Solving Method of Images Problem for Infinite Conducting Cylinder

[natski]

Hello, I am requiring some help on a problem on method of images.

The problem is, you are given a infinitely long conducting cylinder of radius a, with charge density per unit length lambda (I'll use the symbol Y) which is placed with its axis a distance d from an infinite conducting plane at zero potential (I assume it is parallel to it).

Show that two infinitely long line charges of value Y and -Y, parallel to and at a distance of (d^2 - a^2)^0.5 either side of the plane, give rise to the required potential distribution between the cylinder and the plane.

My tried solution...

firstly, the potential due to a line charge is...

V = -[Y/(2*pi*$)]*ln(r) + const.

where pi = 3.14... and $ = epsilon zero = 8.854 x 10^-12 Fm^-1

secondly, the potential due to a cylinder of radius a is...

V = -[Y/(2*pi*$)]*ln(r/a)

I'm interpreting this problem as we have two situations, one with just the cylinder and the plane, and one with the two line charges either side of the plane. By job is to find the distance x from the line charges to the plane which arise to the exact same potential distribution as with the cylinder & plane.

So, consider the potential at a line P defined by being inbetween the axis of the cylinder and the infinitely conducting plane and at a distance p from the infinitely conducting plane.

Looking at the case of the cylinder, the potential along line P is given by...

V = -[Y/(2*pi*$)]*ln[(d-p)/a)] {1}

where i will reiterate that d is the distance from the axis of the cylinder to the plane and p is the distance from the line P to the plane.

Looking at the case of two line charges, the potential along line P is given by:

V = -[Y/(2*pi*$)]*ln(x-p) + const. + [Y/(2*pi*$)]*ln(x+p) - const.

where x is the distance from a line charge to the plane (same for both)
so I'm going to cancel the constants for simplicity (risky I know)

therefore V = [Y/(2*pi*$)]*[ln(x+p) - ln(x-p)] {2}

now {1} and {2} must be equivalent for the potential distribution to be the same in both cases...

so this gives us, cancelling the [Y/(2*pi*$)] term...
ln(x-p) - ln(x+p) = ln[(d-y)/a]

giving ln[(x-p)/(x+p)] = ln[(d-p)/a]
so (x-p)/(x+p) = (d-p)/a

I was kind of hoping at this point the p's would cancel.

Clearly we need to find x, without any p's in it... this is where I cannot progress any further.

Any help or tips would be appreciated.

Natski

 

[vincentchan]

before you write down the formulas of the potential, ask yourself one question... In you formulas, where is the zero potential for the line charge (if any) and cylinderical charge? you will soon realize what did you go wrong...

I can do this problem without messy mathematics. first, imagine you have a cylinder ONLY (let's say the metal sheet is not there yet), how does this potential compare with ONE line charge (when r>a), can you show that? (from your formulas) and what is the constant C in your formulas...

Okay, now we put the metal sheet in the system, put is to the left of the cylinder, the graph is like this

|
| o <---the top view of cylinder
|
| <----------------metal sheet

you know the E-field @ the sheet is perpedicular to the sheet, that's mean Ey and Ez is zero... compare to your line charge model, how can you put an image charge in your line charge model such that Ey and Ez cancel out?... and what is the potential there?

notice that if the potential of the boundary is equal, the potential is equal every where in your system.. what is that therom called? ... and how can you applied it here??

 

[wais]

Hi Natski, thanks for reaching out for help on this problem. It seems like you have a good understanding of the method of images and have made some good progress on the solution. Let me try to guide you through the rest of the steps.

First, let's consider the potential at a point P on the axis of the cylinder, at a distance p from the plane. In the case of just the cylinder and the plane, the potential at P is given by equation {1} in your solution. In the case of the two line charges, the potential at P is given by equation {2}.

Now, as you correctly noted, for the potential distribution to be the same in both cases, equations {1} and {2} need to be equivalent. This means that the potential at P must be the same in both cases. So, we can set equations {1} and {2} equal to each other and solve for x.

This gives us ln(x-p) - ln(x+p) = ln[(d-p)/a]. Using the logarithm property ln(a) - ln(b) = ln(a/b), we can rewrite this as ln[(x-p)/(x+p)] = ln[(d-p)/a]. From here, we can exponentiate both sides to get rid of the natural logarithm, giving us (x-p)/(x+p) = (d-p)/a. Now, we can cross-multiply to get x^2 - p^2 = ax + ap - px - p^2. Simplifying this, we get x^2 - (a+p)x + ap = 0.

This is a quadratic equation in x, which can be solved using the quadratic formula. Once you have the two values for x, you can substitute them into equation {2} to find the potential at P in the case of the two line charges. You will see that it will be the same as the potential given by equation {1} for the cylinder and the plane.

I hope this helps you complete the solution to this problem. Let me know if you have any further questions or need more clarification. Good luck!