Question about spherically symmetric charged objects

In summary, a spherically symmetric object of radius R and charge density rho can be treated at point charges.
  • #1
simpleton
58
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.
 
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  • #2
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?
 
  • #3
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! :)
 
  • #4
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?
 
  • #5
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:

Related to Question about spherically symmetric charged objects

1. What is a spherically symmetric charged object?

A spherically symmetric charged object is an object with a uniform distribution of electric charge on its surface, resulting in a symmetrical electric field in all directions. This can include objects such as spheres, shells, and concentric spheres.

2. How does the charge distribution affect the electric field of a spherically symmetric object?

The charge distribution of a spherically symmetric object determines the strength and direction of the electric field it produces. The electric field is strongest at the surface of the object and decreases in strength as you move away from the surface.

3. What is the difference between a solid and hollow spherically symmetric charged object?

A solid spherically symmetric charged object has a uniform distribution of charge throughout its entire volume, while a hollow spherically symmetric charged object has a uniform distribution of charge only on its surface. This results in different electric fields for each type of object.

4. How does the distance from the center of a spherically symmetric charged object affect the electric field?

The electric field of a spherically symmetric charged object is directly proportional to the distance from the center of the object. As you move further away from the center, the strength of the electric field decreases.

5. How does the charge of a spherically symmetric object affect the electric potential at a point in space?

The electric potential at a point in space is directly proportional to the charge of a spherically symmetric object. A larger charge will result in a higher electric potential, while a smaller charge will result in a lower electric potential.

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