# Surface charge density due to point charge

1. Apr 2, 2013

### NewtonianAlch

1. The problem statement, all variables and given/known data
Using the method of images, solve the following

3. The attempt at a solution

I have very little idea on how to solve this. I looked up various methods online and none seem to be what the lecturer has used.

This is the solution he posted up, but as usual it is missing 300 steps and also seems to differ to the short answer he has given in the actual problem sheet, although I figured because of the 2 sin θ, the 2 and 4 cancel, leaving only 2 at the bottom. And ε disappears because for finding ρ (surface charge density) we multiply the E-field by ε.

However I don't understand how a 2sinθ came to be and then abruptly disappears.

Any help appreciated.

2. Apr 2, 2013

### Staff: Mentor

The electric fields from the two charges (the real and the imaginary one) go along the lines in the sketch (one is marked with "E". The magnitude is calculated with the usual formula, and their vertical component is sin(theta) of their magnitude. You have two charges, both have their vertical component in the same direction, therefore you get a factor of 2 there.

To get rid of the angle, he uses sin(theta)=d/R. You can see the d in the numerator, and R is converted to sqrt(d^2+r^2) in the denominator. The factor of 2 seems to be missing there.
Right.

3. Apr 2, 2013

### rude man

If I may add, and pardon if it's superfluous info:
Two main ideas here:

1. E field just above a conducting surface is εE = σ where σ is surface charge density. The direction of E is normal to, and away from, the conductor if the surface charge density is +. This is easily shown using Gauss's theorem. If the surface charge is -, then the direction is into the plane.

2. The concept of imaging which is as shown on the proffered answer sheet. The basis is that the potetial anywhere along the sheet is constant (equipotential surface) if the image charge is added and the presence of the sheet ignored.

4. Apr 3, 2013

### NewtonianAlch

Thanks for the responses, I understand it clearly now!