# Question #5be02

Apr 8, 2017

If $\textcolor{red}{r \le R}$
${V}_{r} - {V}_{R} = \frac{\Lambda}{2 \pi {\varepsilon}_{0}} \left[1 - {r}^{2} / {R}^{2}\right]$

If $\textcolor{red}{r \ge R}$
${V}_{r} - {V}_{R} = \frac{\Lambda}{2 \pi {\varepsilon}_{0} {R}^{2}} \left[\ln \left(\frac{R}{r}\right)\right]$

#### Explanation:

We are going to use Gauss's Law: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elecyl.html

If $\textcolor{red}{r \le R}$

${V}_{r} - {V}_{R} = {\int}_{r}^{R} {E}_{r} \mathrm{dr}$

(${V}_{r} - {V}_{R}$) is the Electrical potential http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elepe.html

${E}_{r} = \frac{\Lambda r}{2 \pi {\varepsilon}_{0} {R}^{2}}$

${\int}_{r}^{R} \frac{\Lambda r}{2 \pi {\varepsilon}_{0} {R}^{2}} \mathrm{dr}$

$\frac{\Lambda}{2 \pi {\varepsilon}_{0} {R}^{2}} {\int}_{r}^{R} r \mathrm{dr}$

$\frac{\Lambda}{2 \pi {\varepsilon}_{0} {R}^{2}} {\left[{r}^{2} / 2\right]}_{r}^{R}$

$\frac{\Lambda}{2 \pi {\varepsilon}_{0} {R}^{2}} {\left[{r}^{2} / 2\right]}_{r}^{R}$

$\frac{\Lambda}{2 \pi {\varepsilon}_{0} {R}^{2}} \left[{R}^{2} - {r}^{2}\right]$

${V}_{r} - {V}_{R} = \frac{\Lambda}{2 \pi {\varepsilon}_{0}} \left[1 - {r}^{2} / {R}^{2}\right]$

or

If $\textcolor{red}{r \ge R}$

${E}_{r} = \frac{\Lambda}{2 \pi {\varepsilon}_{0} r}$

${V}_{r} - {V}_{R} = {\int}_{r}^{R} {E}_{r} \mathrm{dr}$

${\int}_{r}^{R} \frac{\Lambda}{2 \pi {\varepsilon}_{0} r}$

$\frac{\Lambda}{2 \pi {\varepsilon}_{0} {R}^{2}} {\int}_{r}^{R} \left(\frac{1}{r}\right)$

$\frac{\Lambda}{2 \pi {\varepsilon}_{0} {R}^{2}} \left[\ln {\left(r\right)}_{r}^{R}\right]$

${V}_{r} - {V}_{R} = \frac{\Lambda}{2 \pi {\varepsilon}_{0} {R}^{2}} \left[\ln \left(\frac{R}{r}\right)\right]$