91影视

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.

${鈭�}^2=\frac{1}{r}\frac{鈭�}{鈭�r}\left(r\frac{鈭�}{鈭�r}\right)+\frac{1}{r^2}\frac{{鈭�}^2}{鈭�{胃}^2}$ 鈥�.. (1)

Use equation (1) to find the general solution.

$V=\underset{n=0}{\overset{鈭�}{鈭�}}\left(r^n\left(A_n\sin \left(n胃\right)+B_n\cos \left(n胃\right)\right)+r^{-n}\left(A_n^{\text{'}}\sin \left(n胃\right)+B_n^{\text{'}}\cos \left(n胃\right)\right)\right.$

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

$V=\underset{n=0}{\overset{鈭�}{鈭�}}{r}^{-n}\left(A_n^{\text{'}}\sin \left(n胃\right)+B_n^{\text{'}}\cos \left(n胃\right)\right)$

The uniform electric field in negative x direction is,

$E=-E_0\mathbf{i}$ 鈥�.. (2)

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

$E=-鈭�V^{\text{'}}$ 鈥�.. (3)

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

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

鈥�.. (4)

$E_0\mathbf{i}=\frac{鈭�V}{鈭�x}\mathbf{i}$

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

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

Replace x by cylindrical coordinates.

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

The solution for theLaplace鈥檚 equationis:

$V=\underset{n=0}{\overset{鈭�}{鈭�}}{r}^{-n}\left(A_n^{\text{'}}\sin \left(n胃\right)+B_n^{\text{'}}\cos \left(n胃\right)\right)$ 鈥�.. (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$B_1^{\text{'}}$.

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

Substitute the coefficients in the general solution.

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

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