Question about spherically symmetric charged objects

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Spherically symmetric charged objects can be treated as point charges when considering forces outside their radius, as confirmed by Gauss' Law. To derive the force on a test charge placed at a distance from the center, one can use a Gaussian surface, leading to the expression E = k*Q/(x^2). The discussion clarifies that if the object is a dielectric, the charge density is volume-based, while for conductors, charges reside on the surface. The user also inquires about the appropriate forum for posting such questions, indicating a desire for guidance on where to seek help for similar queries. Understanding these principles is essential for analyzing electrostatic forces in spherical charge distributions.
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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.
 
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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
 
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