capacitance related to applied charge?

[sciencewhiz]

Is capacitance related to the applied charge of the plates?

In a standard physics or E&M class, the equation for capacitance is derived by applying a known charge to the plates, determining the electric field, and then integrating the field to get the voltage. This is relatively easy with a paralel plate or spherical capacitor.

However, in my case, I have a spherical capacitor with a static charge appied to the inner sphere, and time varying charge applied to the outer sphere (and a dielectric material between).

If the capacitance is related to charge (and voltage, by definition), I would need to calculate the electric field given with those two charges, in 3 dimensions. If it's not, I just have to apply the standard equations derived in physics. It seems to me that capacitance should only be related to the geometry and dielectric constants of the material.


Note: This is for a homework problem, in a 5th year EE course in Mechatronics. The professor assumes that you've had advanced Electric and Magnetic fields courses, which aren't required, and I haven't had. When I talked to him, he said to go ahead and treat it like equal and opposite static charges were applied to the capacitor. However, due to his poor english skills and the fact that he doesn't want to help any student, I wasn't able to find out if that is the correct way, or if he was just trying to get me off his back.

 

[TuxDave]

Are the charges equal and opposite charges on the inner and outer sphere?

Anyways, you might want to apply Gauss' law for calculating the electric field since you have a symmetrical system. Gauss' Law says that if your draw an enclosed surface containing a charge, the total electric flux going through the surface is equal to some constant. So if you enclose the inner sphere with another cocentric circle, you can apply symmetry to find the electric field at any point.

 

[sciencewhiz]

The charges are not equal and opposite. On the inside it is 1 microcoulumb (uC) and on the outside it is 10(sin(t)-1) (qi and qo(t), respectively)

When you apply Gauss's law to the middle area, you only enclose the charge of inner sphere, and the fact that the outside is not equal and opposite doesn't matter.

If the outside varied with space, it might matter, but since it only varies with time, it shouldn't factor in, right?

 

[TuxDave]

Originally posted by: sciencewhiz
The charges are not equal and opposite. On the inside it is 1 microcoulumb (uC) and on the outside it is 10(sin(t)-1) (qi and qo(t), respectively)

When you apply Gauss's law to the middle area, you only enclose the charge of inner sphere, and the fact that the outside is not equal and opposite doesn't matter.

If the outside varied with space, it might matter, but since it only varies with time, it shouldn't factor in, right?

Yeah... it shouldn't matter. Anyways, I forgot to include my final answer. I don't think the capacitance will be charge variant.