91影视

The side length of the long square metal pipe is a.

The potential on one side is $V_0$and the other three sides are zero potential.

$$\begin{gathered} V(y=0)=0.....\left(1\right) \\ V(y=a)=V_0.......\left(2\right) \\ V(x=a/2)=0......\left(3\right) \\ V(x=-a/2)......\left(4\right) \end{gathered}$$

The boundary conditions for the square pipe are

$$\begin{gathered} V\left(R,-\frac{p}{4}<f<\frac{p}{4}\right)=0....\left(5\right) \\ V\left(R\left|f\right|>\frac{p}{4}\right)=0.......\left(6\right) \end{gathered}$$

The potential inside the rectangular pipe of sides a, b with $V_0$potential at and zero potential on the other boundaries is

localid="1657628313896" ${\mathit{V}}_{\mathbf{0}}\mathbf{[}\frac{\mathbf{y}}{\mathbf{a}}\mathbf{+}\frac{\mathbf{2}}{\mathbf{蟿}\mathbf{蟿}}\underset{\mathbf{}}{\mathbf{鈭�}}\frac{{\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{)}}^{\mathbf{n}}}{\mathbf{n}}\frac{\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{h}\left({\displaystyle \frac{n蟿蟿x}{a}}\right)\mathbf{s}\mathbf{i}\mathbf{n}\left(\frac{n蟿蟿x}{a}\right)}{\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{h}\left(\frac{n蟿蟿x}{a}\right)}\mathbf{]}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{7}\mathbf{\right)}$

The surface charge density as a function of x is given by

$\mathit{蟽}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathbf{-}{\mathit{蔚}}_{\mathbf{0}}{\overline{)\frac{\mathbf{鈭�}\mathbf{V}}{\mathbf{鈭�}\mathbf{y}}}}_{\mathbf{y}\mathbf{=}\mathbf{0}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{8}\mathbf{\right)}$

The general solution for the potential function in cylindrical coordinates is

$\mathbf{V}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{f}\mathbf{)}\mathbf{=}{\mathbf{a}}_{\mathbf{0}}\mathbf{+}{\mathbf{b}}_{\mathbf{0}}\mathbf{Inr}\mathbf{+}\underset{\mathbf{k}\mathbf{=}\mathbf{1}}{\overset{\mathbf{鈭�}}{\mathbf{鈭�}}}\mathbf{(}{\mathbf{a}}_{\mathbf{k}}{\mathbf{r}}^{\mathbf{k}}\mathbf{+}{\mathbf{b}}_{\mathbf{k}}{\mathbf{r}}^{\mathbf{-}\mathbf{k}}\mathbf{)}\left[c_k\cos \left(\mathrm{kf}\right)+d_k\sin \left(\mathrm{kf}\right)\right]\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{9}\mathbf{\right)}$

Equation (7) for b = a/2 reduces to

$V_0[\frac{\mathrm{y}}{\mathrm{a}}+\frac{2}{\mathrm{蟿蟿}}\underset{}{鈭�}\frac{{(-1)}^{\mathrm{n}}}{\mathrm{n}}\frac{\cosh \left({\displaystyle \frac{\mathrm{苍蟿蟿虫}}{\mathrm{a}}}\right)\sin \left(\frac{\mathrm{苍蟿蟿虫}}{\mathrm{a}}\right)}{\cosh \left(\frac{\mathrm{苍蟿蟿虫}}{\mathrm{a}}\right)}].....\left(7\right)$

The surface charge density in accordance with equation (8) is as follows

$$\begin{gathered} 蟽\left(x\right)=-{蔚}_0{V}_0{V}_0{[\frac{\mathrm{y}}{\mathrm{a}}+\frac{2}{\mathrm{蟿蟿}}\underset{n}{鈭�}\left(\frac{n\mathrm{蟺}}{a}\right)\frac{{(-1)}^{\mathrm{n}}\cosh \left({\displaystyle \frac{\mathrm{苍蟿蟿虫}}{\mathrm{a}}}\right)\cos \left(\frac{\mathrm{苍蟿蟿}y}{\mathrm{a}}\right)}{nc\mathrm{osh}\left(\frac{\mathrm{苍蟿蟿虫}}{\mathrm{a}}\right)}]}_{=0} \\ =\frac{-{蔚}_0{V}_0}{a}\left[1+2\underset{n}{鈭�}\frac{{(-1)}^{\mathrm{n}}\cosh \left({\displaystyle \frac{\mathrm{苍蟿蟿虫}}{\mathrm{a}}}\right)}{c\mathrm{osh}\left(\frac{n\mathrm{蟺}}{2}\right)}\right] \end{gathered}$$

