91Ó°ÊÓ

The uniform electric field is $E_0$

The radius of the cylinder is a.

The susceptibility is${\mathrm{χ}}_e$

The expression for the general equation to calculate the potential of the cylinder symmetry is given as follows.

$V\left(s,\mathrm{ϕ}\right)=a_0+b_{0}lns+\underset{k=1}{\overset{∞}{∑}}{s}^k\left(a_{k}cosk\mathrm{ϕ}+b_{k}sink\mathrm{ϕ}\right)s+s^{-k}\left(c_{k}cosk\mathrm{ϕ}+d_{k}sink\mathrm{ϕ}\right)$

Here,$a_0,b_0,a_x,b_{x,}{c}_k$ and$d_k$ are the constant

Consider the following figure which shows the long cylinder placed in uniform electric field $E_0$.

Let assume the cylinder is uncharged initially and placed into uniform electric field.

In the presence of the dielectric the potential would be $-E_{0}x=-E_{0}s\cos \mathrm{Ï•}$.

To calculate the potential inside and outside the cylinder use the boundary condition.

(i) $V_{in}=V_{out}$ at$s=a$

(ii) $\mathrm{Î\mu }\frac{∂V_{in}}{∂r}={\mathrm{Î\mu }}_0\frac{∂V_{out}}{∂r}$ at$s=a$

(iii) $V_{out}→-E_{0}s\cos \mathrm{ϕ}$ for$s>>a$

The general equation to calculate the potential of cylinder with symmetry is given by,

$V\left(s,\mathrm{ϕ}\right)=a_0+b_0\ln s+\underset{k=1}{\overset{∞}{∑}}{s}^k\left(a_k\cos k\mathrm{ϕ}+b_k\sin k\mathrm{ϕ}\right)+s^{-k}\left(c_k\cos k\mathrm{ϕ}+d_k\sin k\mathrm{ϕ}\right)$

The expression for the potential inside the cylinder is given by,

$V_{in}=\underset{k=1}{\overset{∞}{∑}}{s}^k\left(a_k\cos k\mathrm{ϕ}+b_k\sin k\mathrm{ϕ}\right)$

The expression for the potential outside the cylinder is given by,

$V_{out}=-E_{0}s\cos \mathrm{ϕ}+\underset{k=1}{\overset{∞}{∑}}{s}^{-k}\left(c_k\cos k\mathrm{ϕ}+d_k\sin k\mathrm{ϕ}\right)$

By using the boundary condition (1),

$∑a^k\left(a_k\cos k\mathrm{ϕ}+b_k\sin k\mathrm{ϕ}\right)=-E_{0}a\cos \mathrm{ϕ}+∑a^{-k}\left(c_k\cos k\mathrm{ϕ}+d_k\sin k\mathrm{ϕ}\right)$

By using the boundary condition (2),

$\mathrm{Î\mu }∑k{a}^{k-1}\left(a_k\cos k\mathrm{Ï•}+b_k\sin k\mathrm{Ï•}\right)=-E_{0}a\cos \mathrm{Ï•}-∑k{a}^{-k-1}\left(c_k\cos k\mathrm{Ï•}+d_k\sin k\mathrm{Ï•}\right)$

The above equations will satisfy only when

$b_k=d_k=0$ for all values of k

$a_k=c_k=0$ except for k=1

For$k=1$

$$\begin{gathered} a{a}_1=-E_{0}a+a^{-1}{c}_1 \\ {\mathrm{Î\mu }}_r{a}_1=-E_0-a^{-2}{c}_1 \end{gathered}$$

Solve the above two equations,

$a_1=\frac{-2E_0}{{\mathrm{Î\mu }}_r+1},c_1=a^2{E}_0\frac{{\mathrm{Î\mu }}_r-1}{{\mathrm{Î\mu }}_r+1}$

Therefore,$V_{in}\left(s,\mathrm{Ï•}\right)=-\frac{2E_0}{\left(1+{\mathrm{Î\mu }}_r\right)}s\cos \mathrm{Ï•}$

Let assume $s\cos \mathrm{Ï•}$is X.

Hence the expression for the potential inside the cylinder is$V_{in}\left(s,\mathrm{Ï•}\right)=-\frac{2E_0}{\left(1+{\mathrm{Î\mu }}_r\right)}x$

The electric field inside the cylinder is given by,

$E_{in}\left(s,\mathrm{ϕ}\right)=-\frac{∂V_{in}}{∂x}\hat{x}$

Substitute$-\frac{2E_0}{\left(1+{\mathrm{Î\mu }}_r\right)}x$for $V_{in}$into above equation.

$$\begin{gathered} E_{in}\left(s,\mathrm{Ï•}\right)=-\frac{∂}{∂x}\left(-\frac{2E_0}{\left(1+{\mathrm{Î\mu }}_r\right)}x\right)\hat{x} \\ {E}_{in}\left(s,\mathrm{Ï•}\right)=\frac{2E_0}{\left(1+{\mathrm{Î\mu }}_r\right)}\hat{x} \end{gathered}$$

Substitute $1+{\mathrm{χ}}_e$for ${\mathrm{Î\mu }}_r$into above equation.

$$\begin{gathered} E_{in}\left(s,\mathrm{ϕ}\right)=\frac{2E_0}{\left(1+1+{\mathrm{χ}}_e\right)}\hat{x} \\ {E}_{in}\left(s,\mathrm{ϕ}\right)=\frac{2E_0}{\left(2+{\mathrm{χ}}_e\right)}\hat{x} \\ {E}_{in}\left(s,\mathrm{ϕ}\right)=\frac{E_0}{\left(1+\frac{{\mathrm{χ}}_e}{2}\right)}\hat{x} \end{gathered}$$

Hence, the resultant electrical field inside the cylinder is$E_{in}\left(s,\mathrm{ϕ}\right)=\frac{E_0}{\left(1+\frac{{\mathrm{χ}}_e}{2}\right)}\hat{x}$.