A long conducting cylinder is placed parallel to thez-axis in an originally uniform electric field in the negativexdirection. The cylinder is held at zero potential. Find the potential in the region outside the cylinder.

In an initially uniform electric field in the negative x direction, a long conducting cylinder is placed parallel to the z-axis. The cylinder's potential is held at zero.

The total of the second-order partial derivatives of R, the unknown function, in Cartesian coordinates equals 0, according to Laplace's equation.

Write the Laplace equation in the cylindrical coordinates.

${\nabla }^2=\frac{1}{r}\frac{\partial }{\partial r}\left(r\frac{\partial }{\partial r}\right)+\frac{1}{r^2}\frac{{\partial }^2}{\partial {\theta }^2}$ ….. (1)

Use equation (1) to find the general solution.

$V=\sum _{n=0}^{\infty }\left(r^n\left(A_n\sin \left(n\theta \right)+B_n\cos \left(n\theta \right)\right)+r^{-n}\left(A_n^{\text{'}}\sin \left(n\theta \right)+B_n^{\text{'}}\cos \left(n\theta \right)\right)\right.$

To find the potential outside the cylinder, coefficients of $r^n$should be zero.

$V=\sum _{n=0}^{\infty }{r}^{-n}\left(A_n^{\text{'}}\sin \left(n\theta \right)+B_n^{\text{'}}\cos \left(n\theta \right)\right)$

The uniform electric field in negative x direction is,

$E=-E_0\mathbf{i}$ ….. (2)

The electric field in a negative gradient of the potential is,

$E=-\nabla V^{\text{'}}$ ….. (3)

Compare equation (2) and (3), you get

$-E_0\mathbf{i}=-\nabla {\mathbf{V}}^{\text{'}}$

….. (4)

$E_0\mathbf{i}=\frac{\partial V}{\partial x}\mathbf{i}$

Integrate equation (4) to get$V\text{'}$.

$V^{\text{'}}=E_{0}x$

Replace x by cylindrical coordinates.

$$\begin{gathered} x=r\cos \theta \\ {V}^{\text{'}}=E_{0}r\cos \theta \end{gathered}$$

The solution for theLaplace’s equationis:

$V=\sum _{n=0}^{\infty }{r}^{-n}\left(A_n^{\text{'}}\sin \left(n\theta \right)+B_n^{\text{'}}\cos \left(n\theta \right)\right)$ ….. (5)

Replace the boundary conditions in equation (5).

$$\begin{gathered} A_n^{\text{'}}=0 \\ {B}_n^{\text{'}}=0 \end{gathered}$$

Replace the above values in the general equation to find the value of$B_1^{\text{'}}$.

$$\begin{gathered} 0=E_{0}a\cos \theta +0+B_1^{\text{'}}{a}^{-1}\cos \theta \\ {B}_1^{\text{'}}=-E_0{a}^2 \end{gathered}$$

Substitute the coefficients in the general solution.

$$\begin{gathered} V=E_{0}r\cos \theta +\left(-E_0{a}^2\right)\frac{1}{r}\cos \theta \\ =E_0\cos \theta \left(r-\frac{a^2}{r}\right) \end{gathered}$$

Hence The potential in the region outside the cylinder is$V=E_0\cos \theta \left(r-\frac{a^2}{r}\right)$ .