- #1

neroE

- 4

- 1

- Homework Statement
- N/A

- Relevant Equations
- Coulomb's law mainly.

Hello,

This question, which I found in various electricitiy and magnetism books (e.g. Introduction to electrodynamics grif.).

There are many variations of this question, I am mainly interested in the following setup of it:

-Suppose there is a charged disk of radius R lying in the xy-plane, and the electric field is uniform. What is the electric field at the point P(0,0,z)?

I know this can be solved by first considering a ring, deriving a formula for it, and then summing the infinite number of rings that make up a disk (via integration), and thus, the desired result is yielded. However, I am more interested in how to approach this using surface integration?I started with the parametrization G(u,v) = <ucos(v),u*sin(v),0> and found the normal unit vector to be <0,0,1>

where: 0<u<R and 0<v<2*pi

But, I am not sure how to continue from there, I attached a pdf showing my subsequent steps, which I am not sure if it is valid or not.

Please help, and thank you in advance.

This question, which I found in various electricitiy and magnetism books (e.g. Introduction to electrodynamics grif.).

There are many variations of this question, I am mainly interested in the following setup of it:

-Suppose there is a charged disk of radius R lying in the xy-plane, and the electric field is uniform. What is the electric field at the point P(0,0,z)?

I know this can be solved by first considering a ring, deriving a formula for it, and then summing the infinite number of rings that make up a disk (via integration), and thus, the desired result is yielded. However, I am more interested in how to approach this using surface integration?I started with the parametrization G(u,v) = <ucos(v),u*sin(v),0> and found the normal unit vector to be <0,0,1>

where: 0<u<R and 0<v<2*pi

But, I am not sure how to continue from there, I attached a pdf showing my subsequent steps, which I am not sure if it is valid or not.

Please help, and thank you in advance.