91影视

The radius if the cylinder is $R$.

The uniform surface charge in upper half of the cylinder is ${蟽}_0$.

The uniform surface charge in upper lower of the cylinder is $鈭�{蟽}_0$.

outside the cylinder.

The expression for the potential is given as follows.

$V(s,蠒)=a_0+b_{0}lns+\underset{K=1}{\overset{伪}{鈭�}}[s_p^k(a_{k}cosk蠒+b_{k}sink蠒)+s^{-k}(c_{k}cosk蠒+d_{k}sink蠒)]$鈥︹�� (1)

Here, $a_k$,$b_k$, $c_k$and $d_k$are the constant.

From equation (1)

The potential inside the shell is given as follows.

$V_{in}(s,蠒)=\underset{K=1}{\overset{伪}{鈭�}}{s}^k(a_{k}cosk蠒+b_{k}sink蠒)$ 鈥︹��. (2)

The potential outside the shell is given as follows.

$V_{out}(s,蠒)=\underset{K=1}{\overset{伪}{鈭�}}{s}^{-k}(c_{k}cosk蠒+d_{k}sink蠒)$ 鈥︹�� (3)

At the boundary condition, potential is continuous at .$s=R$Hence, equate the potential inside and outside the cylinder.

$\underset{K=1}{\overset{伪}{鈭�}}{R}^k(a_{k}cosk蠒+b_{k}sink蠒)=\underset{K=1}{\overset{伪}{鈭�}}{R}^{-k}(c_{k}cosk蠒+d_{k}sink蠒)$

Compare the coefficient of$\mathrm{cosk}蠒$and$\sin k蠒$from the above equation.

$\begin{array}{c}{a}_k{R}^k=R^{鈭�k}{c}_k\\ c_k=R^{2k}{a}_k\end{array}$

Similarly,

$\begin{array}{c}{b}_k{R}^k=R^{鈭�k}{d}_k\\ d_k=R^{2k}{b}_k\end{array}$

Consider the equation which relates the normal derivative of the potential with the surface charge density.

${\frac{鈭�V}{鈭�s}|}_{R+}鈭�{\frac{鈭�V}{鈭�s}|}_{R鈭�}=\frac{鈭�蟽}{{蔚}_0}$ 鈥︹��. (4)

Calculate the derivative of potential inside cylinder.

$\begin{array}{l}{\frac{鈭�V}{鈭�s}|}_{R+}=鈭�\frac{鈭�}{鈭�s}{s}^{-k}{(c_{k}cosk蠒+d_{k}sink蠒)}_{s=R}\\ {\frac{鈭�V}{鈭�s}|}_{R+}=鈭�\left(\frac{鈭�k}{s^{k+1}}\right){(c_{k}cosk蠒+d_{k}sink蠒)}_{s=R}\\ {\frac{鈭�V}{鈭�s}|}_{R+}=鈭�\left(\frac{鈭�k}{R^{k+1}}\right)(c_{k}cosk蠒+d_{k}sink蠒)\end{array}$

Calculate the derivative of potential outside cylinder.

$\begin{array}{l}{\frac{鈭�V}{鈭�s}|}_{R鈭�}=鈭�\frac{鈭�}{鈭�s}{s}^k{(a_{k}cosk蠒+b_{k}sink蠒)}_{s=R}\\ {\frac{鈭�V}{鈭�s}|}_{R鈭�}=鈭�\left(k{s}^{k鈭�1}\right){(a_{k}cosk蠒+b_{k}sink蠒)}_{s=R}\\ {\frac{鈭�V}{鈭�s}|}_{R鈭�}=鈭�\left(k{R}^{k鈭�1}\right)(a_{k}cosk蠒+b_{k}sink蠒)\end{array}$

Recall the equation (4),

${\frac{鈭�V}{鈭�s}|}_{R+}鈭�{\frac{鈭�V}{鈭�s}|}_{R鈭�}=\frac{鈭�蟽}{{蔚}_0}$

Substitute $鈭�\left(\frac{鈭�k}{R^{k+1}}\right)(c_{k}cosk蠒+d_{k}sink蠒)$for${\frac{鈭�V}{鈭�s}|}_{R+}$and$鈭�\left(k{R}^{k鈭�1}\right)(a_{k}cosk蠒+b_{k}sink蠒)$for ${\frac{鈭�V}{鈭�s}|}_{R鈭�}$into above equation.

