\overset{\mathrm{n}}{\int }}\mathrm{Edr}\\ {\mathrm{V}}_{\left(\mathrm{r}\right)}-{\mathrm{V}}_{\left(\mathrm{\infty }\right)}=\underset{\mathrm{\eta }}{\overset{\mathrm{n}}{\int }}\frac{\mathrm{A}}{{\mathrm{r}}^4}\mathrm{dr}\end{array}$

As ${\mathrm{V}}_{\mathrm{\infty }}=0$, then

$\begin{array}{c}{\mathrm{V}}_{\left(\mathrm{r}\right)}=-\mathrm{A}{\left[\frac{{\mathrm{r}}^3}{-3}\right]}_{{\mathrm{r}}_1}^{{\mathrm{r}}_2}\\ =\frac{-\mathrm{A}}{3}\left[\frac{1}{{\mathrm{r}}_2^3}-\frac{1}{{\mathrm{r}}_1^3}\right]\\ =\frac{-\mathrm{A}}{3}\left[\frac{1}{3^3}-\frac{1}{2^3}\right]\\ =+2.93\times {10}^{-2}\mathrm{A}\end{array}$

${\mathrm{V}}_{\left(\mathrm{r}\right)}=+\mathrm{A}(\text{0.029}/{\text{m}}^{\text{3}})$

Hence,the magnitude of the electric potential difference is $+2.93\times {10}^{-2}\mathrm{A}$.