Potential Inside a Hollow Conduction Sphere

  • Thread starter Cheetox
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  • #1
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Here is a question I have been pondering on for a while and got rather stuck, could anyone help out?

'A charge is diestributed uniformley with density [tex]\sigma[/tex] over the surface of a hollow conducting sphere of radius a. Show by direct integration that the potential at any point inside it is a[tex]\sigma[/tex] / [tex]\epsilon[/tex]0 and that this is the potential of the sphere itself'

plus could anyone give me a really good defintion (mathmatcially and in words) for potential, I havn't really grasped it,

many thanks
 

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  • #2
Ben Niehoff
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The electric potential is a quantity that represents potential energy per unit charge. It is measured in volts, which are the same as joules per coulomb.

So basically, if the potential difference between points A and B is V, then a charge of magnitude Q, resting at A, has a potential energy of QV to travel from A to B.

Mathematically, the potential of a point charge at [itex]\vec x_0[/itex] is given by

[tex]\phi(\vec x) = \frac{1}{4 \pi \epsilon_0} \, \frac{q}{|\vec x - \vec x_0|} + C[/tex]

where C is an arbitrary constant, usually set to zero.
 
Last edited:
  • #3
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It's a well-known theorem of hollow spheres that the electric field strength on a charge anywhere inside the sphere is 0. Integrating 0 gives a constant, and that is why the potential is constant.

Ie the potential is whatever you set it to.

Teacher probably wants you to assume potential at infinity to be 0, so that the potential at the surface works out to:

V = - 1/4πε Q/r (relative to infinity)

Q is the total charge, ie

Q = σ * surface area = σ 4πr^2

So the potential at any point inside, is the same as the potential at the surface ie

V = - σr/ε

If you don't want to use the theorem of hollow spheres (applies to gravity to), you have to integrate Ben's formula over the entire surface, ie cut the sphere up in infinitesimal charges dq.

And the integral is not trivial.
 
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