$鈭�\left(\frac{鈭�k}{R^{k+1}}\right)(c_{k}cosk蠒+d_{k}sink蠒)+鈭�\left(k{R}^{k鈭�1}\right)(a_{k}cosk蠒+b_{k}sink蠒)=\frac{鈭�蟽}{{蔚}_0}$

Substitute $R^{2k}{a}_k$for$c_k$and $R^{2k}{b}_k$for$d_k$into above equation.

$\begin{array}{l}鈭�\left(\frac{鈭�k}{R^{k+1}}\right)(R^{2k}{a}_{k}cosk蠒+R^{2k}{b}_{k}sink蠒)鈭�鈭�\left(k{R}^{k鈭�1}\right)(a_{k}cosk蠒+b_{k}sink蠒)=\frac{鈭�蟽}{{蔚}_0}\\ 鈭�鈭�\left(k{R}^{k鈭�1}\right)(a_{k}cosk蠒+b_{k}sink蠒)鈭�鈭�\left(k{R}^{k鈭�1}\right)(a_{k}cosk蠒+b_{k}sink蠒)=\frac{鈭�蟽}{{蔚}_0}\\ 鈭�2k{R}^{k鈭�1}(a_{k}cosk蠒+b_{k}sink蠒)=\frac{蟽}{{蔚}_0}\end{array}$ 鈥︹赌�(5)

Noe defines the above equation for the intervals,

$鈭�2k{R}^{k鈭�1}(a_{k}cosk蠒+b_{k}sink蠒)=\{\begin{array}{l}\frac{{蟽}_0}{{蔚}_0}\text{鈥夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌�}(0<蠒<蟺)\\ \frac{鈭�{蟽}_0}{{蔚}_0}\text{鈥夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌�}(蟺<蠒<2蟺)\end{array}$

The value of integral of above angles,

$\begin{array}{l}{鈭�}_0^{2蟺}\mathrm{sink}蠒\cos l蠒d蠒=0\\ {鈭�}_0^{2蟺}cosk蠒\cos l蠒d蠒=\{\begin{array}{l}0\text{鈥夆赌夆赌夆赌夆赌夆赌夆赌夆赌�}k鈮�l\\ 蟺\text{鈥夆赌夆赌夆赌夆赌夆赌夆赌夆赌�}k=l\end{array}\text{鈥夆赌�}\end{array}$

Multiply the equation (5) with $\mathrm{cosl}蠒$.

$\begin{array}{l}2k{R}^{k鈭�1}[{鈭�}_0^{2蟺}{a}_k\cos k蠒\cos l蠒d蠒+{鈭�}_0^{2蟺}{b}_k\sin k蠒\mathrm{cosl}蠒d蠒]=\frac{{蟽}_0}{{蔚}_0}{[{鈭�}_0^{2蟺}\cos l蠒d蠒鈭�{鈭�}_0^{2蟺}\cos l蠒]}_{}\underset{未x鈫�0}{\lim }\\ 2k{R}^{k鈭�1}蟺a_k=\frac{{蟽}_0}{{蔚}_0}{\left[\frac{\sin l蠒}{l}\right]}_0^{蟺}\\ 2k{R}^{k鈭�1}蟺a_k=\frac{{蟽}_0}{{蔚}_0}\left(0\right)\\ a_k=0\end{array}$

The value of$a_k$is becomes zero therefore multiply with$\sin l蠒$and integrate.

$\begin{array}{l}2k{R}^{k鈭�1}[{鈭�}_0^{2蟺}{a}_k\cos k蠒\sin l蠒d蠒+{鈭�}_0^{2蟺}{b}_k\sin k蠒\mathrm{sinl}蠒d蠒]=\frac{{蟽}_0}{{蔚}_0}[{鈭�}_0^{2蟺}\sin l蠒d蠒鈭�{鈭�}_0^{2蟺}\cos l蠒]\\ 2k{R}^{k鈭�1}蟺b_k=\frac{{蟽}_0}{{蔚}_0}[{(鈭�\frac{\cos l蠒}{l})}_0^{蟺}+{\left(\frac{\cos l蠒}{l}\right)}_0^{2蟺}]\\ b_k=\frac{{蟽}_0}{l{蔚}_0}(2鈭�2\cos l蟺)\\ b_k=\frac{{蟽}_0}{2k蟺l{蔚}_0{R}^{k鈭�1}}(2鈭�2\cos l蟺)\end{array}$

The value of$蟺$is obtained at the condition .$k=l$

$b_k=\text{鈥�}\{\begin{array}{l}0\text{鈥夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆�夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌塱f鈥�}l\text{鈥塱蝉鈥塭惫别苍}\\ \frac{2{蟽}_0}{蟺k^2{R}^{k鈭�1}{蔚}_0}\text{鈥夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌塱f鈥�}l\text{鈥塱蝉鈥塷诲诲}\end{array}$

Similarly for the outside of the cylinder.