The charge per unit length is

$$\begin{gathered} 位={鈭�}_{-\frac{a}{2}}^{\frac{a}{2}}蟽\left(x\right)dx \\ =\frac{-{蔚}_0{V}_0}{a}{鈭�}_{-\frac{a}{2}}^{\frac{a}{2}}\left[1+2\underset{n}{鈭�}\frac{{(-1)}^{\mathrm{n}}\cosh \left({\displaystyle \frac{\mathrm{苍蟿蟿虫}}{\mathrm{a}}}\right)}{c\mathrm{osh}\left(\frac{n\mathrm{蟺}}{2}\right)}\right]dx \\ =\frac{-{蔚}_0{V}_0}{a}\left[a+2\underset{n}{鈭�}\frac{{(-1)}^n}{\cos h\left({\displaystyle \frac{n\mathrm{蟺}}{2}}\right)}{鈭�}_{-\frac{a}{2}}^{\frac{a}{2}}\cos h\left(\frac{n\mathrm{蟺虫}}{a}\right)\right] \\ =\frac{-{蔚}_0{V}_0}{a}\left[a+\frac{4\mathrm{蟺}}{\mathrm{蟺}}\underset{n}{鈭�}\frac{{(-1)}^n\tan h\left({\displaystyle \frac{n\mathrm{蟺}}{2}}\right)}{n}\right] \end{gathered}$$

This can be numerically solved to

$位=\frac{-{蔚}_0{V}_0}{\mathrm{蟺}}\mathrm{In}2$

From equation (9), for potential inside the pipe, $b_0$and $b_k$must be zero, otherwise the potential will blow up at r = 0. The configuration is symmetric in $蠒$, hence $d_k=0$. Absorption of $c_k$into $a_k$reduces equation (3) to

$V(r,蠒)=a_0+\underset{k-1}{\overset{鈭�}{鈭�}}{a}_k{r}^k\cos \left(k蠒\right)$

Apply Fourier trick on the boundary equation to get

$$\begin{gathered} \underset{k=0}{\overset{鈭�}{鈭�}}{a}_k{R}^k{鈭�}_{-\mathrm{蟺}}^{\mathrm{蟺}}\cos \left(k蠒\right)\cos \left(k^{\text{'}}蠒\right)d蠒=V_0{鈭�}_{-\mathrm{蟺}/4}^{\mathrm{蟺}/4}\cos \left(k^{\text{'}}蠒\right)d蠒 \\ =\left\{\begin{array}{ll}\frac{V_0}{K^{\text{'}}}\sin \left(\frac{K^{\text{'}}\mathrm{蟺}}{4}\right)& \mathrm{if}{K}^{\text{'}}鈮�0\\ \frac{V_0\mathrm{蟺}}{2}& \mathrm{if}{K}^{\text{'}}=0\end{array}\right. \end{gathered}$$

Also

${鈭�}_{-\mathrm{蟺}}^{\mathrm{蟺}}\cos \left(k蠒\right)\cos \left(k^{\text{'}}蠒\right)d蠒\left\{\begin{array}{ll}2\mathrm{蟺}& k=K^{\text{'}}=0\\ \mathrm{蟺}& k=K^{\text{'}}鈮�0\end{array}\right.$

Use these two solutions to get

$$\begin{gathered} a_0=\frac{V_0}{4} \\ {a}_k=\frac{2V_0}{{\mathrm{蟺办搁}}^{\mathrm{k}}}\sin \left(\frac{k\mathrm{蟺}}{4}\right) \end{gathered}$$

