Surface charge density due to point charge

[NewtonianAlch]

Homework Statement

Using the method of images, solve the following

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.

 

[mfb]

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.

And ε disappears because for finding ρ (surface charge density) we multiply the E-field by ε.

Right.

 

[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.

 

[NewtonianAlch]

Thanks for the responses, I understand it clearly now!

 

[Nickie]

The method of images is a powerful tool in solving problems involving point charges and conductors. It involves creating a "mirror image" of the original point charge, called an image charge, that is located at a specific distance from the conductor. This image charge creates an electric field that cancels out the electric field of the original charge at the surface of the conductor, resulting in a net zero electric field inside the conductor.

In this specific problem, we are trying to find the surface charge density (ρ) due to a point charge (q) located at a distance (d) from a conducting plane. The surface charge density is defined as the charge per unit area, and is given by ρ = q/A, where A is the area of the conducting plane.

To solve this problem, we first need to find the electric field at the surface of the conductor due to the point charge. This can be done using Coulomb's law, which states that the electric field (E) at a distance (r) from a point charge (q) is given by E = kq/r^2, where k is the Coulomb's constant.

Next, we use the method of images to create an image charge (q') that is located at a distance (d') from the conductor, such that the electric field due to this image charge at the surface of the conductor is equal and opposite to the electric field due to the original charge. This results in a net zero electric field at the surface of the conductor.

Now, using the superposition principle, we can add the electric fields due to the original charge and the image charge to find the total electric field at the surface of the conductor. This is given by E = kq/r^2 - kq'/r'^2, where r' is the distance from the image charge to the surface of the conductor.

To find the surface charge density, we multiply the electric field by the permittivity of free space (ε) and take the limit as the distance from the conductor goes to zero. This is because the surface charge density is defined as the charge per unit area at the surface of the conductor, and as the distance goes to zero, the area also goes to zero.

In this specific problem, the 2sinθ term comes from the geometry of the problem, and it disappears because as the distance from the conductor goes to zero, the angle θ also goes to zero, making the sinθ term equal