Solving Capacitance Problem: Identical Answers Explained

[erisedk]

Homework Statement

Why are both these problems identical?
In both cases, c>b>a.
Q1 Three concentric conducting shells A, B and C of radii a, b and c are arranged as shown. A dielectric of dielectric constant K is filled between A and B. Find the capacitance between A and C.

Q2 A spherical capacitor is made of two conducting spherical shells of radii a and c. The space between the shells is filled with a dielectric of dielectric constant K unto a radius b as shown. Find the capacitance of the system.

Homework Equations

The Attempt at a Solution

Both have answers (4πε_o.Kabc)/(Ka(c-b)+c(b-a))
I know how to do the first one, but I don't understand why both have the same answer, ie. why the capacitance of the system is equal to the capacitance between A and C.

 

[vela]

Assume a charge of Q on the inner conductor. Does the presence of the conductor B change the electric field in the two regions between the two cases?

 

[erisedk]

Assume a charge of Q on the inner conductor. Does the presence of the conductor B change the electric field in the two regions between the two cases?

I know it doesn't mathematically, but I'm not getting a feel for it.

 

[ehild]

The potential is constant on a metal surface. Because of spherical symmetry, the equipotential surfaces are concentric spheres between A and C.
Replacing an equipotential surface with a very thin metal foil does not change anything, the electric field stays the same. That thin metal shell can be placed at radius b, and nothing is changed, but the set-up becomes identical with the first one.

 

[erisedk]

The potential is constant on a metal surface. Because of spherical symmetry, the equipotential surfaces are concentric spheres between A and C.
Replacing an equipotential surface with a very thin metal foil does not change anything, the electric field stays the same. That thin metal shell can be placed at radius b, and nothing is changed, but the set-up becomes identical with the first one.

But won't the metal surface induce charges on it's inner and outer surfaces, and thus alter the electric field?

 

[ehild]

The electric field induce charge on both surfaces of the metal, but the net charge of the shell remains zero.
Well, calculate the electric field below and above the middle metal shell, applying Gauss' Law.

 

[erisedk]

Got it! Thanks :)