The potential function thus becomes

$V(r,蠒)=V_0\left[\frac{1}{4}+\frac{2}{\mathrm{蟺}}\underset{k=1}{\overset{鈭�}{鈭�}}\frac{\sin (k\mathrm{蟺}/4)}{k}{\left(\frac{r}{k}\right)}^k\cos \left(k蠒\right)\right]$

To find the line charge, first differentiate with respect to r, integrate with respect to $蠒$and finally put the limit $r鈫�R$.

role="math" localid="1657688183512" $$\begin{gathered} 蟽={蔚}_0\frac{鈭�V}{鈭�r} \\ =\frac{2{蔚}_0{V}_0}{\mathrm{蟺搁}}\underset{k=1}{\overset{鈭�}{鈭�}}\frac{\sin (k\mathrm{蟺}/4)}{k}k{\left(\frac{r}{R}\right)}^{k-1}\cos \left(k蠒\right) \\ =\frac{2{蔚}_0{V}_0}{\mathrm{蟺搁}}\underset{k=1}{\overset{鈭�}{鈭�}}{\left(\frac{r}{R}\right)}^{k-1}\sin (k\mathrm{蟺}/4)\cos \left(k蠒\right) \end{gathered}$$

The line charge on the opposite wall to that with potential $V_0$is

$$\begin{gathered} 位=2{鈭�}_{\frac{3\mathrm{蟺}}{4}}^{\mathrm{蟺}}蟽Rd蠒 \\ =2{鈭�}_{\frac{3\mathrm{蟺}}{4}}^{\mathrm{蟺}}\left[\frac{2{蔚}_0{V}_0}{\mathrm{蟺搁}}\underset{k=1}{\overset{鈭�}{鈭�}}{\left(\frac{r}{R}\right)}^{k-1}\sin (k\mathrm{蟺}/4)\cos \left(k蠒\right)\right]Rd蠒 \\ =\frac{4{蔚}_0{V}_0}{蟺}\underset{k=1}{\overset{鈭�}{鈭�}}{\left(\frac{r}{R}\right)}^{k-1}\sin (k\mathrm{蟺}/4){鈭�}_{\frac{3\mathrm{蟺}}{4}}^{\mathrm{蟺}}\cos \left(k蠒\right)d蠒 \\ =-\frac{4{蔚}_0{V}_0}{蟺}\underset{k=1}{\overset{鈭�}{鈭�}}{\left(\frac{r}{R}\right)}^{k-1}\sin (k\mathrm{蟺}/4){鈭�}_{\frac{3\mathrm{蟺}}{4}}^{\mathrm{蟺}}\sin (3k\mathrm{蟺}/4) \end{gathered}$$

This can be expanded in terms of x = r/ R as

$$\begin{gathered} 位=-\frac{4{蔚}_0{V}_0}{\mathrm{蟺}}\left[\frac{1}{2x}\left(x+\frac{x^3}{3}+\frac{x^3}{5}+....\right)-\frac{1}{x}\left(\frac{x^2}{2}+\frac{x^6}{6}+\frac{x^{10}}{10}+...\right)\right] \\ =-\frac{2{蔚}_0{V}_0}{\mathrm{蟺虫}}\left[\frac{1}{2}\mathrm{In}\left(\frac{1+x}{1-x}\right)-\frac{1}{2}\mathrm{In}\left(\frac{1+x^2}{1-x^2}\right)\right] \\ =-\frac{{蔚}_0{V}_0}{\mathrm{蟺虫}}\mathrm{In}\left(\frac{1+x}{1-x}脳\frac{1+x^2}{1-x^2}\right) \\ =-\frac{{蔚}_0{V}_0}{\mathrm{蟺虫}}\mathrm{In}\left(\frac{(1+x^2)}{1-x^2}\right) \end{gathered}$$

Thus, the charge for the limit $r鈫�R(x鈫�1)$is

$$\begin{gathered} {\overline{)位}}_{x鈫�1}=-\frac{{蔚}_0{V}_0}{\mathrm{蟺}}\mathrm{In}\left(\frac{{(1+1)}^2}{1+1^2}\right) \\ =-\frac{{蔚}_0{V}_0}{\mathrm{蟺}}\mathrm{In}2 \end{gathered}$$