Question: A rare case in which the electrostatic field E for a circuit can actually be calculated is the following: Imagine an infinitely long cylindrical sheet, of uniform resistivity and radius a . A slot (corresponding to the battery) is maintained at $\mathbf{\pm }\raisebox{1ex}{${\mathbf{V}}_{\mathbf{0}}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{2}$}\right.\mathbf{}\mathbf{}\mathit{a}\mathit{t}\mathbf{}\mathit{\varphi }\mathbf{=}\mathbf{\pm }\mathbf{\pi }$, and a steady current flows over the surface, as indicated in Fig. 7.51. According to Ohm's law, then,

$\mathbf{V}\left(a,\varphi \right)\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{0}}\mathbf{\varphi }}{\mathbf{2}\mathbf{\pi }}\mathbf{,}\left(-\pi <\varphi <+\pi \right)$

Figure 7.51

(a) Use separation of variables in cylindrical coordinates to determine $\mathbf{V}\left(s,\varphi \right)$ inside and outside the cylinder.

(b) Find the surface charge density on the cylinder.

Write the formula of the potential inside and outside the cylinder.

${\mathbf{V}}_{\mathbf{1}}\left(\mathbf{s}\mathbf{,}\varphi \right)=-\frac{{\mathbf{V}}_{\mathbf{0}}}{\pi }\sum _{\mathbf{n}=\mathbf{1}}^{\infty }\frac{\mathbf{1}}{\mathbf{n}}{\left(\frac{-\mathbf{a}}{\mathbf{s}}\right)}^{\mathbf{n}}$ …… (1)

Here, v is output voltage, is surface density, a is radius.

Write the formula of the potential outside the cylinder.

${\mathbf{V}}_{\mathbf{2}}\left(\mathbf{s}\mathbf{,}\varphi \right)=-\frac{{\mathbf{V}}_{\mathbf{0}}}{\pi }\sum _{\mathbf{n}=\mathbf{1}}^{\infty }\frac{\mathbf{1}}{\mathbf{n}}{\left(\frac{-\mathbf{s}}{\mathbf{a}}\right)}^{\mathbf{n}} \mathbf{sin}\left(\mathbf{n}\varphi \right)$ …… (2)

Here, $v_0$ is output voltage, $\sigma$ is surface density, a is radius.

Write the formula of surface charge density on the cylinder.

$\sigma ={\mathbf{\epsilon }}_{\mathbf{0}}\left({\mathbf{E}}_{\perp \mathbf{,}\mathbf{2}}-{\mathbf{E}}_{\perp \mathbf{,}\mathbf{1}}\right)$ …… (3)

Here,${\epsilon }_0$ is relative permittivity, ${\mathbf{E}}_{\perp \mathbf{,}\mathbf{1}}$and ${\mathbf{E}}_{\perp \mathbf{,}\mathbf{2}}$are perpendicular components of the electric fields.

Determine the Laplace equation in cylindrical coordinates, independent of . The boundary conditions are:

$V\left(a,\varphi \right)=\frac{V_0\varphi }{2\pi }$ …... (4)

Determine the boundary conditions of inside the cylinder.

$V_2{\left|{}_{s=a}=V_1\right|}_{s=a}$

Determine the boundary conditions of outside the cylinder.

$V_2{\left(s,\varphi \right)}_{s\to \infty }=0$ …… (5)

Where, 2 denotes “outside” and 1 “inside”. The general solution is then:

$V\left(s,\varphi \right)=A_0+B_{0}Ins+\sum _{n=1}^{\infty }\left(A_n{s}^n+B_n{s}^{-n}\right)\left(C_n\cos \left(n\varphi \right)+D_n\sin \left(n\varphi \right)\right)$

Since the solution will also be antisymmetric around because of the potential on the cylinder, . , within, and outside An-0 prevent the solution from exploding. Because there is no component in the potential on the cylinder, as follows:

$$\begin{gathered} V_1\left(s,\varphi \right)=\sum _{n=1}^{\infty }{A}_n{s}^n\sin \left(n\varphi \right) \\ {V}_2\left(s,\varphi \right)=\sum _{n=1}^{\infty }{B}_n{s}^{-n}\sin \left(n\varphi \right) \end{gathered}$$

Using equation (5).

$B_n{a}^{-n}=A_n{a}^{-n}\Rightarrow B_n=a^{-2n}{A}_n$

We now apply the boundary condition to the cylinder's surface (i). Let's say we utilise its inside potential.

$$\begin{gathered} \frac{V_0\varphi }{2\pi }=\sum _{n-1}^{\infty }{A}_n{a}^n\sin \varphi \\ =\sum _{n-1}^{\infty }{A\text{'}}_n\sin \varphi \\ {A\text{'}}_n=a^n{A}_n \\ {A\text{'}}_n=\frac{1}{\pi }{\int }_{-\pi }^{\pi }\frac{V_0\varphi }{2\pi }\sin \left(n\varphi \right)d\varphi \end{gathered}$$

The second integrated produces a sine term that disappears at and, and the coefficient is as follows:

$$\begin{gathered} {A\text{'}}_n=-\frac{V_0}{n{\pi }^2}\pi {\left(-q\right)}^n \\ {A}_n=-\frac{V_0}{n\pi }\frac{{\left(-1\right)}^n}{a^n} \end{gathered}$$

The potentials are then, using (4) and (5):

Let y = a/s outside and y = s/a inside. Then both inside and outside we have a sum of shape:

$\sum _{n=1}^{\infty }{\left(-1\right)}^n\frac{y^n}{n}\sin \left(n\varphi \right)=\mathfrak{I}\sum _{n=1}^{\infty }{\left(-1\right)}^n\frac{y^n}{n}\left(\exp \left(in\varphi \right)\right)$

Of course, we only want the imaginary part of this sum.

