Proving Charge Distribution Inside a Sphere Cavity

In summary, the problem at hand is proving that inside a sphere with a cavity of arbitrary shape and a charge q kept inside, there will be a uniform distribution of charge on the outer surface of the sphere. To do so, one can use Gauss's Law by taking a Gaussian surface inside the conductor to show that the total charge enclosed is zero, meaning there must be a surface charge on the cavity surface cancelling the enclosed charge. Then, by taking a Gaussian surface outside the conductor, it can be argued that the charge distribution on the outside of the sphere is uniform since there is no field coming through the conductor to the outer surface to disturb it.
  • #1
gandharva_23
61
0
inside a sphere there is a cavity of an arbitrary shape . a chage q is kept inside the cavity . there will be a uniform distribution of charge on the outer surface of the sphere . how can i prove that ?
 
Physics news on Phys.org
  • #2
Gauss's Law
 
  • #3
If you are actually going to use Gauss's law to prove this you'd better state the problem a lot more precisely. Is the sphere conducting? Does the sphere have any net charge on it? Otherwise, it's not even true.
 
  • #4
Yes, I guess I made some assumptions about the problem. So can you tell us more?
 
  • #5
YES THE SPHERE IS CONDUCTING ... How can we prove using gauss law that the charge distribution will be uniform ?
 
  • #6
Any ideas? We are only supposed to help you, not work the problem for you. If nothing else comes to mind, state Gauss' law.
 
  • #7
well it can not be proved using gauss law ... the cavity is of any arbitrary shape ... All i can say is that the potential of the metallic sphere will be constant . after that which gaussian surface do i take ? i definitely have a non uniform charge distribuition on that irregular shaped cavity ... so how do i procceed now ?
 
  • #8
Take a gaussian surface inside the conductor. Tells you total charge enclosed is zero - so there's a surface charge on the cavity surface cancelling the enclosed charge. Now take a gaussian outside the conductor. The field must correspond to the charge in the cavity since the conductor is net neutral. So there's a surface charge on the outside equal to the charge in the cavity. You can argue that the charge distribution on the outside of the sphere is uniform since there is no field coming through the conductor to the outer surface to disturb it.
 

FAQ: Proving Charge Distribution Inside a Sphere Cavity

1. What is charge distribution inside a sphere cavity?

Charge distribution inside a sphere cavity refers to the distribution of electric charge within a hollow spherical cavity. This can be measured by determining the amount and location of electric charge present within the cavity.

2. How is charge distribution inside a sphere cavity determined?

The charge distribution inside a sphere cavity can be determined through various methods, such as using an electric field meter or by performing calculations based on the size and shape of the cavity and the amount of charge present within it.

3. Why is it important to prove charge distribution inside a sphere cavity?

Proving charge distribution inside a sphere cavity is important because it can provide valuable information about the electric field and potential within the cavity. This information can be used in various applications, such as in the design of electrical components.

4. What factors affect charge distribution inside a sphere cavity?

The charge distribution inside a sphere cavity can be affected by various factors, such as the size and shape of the cavity, the amount of charge present, and the presence of any conductive material surrounding the cavity.

5. How does proving charge distribution inside a sphere cavity relate to Gauss's law?

Gauss's law states that the electric flux through a closed surface is proportional to the enclosed electric charge. Proving charge distribution inside a sphere cavity relates to this law as the distribution of charge within the cavity can affect the electric flux through its surface.

Back
Top