91影视

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

${\mathbf{V}}_{\mathbf{1}}\left(\mathbf{s}\mathbf{,}蠒\right)=-\frac{{\mathbf{V}}_{\mathbf{0}}}{蟺}\underset{\mathbf{n}=\mathbf{1}}{\overset{鈭�}{鈭�}}\frac{\mathbf{1}}{\mathbf{n}}{\left(\frac{-\mathbf{a}}{\mathbf{s}}\right)}^{\mathbf{n}}$ 鈥︹�� (1)

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

Write the formula of the potential outside the cylinder.

${\mathbf{V}}_{\mathbf{2}}\left(\mathbf{s}\mathbf{,}蠒\right)=-\frac{{\mathbf{V}}_{\mathbf{0}}}{蟺}\underset{\mathbf{n}=\mathbf{1}}{\overset{鈭�}{鈭�}}\frac{\mathbf{1}}{\mathbf{n}}{\left(\frac{-\mathbf{s}}{\mathbf{a}}\right)}^{\mathbf{n}}鈥�\mathbf{sin}\left(\mathbf{n}蠒\right)$ 鈥︹�� (2)

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

Write the formula of surface charge density on the cylinder.

$蟽={\mathbf{蔚}}_{\mathbf{0}}\left({\mathbf{E}}_{鈯�\mathbf{,}\mathbf{2}}-{\mathbf{E}}_{鈯�\mathbf{,}\mathbf{1}}\right)$ 鈥︹�� (3)

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

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

$V\left(a,蠒\right)=\frac{V_0蠒}{2蟺}$ 鈥�... (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.

$V_2{\left(s,蠒\right)}_{s鈫�鈭�}=0$ 鈥︹�� (5)

Where, 2 denotes 鈥渙utside鈥� and 1 鈥渋nside鈥�. The general solution is then:

$V\left(s,蠒\right)=A_0+B_{0}Ins+\underset{n=1}{\overset{鈭�}{鈭�}}\left(A_n{s}^n+B_n{s}^{-n}\right)\left(C_n\cos \left(n蠒\right)+D_n\sin \left(n蠒\right)\right)$

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:

$$\begin{gathered} V_1\left(s,蠒\right)=\underset{n=1}{\overset{鈭�}{鈭�}}{A}_n{s}^n\sin \left(n蠒\right) \\ {V}_2\left(s,蠒\right)=\underset{n=1}{\overset{鈭�}{鈭�}}{B}_n{s}^{-n}\sin \left(n蠒\right) \end{gathered}$$

Using equation (5).

$B_n{a}^{-n}=A_n{a}^{-n}鈬�B_n=a^{-2n}{A}_n$

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

$$\begin{gathered} \frac{V_0蠒}{2蟺}=\underset{n-1}{\overset{鈭�}{鈭�}}{A}_n{a}^n\sin 蠒 \\ =\underset{n-1}{\overset{鈭�}{鈭�}}{A\text{'}}_n\sin 蠒 \\ {A\text{'}}_n=a^n{A}_n \\ {A\text{'}}_n=\frac{1}{蟺}{鈭�}_{-蟺}^{蟺}\frac{V_0蠒}{2蟺}\sin \left(n蠒\right)d蠒 \end{gathered}$$

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

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

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:

$\underset{n=1}{\overset{鈭�}{鈭�}}{\left(-1\right)}^n\frac{y^n}{n}\sin \left(n蠒\right)=\mathfrak{I}\underset{n=1}{\overset{鈭�}{鈭�}}{\left(-1\right)}^n\frac{y^n}{n}\left(\exp \left(in蠒\right)\right)$

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

Determine the complex sum then:

$$\begin{gathered} \underset{n=1}{\overset{鈭�}{鈭�}}{\left(-1\right)}^n\frac{y^n}{n}\left(\exp \left(in蠒\right)\right)=\underset{n=1}{\overset{鈭�}{鈭�}}\frac{{\left(-y\exp \left(i蠒\right)\right)}^n}{n} \\ =\underset{n=1}{\overset{鈭�}{鈭�}}鈭�{\left(-y\exp \left(i蠒\right)\right)}^{n-1}d\left(-y\exp \left(i蠒\right)\right) \end{gathered}$$

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

$$\begin{gathered} 鈭�d\left(ycxp\left(i蠒\right)\right)\underset{n-0}{\overset{鈭�}{鈭�}}{\left(yexp\left(i蠒\right)\right)}^n=鈭�d\left(yexp\left(i蠒\right)\right)\frac{1}{1-\left(-y\exp \left(i蠒\right)\right)} \\ =鈭�d\left(1+y\exp \left(i蠒\right)\right)\frac{1}{1+y\exp \left(i蠒\right)} \\ =\ln \left(1+y\exp \left(i蠒\right)\right) \end{gathered}$$

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

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

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

$$\begin{gathered} \underset{n=1}{\overset{鈭�}{鈭�}}{\left(-1\right)}^n\frac{y^n}{n}\sin \left(n蠒\right)=-\backslash \mathrm{rlap}--蠒 \\ =-{\tan }^{-1}\left(\frac{y\sin 蠒}{1+y\cos 蠒}\right) \end{gathered}$$

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

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

Determine the surface charge is calculated:

$蟽={蔚}_0\left(E_{鈯�,2}-E_{鈯�,1}\right)$

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

Determine the electric fields at the boundary.

$$\begin{gathered} E_{鈯�,2}=-{\left|\frac{鈭�V_2}{鈭�s}\right|}_{s-a} \\ =-\frac{V_0}{蟺}\frac{1}{1+{\left[\frac{a\sin 蠒}{s+a\cos 蠒}\right]}^2}\frac{a\sin 蠒}{{\left(s+a\cos 蠒\right)}^2}\left(-1\right) \\ ={\left|\frac{V_0}{蟺}\frac{a\sin 蠒}{{\left(s+a\cos 蠒\right)}^2+a^2{\sin }^2蠒}\right|}_{s-0} \\ =\frac{V_0}{2蟺a}\frac{\sin 蠒}{1+\cos 蠒} \end{gathered}$$

Determine the electric fields at the boundary.

$$\begin{gathered} E_{鈯�,1}=-{\left|\frac{鈭�V_1}{鈭�s}\right|}_{s-1} \\ =-\frac{V_0}{蟺}\frac{1}{1+{\left[\frac{s\sin 蠒}{a+s\cos 蠒}\right]}^2}\frac{\sin 蠒\left(a+s\cos 蠒\right)-\cos 蠒\sin 蠒}{{\left(s+a\cos 蠒\right)}^2} \\ =-\frac{V_0}{蟺}\frac{a\sin 蠒}{{\left(a+a\cos 蠒\right)}^2+a^2{\sin }^2蠒} \\ =-\frac{V_0}{2蟺a}\frac{\sin 蠒}{1+\cos 蠒} \end{gathered}$$

Therefore, the value of surface charge density on the cylinder is$蟽=\frac{{蔚}_0{V}_0}{蟺a}\tan \frac{蠒}{2}$.