Determine the complex sum then:

$$\begin{gathered} \sum _{n=1}^{\infty }{\left(-1\right)}^n\frac{y^n}{n}\left(\exp \left(in\varphi \right)\right)=\sum _{n=1}^{\infty }\frac{{\left(-y\exp \left(i\varphi \right)\right)}^n}{n} \\ =\sum _{n=1}^{\infty }\int {\left(-y\exp \left(i\varphi \right)\right)}^{n-1}d\left(-y\exp \left(i\varphi \right)\right) \end{gathered}$$

Since both inside and outside of y1 are convergent, the total becomes:

$$\begin{gathered} \int d\left(ycxp\left(i\varphi \right)\right)\sum _{n-0}^{\infty }{\left(yexp\left(i\varphi \right)\right)}^n=\int d\left(yexp\left(i\varphi \right)\right)\frac{1}{1-\left(-y\exp \left(i\varphi \right)\right)} \\ =\int d\left(1+y\exp \left(i\varphi \right)\right)\frac{1}{1+y\exp \left(i\varphi \right)} \\ =\ln \left(1+y\exp \left(i\varphi \right)\right) \end{gathered}$$

Now, write,$1+y\exp \left(i\varphi \right)=\rho \exp \left(i\varphi \text{'}\right)$where $\rho and\varphi \text{'}$ are the magnitude and phase angle of the complex number in the polar representation. Then:

$$\begin{gathered} -\ln \left(1+y\exp \left(i\varphi \right)\right)=-\ln \rho -i\varphi \text{'} \\ \rho =\left|1+y\exp \left(i\varphi \right)\right| \\ \varphi \text{'}={\tan }^{-1}\left(\frac{\mathfrak{I}\left(1+y\exp \left(i\varphi \right)\right)}{\mathfrak{R}\left(1+y\exp \left(i\varphi \right)\right)}\right) \end{gathered}$$

As we said, we only want the imaginary part of the sum and thus:

$$\begin{gathered} \sum _{n=1}^{\infty }{\left(-1\right)}^n\frac{y^n}{n}\sin \left(n\varphi \right)=-\backslash \mathrm{rlap}--\varphi \\ =-{\tan }^{-1}\left(\frac{y\sin \varphi }{1+y\cos \varphi }\right) \end{gathered}$$

As a result, given the concept of y, the potential within and outside is as follows:

$$\begin{gathered} V_2\left(s,\varphi \right)=\frac{V_0}{\pi }{\tan }^{-1}\left[\frac{\left({\displaystyle \raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$s$}\right.}\right)\sin \varphi }{1+\left(\raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\right)\cos \varphi }\right] \\ {V}_1\left(s,\varphi \right)=\frac{V_0}{\pi }{\tan }^{-1}\left[\frac{\left({\displaystyle \raisebox{1ex}{$s$}\!\left/ \!\raisebox{-1ex}{$a$}\right.}\right)\sin \varphi }{1+\left(\raisebox{1ex}{$s$}\!\left/ \!\raisebox{-1ex}{$a$}\right.\right)\cos \varphi }\right] \end{gathered}$$

Determine the surface charge is calculated:

$\sigma ={\epsilon }_0\left(E_{\perp ,2}-E_{\perp ,1}\right)$

Where, the perpendicular components of the electric fields at the boundary are:

Determine the electric fields at the boundary.

$$\begin{gathered} E_{\perp ,2}=-{\left|\frac{\partial V_2}{\partial s}\right|}_{s-a} \\ =-\frac{V_0}{\pi }\frac{1}{1+{\left[\frac{a\sin \varphi }{s+a\cos \varphi }\right]}^2}\frac{a\sin \varphi }{{\left(s+a\cos \varphi \right)}^2}\left(-1\right) \\ ={\left|\frac{V_0}{\pi }\frac{a\sin \varphi }{{\left(s+a\cos \varphi \right)}^2+a^2{\sin }^2\varphi }\right|}_{s-0} \\ =\frac{V_0}{2\pi a}\frac{\sin \varphi }{1+\cos \varphi } \end{gathered}$$

Determine the electric fields at the boundary.

$$\begin{gathered} E_{\perp ,1}=-{\left|\frac{\partial V_1}{\partial s}\right|}_{s-1} \\ =-\frac{V_0}{\pi }\frac{1}{1+{\left[\frac{s\sin \varphi }{a+s\cos \varphi }\right]}^2}\frac{\sin \varphi \left(a+s\cos \varphi \right)-\cos \varphi \sin \varphi }{{\left(s+a\cos \varphi \right)}^2} \\ =-\frac{V_0}{\pi }\frac{a\sin \varphi }{{\left(a+a\cos \varphi \right)}^2+a^2{\sin }^2\varphi } \\ =-\frac{V_0}{2\pi a}\frac{\sin \varphi }{1+\cos \varphi } \end{gathered}$$

Therefore, the value of surface charge density on the cylinder is$\sigma =\frac{{\epsilon }_0{V}_0}{\pi a}\tan \frac{\varphi }{2}$.