Question

A long cylindrical shell of radius $R$carries a uniform surface charge on ${\sigma }_0$the upper half and an opposite charge $-{\sigma }_0$on the lower half (Fig. 3.40). Find the electric potential inside and outside the cylinder.

Short answer

The expression for the potential inside the cylinder is$\frac{2{\sigma }_{0}R}{\pi {\epsilon }_0}\sum _{K=1,3,5}\frac{1}{k^2}\sin k\varphi {\left(\frac{s}{R}\right)}^k$ and outside the cylinder is $\frac{2{\sigma }_{0}R}{\pi {\epsilon }_0}\sum _{K=1,3,5}\frac{1}{k^2}\sin k\varphi {\left(\frac{R}{s}\right)}^k$.

Step-by-step solution

Write the given data from the question.

The radius if the cylinder is $R$.

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

The uniform surface charge in upper lower of the cylinder is $-{\sigma }_0$.

Determine the formulas to calculate the electric potential inside and outside the cylinder.

outside the cylinder.

The expression for the potential is given as follows.

$V(s,\varphi )=a_0+b_{0}lns+\sum _{K=1}^{\alpha }[s_p^k(a_{k}cosk\varphi +b_{k}sink\varphi )+s^{-k}(c_{k}cosk\varphi +d_{k}sink\varphi )]$…… (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.

$V_{in}(s,\varphi )=\sum _{K=1}^{\alpha }{s}^k(a_{k}cosk\varphi +b_{k}sink\varphi )$ ……. (2)

The potential outside the shell is given as follows.

$V_{out}(s,\varphi )=\sum _{K=1}^{\alpha }{s}^{-k}(c_{k}cosk\varphi +d_{k}sink\varphi )$ …… (3)

Calculate the electric potential inside and outside the cylinder.

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

$\sum _{K=1}^{\alpha }{R}^k(a_{k}cosk\varphi +b_{k}sink\varphi )=\sum _{K=1}^{\alpha }{R}^{-k}(c_{k}cosk\varphi +d_{k}sink\varphi )$

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

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

Similarly,

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

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

${\frac{\partial V}{\partial s}|}_{R+}-{\frac{\partial V}{\partial s}|}_{R-}=\frac{-\sigma }{{\epsilon }_0}$ ……. (4)

Calculate the derivative of potential inside cylinder.

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

Calculate the derivative of potential outside cylinder.

$\begin{array}{l}{\frac{\partial V}{\partial s}|}_{R-}=\sum \frac{\partial }{\partial s}{s}^k{(a_{k}cosk\varphi +b_{k}sink\varphi )}_{s=R}\\ {\frac{\partial V}{\partial s}|}_{R-}=\sum \left(k{s}^{k-1}\right){(a_{k}cosk\varphi +b_{k}sink\varphi )}_{s=R}\\ {\frac{\partial V}{\partial s}|}_{R-}=\sum \left(k{R}^{k-1}\right)(a_{k}cosk\varphi +b_{k}sink\varphi )\end{array}$

Recall the equation (4),

${\frac{\partial V}{\partial s}|}_{R+}-{\frac{\partial V}{\partial s}|}_{R-}=\frac{-\sigma }{{\epsilon }_0}$

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

$\sum \left(\frac{-k}{R^{k+1}}\right)(c_{k}cosk\varphi +d_{k}sink\varphi )+\sum \left(k{R}^{k-1}\right)(a_{k}cosk\varphi +b_{k}sink\varphi )=\frac{-\sigma }{{\epsilon }_0}$

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

$\begin{array}{l}\sum \left(\frac{-k}{R^{k+1}}\right)(R^{2k}{a}_{k}cosk\varphi +R^{2k}{b}_{k}sink\varphi )-\sum \left(k{R}^{k-1}\right)(a_{k}cosk\varphi +b_{k}sink\varphi )=\frac{-\sigma }{{\epsilon }_0}\\ -\sum \left(k{R}^{k-1}\right)(a_{k}cosk\varphi +b_{k}sink\varphi )-\sum \left(k{R}^{k-1}\right)(a_{k}cosk\varphi +b_{k}sink\varphi )=\frac{-\sigma }{{\epsilon }_0}\\ \sum 2k{R}^{k-1}(a_{k}cosk\varphi +b_{k}sink\varphi )=\frac{\sigma }{{\epsilon }_0}\end{array}$ ……(5)

Noe defines the above equation for the intervals,

$\sum 2k{R}^{k-1}(a_{k}cosk\varphi +b_{k}sink\varphi )=\{\begin{array}{l}\frac{{\sigma }_0}{{\epsilon }_0}(0<\varphi <\pi )\\ \frac{-{\sigma }_0}{{\epsilon }_0}(\pi <\varphi <2\pi )\end{array}$

The value of integral of above angles,

$\begin{array}{l}{\int }_0^{2\pi }\mathrm{sink}\varphi \cos l\varphi d\varphi =0\\ {\int }_0^{2\pi }cosk\varphi \cos l\varphi d\varphi =\{\begin{array}{l}0k\ne l\\ \pi k=l\end{array}\end{array}$

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

$\begin{array}{l}2k{R}^{k-1}[{\int }_0^{2\pi }{a}_k\cos k\varphi \cos l\varphi d\varphi +{\int }_0^{2\pi }{b}_k\sin k\varphi \mathrm{cosl}\varphi d\varphi ]=\frac{{\sigma }_0}{{\epsilon }_0}{[{\int }_0^{2\pi }\cos l\varphi d\varphi -{\int }_0^{2\pi }\cos l\varphi ]}_{}\underset{\delta x\to 0}{\lim }\\ 2k{R}^{k-1}\pi a_k=\frac{{\sigma }_0}{{\epsilon }_0}{\left[\frac{\sin l\varphi }{l}\right]}_0^{\pi }\\ 2k{R}^{k-1}\pi a_k=\frac{{\sigma }_0}{{\epsilon }_0}\left(0\right)\\ a_k=0\end{array}$

