Coulomb triple integral for a sphere

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
11 replies · 3K views
Luke Bower
Messages
6
Reaction score
0

Homework Statement


Find the electric field of a sphere of radius R and charge Q outside sphere. Use only a Coulomb integral to do this.

Homework Equations


I know that I have to use a triple integral to find the E-field. I am just unsure of my whole setup really.

The Attempt at a Solution


I tried setting up a triple integral of (0-2pi),(0,R) and then I don't know what to put for the the bounds of the third.
 
Physics news on Phys.org
You need to assume some more information. Is the charge distribution spherically symmetric?
In spherical coordinates, your three coordinates are ##(r,\theta,\phi)##. Do you know what the valid range is for the ##\theta## coordinate? If not, you'd better study up on the spherical coordinate system.
 
Luke Bower said:

Homework Statement


Find the electric field of a sphere of radius R and charge Q outside sphere. Use only a Coulomb integral to do this.

Homework Equations


I know that I have to use a triple integral to find the E-field. I am just unsure of my whole setup really.

The Attempt at a Solution


I tried setting up a triple integral of (0-2pi),(0,R) and then I don't know what to put for the the bounds of the third.
Could you please provide the actual problem statement? Your version is kind of ambiguous.
 
Problem statement: find th electric field of a sphere of radius R and charge Q outside the sphere. Do this using BOTH a gauss's law and coulomb integral.

I have done the gaussian one, which was pretty easy. I just can't figure out the coulomb integral.

No i have not used spherical coordinates before and it is not a prereq for this course apparently. But if i have to do it then i have to ha

I appreciate any help.
 
Is that the problem statement as given to you word for word? It still seems kind of vague. Charge Q outside the sphere? Q could be anywhere then, except on or inside the sphere. Do you mean Q is on the surface of the sphere? If so, is it distributed uniformly? Is the sphere conducting or is it an insulator?
 
vela said:
Is that the problem statement as given to you word for word? It still seems kind of vague. Charge Q outside the sphere? Q could be anywhere then, except on or inside the sphere. Do you mean Q is on the surface of the sphere? If so, is it distributed uniformly? Is the sphere conducting or is it an insulator?

My teacher said the sphere is conducting. And yes that is word for word. He is very vague in all of his problem statements.
 
Wow TCC Engineering Physics II I'm guessing.
 
If it's a conducting sphere, the charge is going to distributed on the surface, so you only have to integrate over the surface.

You've probably worked out or seen worked out the electric field due to a ring of charge. Try using that result by splitting up the sphere into a bunch of infinitesimal rings.
 
jugglar456 said:
Wow TCC Engineering Physics II I'm guessing.

jugglar456 said:
Wow TCC Engineering Physics II I'm guessing.
yeah pretty much ha
 
vela said:
If it's a conducting sphere, the charge is going to distributed on the surface, so you only have to integrate over the surface.

You've probably worked out or seen worked out the electric field due to a ring of charge. Try using that result by splitting up the sphere into a bunch of infinitesimal rings.
Ok so it would only be a double integral and not a triple integral right? because it would just be the area and the volume would not be taken into account.
 
well i asked my teacher today and he said that would have to assume a non-conducting sphere, (was told by classmate that he said conducting ha). So he suggested the same thing as you to find the E-field due to a coin shape of charge. Which I have, and then assume the sphere is composed of infitesimal rings like that

When I work it out I end up getting a different answer than my gaussian surface. Its close but only off by 1/(4R)