2
$\begingroup$

I was looking at some problems/questions about electrostatics and came across this question:

An infinitely long cylinder of radius $a$ and with charge density per unit length $\lambda$ is placed with its axis a distance $d$ from an infinite conducting plane which is at zero potential. Show that two line charges with charges $\lambda$ and $-\lambda$ parallel to and at distance $\sqrt{{d^2 - a^2}}$ either side of the plane, give rise to the potential distribution between the cylinder and plane.

Hence show that the potential of the cylinder is given by $$ V = \frac{\lambda}{2\pi\epsilon_0}ln\left(\frac{d + \sqrt(d+a)}{a}\right) \ . $$

The main issues I am having are to do with the first part and showing rather than verifying the configuration. Is there a way of deriving this and how would you go about deriving it (without simply 'guessing')?
I have also tried to think of the cylinder as many, infinitely-long line charges arranged in a circle, and then reflected each one in the plane to produce an image of a corresponding cylinder. Is it valid to represent the cylinder with a line of charge at its centre?

$\endgroup$
  • $\begingroup$ Is $a$ the radius of the cylinder? $\endgroup$ – Floris Jan 30 '16 at 19:41
  • $\begingroup$ yes sorry I forgot to mention that, I'll edit it now $\endgroup$ – Hakkihan Tunbak Jan 30 '16 at 22:52
1
$\begingroup$

Assuming the cylinders are conducting.

The method of images is derived by showing that the equipotentials of two lines of opposite charge are cylinders or a plane. You then have to do a bit of algebra to figure out where to put the charges so the equipotentials coincide with the given cylinders. (The charge will not be at the centre of the cylinder and the surface charge will not be distributed uniformly around it.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.