Question

A long cylinder has been cut into quarter cylinders which are insulated from each other; alternate quarter cylinders are held at potentials +100 and -100. Find the electrostatic potential inside the cylinder. Hints: Do you see a relation to Problem 12 above? Also see Problem 5.12.

Short answer

The electrostatic potential inside the cylinder is$V(r,\theta )=\frac{400}{\pi }\sum _{\text{odd}k}\frac{1}{k}{\left(\frac{r}{a}\right)}^{2k}\sin 2k\theta$.

Step-by-step solution

Given Information:

It has been given that a long cylinder is cut into quarter which are insulated from each other; alternate quarter cylinders are held at potentials +100 and −100.

Definition of Laplace’s equation:

Laplace’s equation in cylindrical coordinates is,

$\begin{array}{c}{\mathbf{\nabla }}^{\mathbf{2}}\mathbf{u}\mathbf{=}\frac{\mathbf{1}}{\mathbf{r}}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{r}}\mathbf{\left(}\mathbf{r}\frac{\mathbf{\partial }\mathbf{u}}{\mathbf{\partial }\mathbf{r}}\mathbf{\right)}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{r}}^{\mathbf{2}}}\frac{{\mathbf{\partial }}^{\mathbf{2}}\mathbf{u}}{\mathbf{\partial }{\mathbf{\theta }}^{\mathbf{2}}}\mathbf{+}\frac{{\mathbf{\partial }}^{\mathbf{2}}\mathbf{u}}{\mathbf{\partial }{\mathbf{z}}^{\mathbf{2}}}\\ \mathbf{=}\mathbf{0}\end{array}$

And to separate the variable the solution assumed is of the form$\mathit{u}\mathbf{=}\mathit{R}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathit{\Theta }\mathbf{\left(}\mathbf{\theta }\mathbf{\right)}\mathit{Z}\mathbf{\left(}\mathbf{z}\mathbf{\right)}$.

Use Laplace equation:

The electrostatic potential inside the cylinder.

$V(a,\theta )=\left\{\begin{array}{l}100,0<\theta <\frac{{\displaystyle \pi }}{{\displaystyle 2}},\pi <\theta <\frac{{\displaystyle 3\pi }}{{\displaystyle 2}}\\ -100,\frac{{\displaystyle \pi }}{{\displaystyle 2}}<\theta <\pi ,\frac{{\displaystyle 3\pi }}{{\displaystyle 2}}<\theta <2\pi \end{array}\right.$

The Laplace equation is${\nabla }^{2}V(x,y)=0.$

Since the cylinder is very long thus there is no dependence on the z-axis. Therefore, the derivative operator${\nabla }^2$in two dimensions is as mentioned below.

${\nabla }^2=\frac{1}{r}\frac{\partial }{\partial r}\left(r\frac{\partial }{\partial r}\right)+\frac{1}{r^2}\frac{{\partial }^2}{\partial {\theta }^2}$

Solve the equation by separating into a radial and angular equation.

$V(r,\Theta )=R\left(r\right)\Theta \left(\theta \right)$

Replace this into the differential equation.

$0=\frac{r}{R}\frac{\partial }{\partial r}\left(r{R}^{\text{'}}\right)+\frac{1}{\Theta }{\Theta }^{\text{'}\text{'}}$

$\begin{array}{c}\frac{r}{R}{\left(r{R}^{\text{'}}\right)}^{\text{'}}=-\frac{{\Theta }^{\text{'}\text{'}}}{\Theta }\\ =n^2\end{array}$

Write the general solution.

$V=\sum _{n=0}^{\infty }(r^n(A_n\sin \left(n\theta \right)+B_n\cos \left(n\theta \right))+r^{-n}(A_n^{\text{'}}\sin \left(n\theta \right)+B_n^{\text{'}}\cos \left(n\theta \right))$

Here $A_n,A_{n^{\text{'}}}^{\text{'}}{B}_n$and $B_n^{\text{'}}$are constants. Set $A_n^{\text{'}}$and $B_n^{\text{'}}$to zero in order to avoid divergent behaviour.

Use boundary condition:

With the first boundary condition$\theta =0$,$B_n=0$is derived.

$V(a,\theta )=\sum _{n=0}^{\infty }{a}^n{A}_n\sin \left(n\theta \right)$

The Fourier coefficient.

$a^n{A}_n=\frac{2}{2\pi }\underset{0}{\overset{2\pi }{\int }}V(a,\theta )\sin \left(n\theta \right)d\theta$

Split the limits in the given intervals and by replacing the boundary conditions.

$a^n{A}_n=\frac{1}{\pi }\left(\begin{array}{l}\underset{0}{\overset{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\int }}100\sin \left(n\theta \right)d\theta +{\int }_{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^{\pi }-100\sin \left(n\theta \right)d\theta \\ +{\int }_{\pi }^{3\pi /2}100\sin \left(n\theta \right)d\theta +{\int }_{3\pi /2}^{2\pi }-100\sin \left(n\theta \right)d\theta \end{array}\right)$

The above equation gives$a^n{A}_n=\frac{200}{n\pi }(1+\cos n\pi )(1-\cos \left(\frac{n\pi }{2}\right))$.

Derive final solution.

Replace 2k for in the equation above.

$a^n{A}_n=\frac{400}{2k\pi }(1-{(-1)}^k)$

In the above expression k takes only odd values because for even values there is trivial solution$a^n{A}_n=0$.

$\begin{array}{c}V(r,\theta )=\sum _{\text{oddk}}\frac{400}{2k\pi }2{\left(\frac{r}{a}\right)}^{2k}\sin 2k\theta \\ =\frac{400}{\pi }\sum _{\text{oddk}}\frac{1}{k}{\left(\frac{r}{a}\right)}^{2k}\sin 2k\theta \end{array}$

Hence, the electrostatic potential inside the cylinder is$V(r,\theta )=\frac{400}{\pi }\sum _{\text{odd}k}\frac{1}{k}{\left(\frac{r}{a}\right)}^{2k}\sin 2k\theta$.