$\begin{array}{l}2k{R}^{k鈭�1}[{鈭�}_0^{2蟺}{c}_k\cos k蠒\sin l蠒d蠒+{鈭�}_0^{2蟺}{d}_k\sin k蠒\mathrm{sinl}蠒d蠒]=\frac{{蟽}_0}{{蔚}_0}[{鈭�}_0^{2蟺}\sin l蠒d蠒鈭�{鈭�}_0^{2蟺}\cos l蠒]\\ 2k{R}^{k鈭�1}蟺d_k=\frac{{蟽}_0}{{蔚}_0}[{(鈭�\frac{\cos l蠒}{l})}_0^{蟺}+{\left(\frac{\cos l蠒}{l}\right)}_0^{2蟺}]\\ d_k=\frac{{蟽}_0}{l{蔚}_0}(2鈭�2\cos l蟺)\\ d_k=\frac{{蟽}_0}{2k蟺l{蔚}_0{R}^{k鈭�1}}(2鈭�2\cos l蟺)\end{array}$

The value of $蟺$is obtained at the condition .$k=l$

$d_k=\text{鈥�}\{\begin{array}{l}0\text{鈥夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆�夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌塱f鈥�}l\text{鈥塱蝉鈥塭惫别苍}\\ \frac{2{蟽}_0}{蟺k^2{R}^{k鈭�1}{蔚}_0}\text{鈥夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌塱f鈥�}l\text{鈥塱蝉鈥塷诲诲}\end{array}$

Substitute $\frac{2{蟽}_0}{蟺k^2{R}^{k鈭�1}{蔚}_0}$for$b_k$ and $0$for$a_k$into equation (2).

$\begin{array}{l}{V}_{in}(s,蠒)=\underset{K=1}{\overset{伪}{鈭�}}{s}^k(\left(0\right)cosk蠒+\frac{2{蟽}_0}{蟺k^2{R}^{k鈭�1}{蔚}_0}sink蠒)\\ V_{in}(s,蠒)=\underset{K=1}{\overset{伪}{鈭�}}{s}^k\left(\frac{2{蟽}_0}{蟺k^2{R}^{k鈭�1}{蔚}_0}sink蠒\right)\\ V_{in}(s,蠒)=\frac{2{蟽}_{0}R}{蟺{蔚}_0}\underset{K=1,3,5}{鈭�}\frac{1}{k^2}\sin k蠒{\left(\frac{s}{R}\right)}^k\end{array}$

Substitute$\frac{2{蟽}_0}{蟺k^2{R}^{k鈭�1}{蔚}_0}\underset{未x鈫�0}{\lim }$for$d_k$and$0$for$c_k$into equation (3).

$\begin{array}{l}{V}_{out}(s,蠒)=\underset{K=1}{\overset{伪}{鈭�}}{s}^{-k}(\left(0\right)cosk蠒+\frac{2{蟽}_0}{蟺k^2{R}^{k鈭�1}{蔚}_0}sink蠒)\\ V_{out}(s,蠒)=\underset{K=1}{\overset{伪}{鈭�}}{s}^{-k}\left(\frac{2{蟽}_0}{蟺k^2{R}^{k鈭�1}{蔚}_0}sink蠒\right)\\ V_{out}(s,蠒)=\frac{2{蟽}_{0}R}{蟺{蔚}_0}\underset{K=1,3,5}{鈭�}\frac{1}{k^2}\sin k蠒{\left(\frac{R}{s}\right)}^k\end{array}$

Hence, the expression for the potential inside the cylinder is and outside the cylinder is .

$\frac{2{蟽}_{0}R}{蟺{蔚}_0}\underset{K=1,3,5}{鈭�}\frac{1}{k^2}\sin k蠒{\left(\frac{s}{R}\right)}^k$and outside the cylinder is$\frac{2{蟽}_{0}R}{蟺{蔚}_0}\underset{K=1,3,5}{鈭�}\frac{1}{k^2}\sin k蠒{\left(\frac{R}{s}\right)}^k$