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

$\begin{array}{l}2k{R}^{k-1}[{\int }_0^{2\pi }{a}_k\cos k\varphi \sin l\varphi d\varphi +{\int }_0^{2\pi }{b}_k\sin k\varphi \mathrm{sinl}\varphi d\varphi ]=\frac{{\sigma }_0}{{\epsilon }_0}[{\int }_0^{2\pi }\sin l\varphi d\varphi -{\int }_0^{2\pi }\cos l\varphi ]\\ 2k{R}^{k-1}\pi b_k=\frac{{\sigma }_0}{{\epsilon }_0}[{(-\frac{\cos l\varphi }{l})}_0^{\pi }+{\left(\frac{\cos l\varphi }{l}\right)}_0^{2\pi }]\\ b_k=\frac{{\sigma }_0}{l{\epsilon }_0}(2-2\cos l\pi )\\ b_k=\frac{{\sigma }_0}{2k\pi l{\epsilon }_0{R}^{k-1}}(2-2\cos l\pi )\end{array}$

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

$b_k=\{\begin{array}{l}0\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}if\hspace{0.17em}}l\text{\hspace{0.17em}is\hspace{0.17em}even}\\ \frac{2{\sigma }_0}{\pi k^2{R}^{k-1}{\epsilon }_0}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}if\hspace{0.17em}}l\text{\hspace{0.17em}is\hspace{0.17em}odd}\end{array}$

Similarly for the outside of the cylinder.

$\begin{array}{l}2k{R}^{k-1}[{\int }_0^{2\pi }{c}_k\cos k\varphi \sin l\varphi d\varphi +{\int }_0^{2\pi }{d}_k\sin k\varphi \mathrm{sinl}\varphi d\varphi ]=\frac{{\sigma }_0}{{\epsilon }_0}[{\int }_0^{2\pi }\sin l\varphi d\varphi -{\int }_0^{2\pi }\cos l\varphi ]\\ 2k{R}^{k-1}\pi d_k=\frac{{\sigma }_0}{{\epsilon }_0}[{(-\frac{\cos l\varphi }{l})}_0^{\pi }+{\left(\frac{\cos l\varphi }{l}\right)}_0^{2\pi }]\\ d_k=\frac{{\sigma }_0}{l{\epsilon }_0}(2-2\cos l\pi )\\ d_k=\frac{{\sigma }_0}{2k\pi l{\epsilon }_0{R}^{k-1}}(2-2\cos l\pi )\end{array}$

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

$d_k=\{\begin{array}{l}0\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}if\hspace{0.17em}}l\text{\hspace{0.17em}is\hspace{0.17em}even}\\ \frac{2{\sigma }_0}{\pi k^2{R}^{k-1}{\epsilon }_0}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}if\hspace{0.17em}}l\text{\hspace{0.17em}is\hspace{0.17em}odd}\end{array}$

Substitute $\frac{2{\sigma }_0}{\pi k^2{R}^{k-1}{\epsilon }_0}$for$b_k$ and $0$for$a_k$into equation (2).

$\begin{array}{l}{V}_{in}(s,\varphi )=\sum _{K=1}^{\alpha }{s}^k(\left(0\right)cosk\varphi +\frac{2{\sigma }_0}{\pi k^2{R}^{k-1}{\epsilon }_0}sink\varphi )\\ V_{in}(s,\varphi )=\sum _{K=1}^{\alpha }{s}^k\left(\frac{2{\sigma }_0}{\pi k^2{R}^{k-1}{\epsilon }_0}sink\varphi \right)\\ V_{in}(s,\varphi )=\frac{2{\sigma }_{0}R}{\pi {\epsilon }_0}\sum _{K=1,3,5}\frac{1}{k^2}\sin k\varphi {\left(\frac{s}{R}\right)}^k\end{array}$

Substitute$\frac{2{\sigma }_0}{\pi k^2{R}^{k-1}{\epsilon }_0}\underset{\delta x\to 0}{\lim }$for$d_k$and$0$for$c_k$into equation (3).

$\begin{array}{l}{V}_{out}(s,\varphi )=\sum _{K=1}^{\alpha }{s}^{-k}(\left(0\right)cosk\varphi +\frac{2{\sigma }_0}{\pi k^2{R}^{k-1}{\epsilon }_0}sink\varphi )\\ V_{out}(s,\varphi )=\sum _{K=1}^{\alpha }{s}^{-k}\left(\frac{2{\sigma }_0}{\pi k^2{R}^{k-1}{\epsilon }_0}sink\varphi \right)\\ V_{out}(s,\varphi )=\frac{2{\sigma }_{0}R}{\pi {\epsilon }_0}\sum _{K=1,3,5}\frac{1}{k^2}\sin k\varphi {\left(\frac{R}{s}\right)}^k\end{array}$

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

$\frac{2{\sigma }_{0}R}{\pi {\epsilon }_0}\sum _{K=1,3,5}\frac{1}{k^2}\sin k\varphi {\left(\frac{s}{R}\right)}^k$and outside the cylinder is$\frac{2{\sigma }_{0}R}{\pi {\epsilon }_0}\sum _{K=1,3,5}\frac{1}{k^2}\sin k\varphi {\left(\frac{R}{s}\right)}^k$