# General Theory of Capacitance

Hello,

I came across this topic while reading Principles of Electrodynamics which can be read from this link starting at page 54.

Looking at page 56, my question is how could there be induced charge on a grounded conductor ? what happens then to the equation Q = C * V, where Q is charge on the said conductor, C is capacitance of that conductor, and V is potential on that conductor.

I would greatly appreciate it if someone provided a rigorous proof of his/her answer.

Thanks

C = Q * V is only valid for a single conductor.

For two conductors You have

$$Q_1 = C_{11} V_1 + C_{21} V_2$$

and

$$Q_2 = C_{12} V_1 + C_{22} V_2$$

If the second conductor is grounded you get:

$$Q_1 = C_{11} V_1$$

$$Q_2 = C_{12} V_1$$

and nowhere is it said that $C_{12}$ must be positive.

How does the existence of charge on the surface of a grounded conductor don't contradict Gauss's Law ?

How does the existence of charge on the surface of a grounded conductor don't contradict Gauss's Law ?

Gauss law relates the charge enclosed by a surface to the electric field perpendicular to the surface integrated across the entire surface. It does not say anything about the potential. To find the potential, you have to integrate the electric field due to BOTH charges from the surface of the grounded charge to infinity.

I understand what Gauss's law is, my question is when there is charge on a grounded conductor, the charge will emanate electric field which will give a rise to a non zero potential when a line integral is taken from infinity to the surface of that charge. This is the contradiction to Gauss's law that confuses me.

I understand what Gauss's law is, my question is when there is charge on a grounded conductor, the charge will emanate electric field which will give a rise to a non zero potential when a line integral is taken from infinity to the surface of that charge. This is the contradiction to Gauss's law that confuses me.

But if you take that line integral, you have to integrate the field of BOTH charges. one of them positive, the other negative. There actually HAS to be an induced charge to make that integral come out to 0.

I don't think Gauss's law is any help computing that integral, but if you do, please state how.

Thanks willem2 for answering my question. I apologize for creating the confusion by referring to induced charge. I'll try to break the issue into readable points.

1- The author defines capacitance as the proportionality constant relating the potential at surface of the conductor relative to infinity to the free charge deposited on that conductor.

2- The author now wants to extend this concept to relate the free charge on the ith conductor, to the potentials on several N conductors.

3- To do that, the author says that all n≠1 conductors are grounded = having 0 potential. Except n=1 conductor.

4- Now, the author starts to relate the free charge on conductors n= 2,3,.., N to the potential on conductor n = 1.

My question is

How could there be free charge on a conductor with potential equal to zero?

My understanding is that a free charge will emit a field (assuming positive charge) that will give rise to a non-zero potential when an line integral of the electric field is carried out from infinity to the location of that charge whether its free space or a surface of a conductor.

My question is

How could there be free charge on a conductor with potential equal to zero?

My understanding is that a free charge will emit a field (assuming positive charge) that will give rise to a non-zero potential when an line integral of the electric field is carried out from infinity to the location of that charge whether its free space or a surface of a conductor.

This is only valid for single isolated, spherically symmetric charges. You have to consider the fields of all the charges. These can give both positive and negative contributions to the line integral.

If there were no induced charges, there would only be the field from the first charge, and this will definitely make the line integral from the second conductor to infinity nonzero, so there have to be induced charges.