# Electric field from a non-uniformly charged disk

• mshahi
In summary, the problem at hand involves finding the electric field components at the location of a point charge located at [0.5, 0.5, 2] in Cartesian coordinates. This is done by using the equation d##\vec E## = σdS/4πε0 ⋅ ##\hat r##/r2 and setting up an integral to calculate the magnitude of the electric field. However, it is important to note that the ##\vec r## in ##d\vec E## is different from the ##r## in ##dS##. Additionally, the relationship to find r in terms of r' and z only holds if the point charge is directly above the disk. Making a drawing
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!

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.

## 1. What is an electric field from a non-uniformly charged disk?

An electric field from a non-uniformly charged disk refers to the force per unit charge exerted on a charged particle placed near the disk. This field is caused by the distribution of electric charges on the surface of the disk.

## 2. How is the electric field calculated for a non-uniformly charged disk?

The electric field is calculated using Coulomb's law, which states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. For a non-uniformly charged disk, the field is calculated by summing the contributions of each small element of charge on the disk.

## 3. What is the difference between a uniform and non-uniformly charged disk?

A uniform charged disk has a constant distribution of charge on its surface, while a non-uniformly charged disk has varying amounts of charge distributed across its surface. This difference affects the strength and direction of the electric field around the disk.

## 4. How does the electric field from a non-uniformly charged disk affect nearby charged particles?

The electric field from a non-uniformly charged disk will exert a force on nearby charged particles, causing them to either be attracted or repelled depending on the direction of the electric field. The strength of the force will also depend on the distance between the particles and the disk, as well as the amount and distribution of charge on the disk.

## 5. Can the electric field from a non-uniformly charged disk be manipulated?

Yes, the electric field from a non-uniformly charged disk can be manipulated by altering the distribution of charge on the disk. This can be done by adding or removing charges, changing the shape of the disk, or changing the material properties of the disk. These manipulations can result in changes to the strength and direction of the electric field around the disk.

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