Electric field from a non-uniformly charged disk

[mshahi]

Homework Statement

We are given a disk with negligible thickness, a radius of 1m, and a surface charge density of σ(x,y) = 1 + cos(π√x²+y²). The disk is centered at the origin of the xy plane. We are also given the location of a point charge in Cartesian coordinates, for example [0.5,0.5,2]. We need to find the electric field components (x,y,z) at the location of the point charge.

Homework Equations

d$\vec E$ = σdS/4πε₀ ⋅ $\hat r$/r²

The Attempt at a Solution

My first thought was to make dS a thin ring centre origin. this would give dS=2πr'dr' where r' is the radius from the origin to the ring. I then thought the I could write r² as r'²+z², and r' would also be equal to √x²+y². Plugging this and σ into the integral just to find the magnitude of $\vec E$ we get:
∫ (1+cos(πr'))r'/(r'²+z²) dr'
However I have realized that this is wrong because a) the answer it gives is far to small to be reasonable, b) I think this calculation assumes a spherical field which it clearly isn't, and c) the relationship to find r in terms of r' and z only holds if the point charge is over the disk, which it isn't necessarily. I think I am meant to split the integral into x, y, and z components but am unsure of whether this is the correct approach. Now i am completely stuck, none of the notes I can find explain this, I even took out a book from the 1950s on electrostatics to try and find the way to solve this and I just cannot for the life of me find it anywhere! Any help would be very much appreciated!

 

[BvU]

Hello mshahi,

Are you aware that the $\vec r$ in $d\vec E$ is a different one from the $r$ in $dS$ ? How ? Writing $r^2$ as $r'\,^2 + z^2$ makes $x$ and $y$ disappear. Or are you only interested in the field on the z axis ?

Make a drawing to oversee the situation and set up an expression for the integration.