Question about spherically symmetric charged objects

  • Context: Graduate 
  • Thread starter Thread starter simpleton
  • Start date Start date
  • Tags Tags
    Charged Symmetric
Click For Summary

Discussion Overview

The discussion revolves around the treatment of spherically symmetric charged objects, specifically how to calculate the force experienced by a test charge placed at a distance from the center of such an object. The focus includes theoretical aspects and the application of Gauss' Law.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant questions how to prove that the force on a test charge at a distance from a spherically symmetric charged object can be expressed as k*(4/3*pi*R^2*rho)*q/R^2.
  • Another participant suggests using Gauss' Law and proposes considering a Gaussian surface to simplify the integral calculation.
  • A participant confirms the use of symmetry arguments to pull the electric field E outside of the integral when applying Gauss' Law.
  • There is a query regarding whether the spherically symmetric object is a dielectric or a conductor, and whether the charge density is surface or volume charge density.
  • A participant clarifies that the object in question is an insulator, as charges would accumulate on the surface if it were a conductor.
  • A participant expresses uncertainty about whether such questions are appropriate for this forum or if they should be directed to a homework forum.

Areas of Agreement / Disagreement

Participants generally agree on the applicability of Gauss' Law for points outside the sphere, but there are unresolved questions regarding the nature of the charge distribution and the appropriate forum for such inquiries.

Contextual Notes

There are limitations in the discussion regarding the assumptions about the charge distribution (surface vs. volume) and the implications of the object's material properties (dielectric vs. conductor). These aspects remain unresolved.

simpleton
Messages
56
Reaction score
0
Hi,

I would like to ask a question about spherically symmetric charged objects. My teacher told me that you can treat spherically symmetric charged objects at point charges. However, my teacher did not prove it. I guess you have to integrate every small volume on the spherically symmetric charge and find its contribution to the force, but I am not sure how to do that.

Therefore, does anyone know how to prove that, given a spherically symmetric object of radius R and charge density rho, if I place a test charge q x metres away from the centre of the charge, the force experienced by the test charge is k*(4/3*pi*R^2*rho)*q/R^2, where k is 1/(4*pi*epilson-nought).

Thanks in advance.
 
Physics news on Phys.org
Assuming you are only interested in points outside the sphere, then your teacher is correct.

To show it, just use Gauss' Law...can you think of a Gaussian surface that will allow you to pull |E| outside of the integral?
 
Oh right!

I think you mean integral(E dA) = Q/epilson-nought.

If my imaginary surface is a sphere, then I can use symmetry arguments to say that E is constant, so I can pull E out.

E*integral (dA) = Q/epilson-nought

If I take the imaginary sphere to have a radius x and has its centre at its origin, then the area of the imaginary surface is 4*pi*x^2. Thus:

E*4*pi*x^2 = Q/epilson-nought

E = Q/(4*pi*epilson-nought*x^2) = k*Q/(x^2)

So the force on a test charge q' is k*q*q'/(x^2).

Thanks a lot! :)
 
simpleton said:
Therefore, does anyone know how to prove that, given a spherically symmetric object of radius R and charge density rho, if I place a test charge q x metres away from the centre of the charge, the force experienced by the test charge is k*(4/3*pi*R^2*rho)*q/R^2, where k is 1/(4*pi*epilson-nought). .
Is your spherically symmetric object a dielectric or a conductor? Is the charge density a surface charge density or volume charge density?
 
If I understand correctly, dielectric is an insulator, and yes, the object I am talking about is an insulator, because if it is a conductor, all the charges will accumulate on the surface.

EDIT: Can I know whether I should post such questions on this forum or on the homework forum? I have many more such questions
 
Last edited:

Similar threads

Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
Replies
92
Views
6K
  • · Replies 3 ·
Replies
3
Views
988
  • · Replies 6 ·
Replies
6
Views
2K