Question involving potentials and spherical conductors/capacitors/things

In summary, the potential inside a spherical conductor of radius a located inside the cavity of a larger spherical conductor of radius b, and that the larger sphere's outer radius is c, and charged at a to a charge of Q is the sum of potentials between a and b, kQ/r, minus the potential at the surface b plus the potential at c. This is due to the magic of 1/r^2 and Gauss's Law, which shows that the electric field inside a spherically symmetric charge distribution is zero, making the potential constant in that region. The potential is additive and it is easier to work from the outside in when adding up contributions.
  • #1
schattenjaeger
178
0
Say you have a spherical conductor of radius a located inside the cavity of a larger spherical conductor of radius b, and that the larger sphere's outer radius is c

If you charge the inner conductor at a to a charge Q, the inside part of the outer conductor becomes charged -Q and the outer outer part with Q again, ok, I got that

then it wanted the potential anywhere. I understand the potential outside the whole thing, the potential between b and c, but when r is from a to b I had trouble. I had the answer beforehand, which is kQ/(rbc)*[bc-r(b-c)] or I may've mixed that up

I see that if you unsimplify that it's like a sum of potentials. It's the potential between a and b, kQ/r, minus the potential AT the surface b(well, the Q is negative there so I guess you can just say plus the potential at b)plus the potential at c

Simple question: Why is it like that? Is there a more systematic way of reaching that answer? oh and of course the answer from a to b is like that one with r=a
 
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  • #2
schattenjaeger said:
Say you have a spherical conductor of radius a located inside the cavity of a larger spherical conductor of radius b, and that the larger sphere's outer radius is c

If you charge the inner conductor at a to a charge Q, the inside part of the outer conductor becomes charged -Q and the outer outer part with Q again, ok, I got that

then it wanted the potential anywhere. I understand the potential outside the whole thing, the potential between b and c, but when r is from a to b I had trouble. I had the answer beforehand, which is kQ/(rbc)*[bc-r(b-c)] or I may've mixed that up

I see that if you unsimplify that it's like a sum of potentials. It's the potential between a and b, kQ/r, minus the potential AT the surface b(well, the Q is negative there so I guess you can just say plus the potential at b)plus the potential at c

Simple question: Why is it like that? Is there a more systematic way of reaching that answer? oh and of course the answer from a to b is like that one with r=a
It is the magic of 1/r^2 and Gauss and all that. Inside any spherically symmetric charge distribution the electric field is zero. That can be shown by direct integration, or by Gauss's Law. That means it takes zero work to move a test charge from one place to another inside a cavity surrounded by such a charge distributiuon, which means the potential due to that charge distribution is constant in that region.

The potential is additive (superposition). It's usually easier to work from the outside into add things up because the outer charges contribute only a constant potential to the inner regions.
 

Related to Question involving potentials and spherical conductors/capacitors/things

1. What is the difference between potential and voltage?

Potential is a measure of the electric potential energy per unit charge, while voltage is the potential difference between two points in an electric field. In other words, potential is a scalar quantity while voltage is a measure of potential difference between two points.

2. How does a spherical conductor affect electric potential?

A spherical conductor has a uniform electric field on its surface and zero electric field inside. This means that the electric potential is constant on the surface and decreases as you move away from the conductor. This is known as the "Faraday cage" effect, where the conductor shields the inside from external electric fields.

3. How does the size of a conductor affect its capacitance?

The capacitance of a conductor is directly proportional to its size. This means that the larger the conductor, the higher its capacitance. This is because a larger conductor can hold more charge, resulting in a higher potential difference between its two ends.

4. Can a spherical capacitor have a non-uniform electric field?

Yes, a spherical capacitor can have a non-uniform electric field if the two conductors have different radii or if the spacing between them is not uniform. In this case, the electric potential will also vary between the two conductors.

5. How do you calculate the capacitance of a spherical capacitor?

The capacitance of a spherical capacitor can be calculated using the formula C = 4πε0ab/(b-a), where a is the radius of the inner conductor, b is the radius of the outer conductor, and ε0 is the permittivity of free space. This formula assumes that the conductors are concentric and have a uniform electric field between them.

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