Question

Find the method of images for problem 4.

Short answer

The image is a uniform charge along an infinite line with a charge per unit length $-\lambda ,$that passes through the point $(\raisebox{1ex}{$R^2$}\!\left/ \!\raisebox{-1ex}{$a$}\right.,0)$in the plane perpendicular to the cylinder.

Step-by-step solution

Given Information:

It has been given that for reference follow example 1.

Uses of the Green function:

The green function is a function that is used to solve a corresponding partial differential equation in three dimensions, namely Poisson’s equation.

$\begin{array}{c}{\mathbf{\nabla }}^{\mathbf{2}}\mathbf{u}\mathbf{=}\mathbf{f}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\\ \mathbf{=}\mathbf{f}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{,}\mathbf{z}\mathbf{)}\end{array}$

Understand the terms to solve:

To find the method of images for Problem 4. In Problem 4 there is a uniform charge along an infinite line, which is outside of an infinite grounded cylinder (the potential on the surface of the cylinder is zero).

The line and the cylinder lay parallel. Because all the cross-sections of the parallel line and cylinder are the same, the problem is in effect a two-dimensional variant of Example l (from the two-dimensional point of view, see the cylinder as a circle and the line as a point).

Restrict ourselves to a single cross-section and set the origin of the xy coordinate system at the center of the cylinder. If the line pierces the plane (the cross-section) through a point on the x-axis a distance from the origin, which is point $(a,0),$the total potential which is the solution to Problem 4.

$V=\lambda \ln a^2-\lambda \ln R^2-\lambda \ln (r^2-2ra\left(\cos \theta \right)+a^2)+\lambda \ln (r^2-2r\frac{R^2}{a}\cos \theta +{\left(\frac{R^2}{a}\right)}^2)$

Here, $\lambda$is the line charge per unit of length and $\theta$is the azimuthal angle.

This result gives the potential in the region outside of the grounded cylinder, and it is observed that, as expected, it doesn't depend on z (the axis parallel to the line and the cylinder).

State the Image method.

The first two terms in result (I) are constants. The third term corresponds to the real uniform charge along an infinite line with charge per unit length $\lambda ,$which passes through the point $(a,0)$in the plane.

The fourth term corresponds to the image charge. Comparing the third and fourth terms it is seen that when $\lambda$is changed with $-\lambda$and a with $R^2/a$in the third term, the fourth term is obtained.

Thus, the image is a uniform charge along an infinite line with a charge per unit length $-\lambda ,$which passes through the point $(R^2/a,0)$in the plane. Because $R^2/a=(R/a)R<R$(the real charged line is outside of the cylinder, meaning that $a>R,$and thus $R/a<1$), the image charged line is located inside the cylinder (parallel to the cylinder and the real charged line). The effect of the real charged line and the image charged line on the area outside of the cylinder is the same as the effect if there is just the real charged line and the grounded cylinder.

Hence, the image is a uniform charge along an infinite line with a charge per unit length $-\lambda ,$which passes through the point $(R^2/a,0)$in the plane perpendicular to the cylinder.