The effects of bound and free charge on a metal surface

[TheCanadian]

I was just looking through a few different solutions in Griffiths EM and I must have not realized it, but do bound and free charges both contribute to the overall electric field?

For example: when dealing with a capacitor with a dielectric between it, one of the solutions wants to find the capacitance through ##C = \frac{Q}{V}##. But looking through the solution, it states ## Q = \sigma_f A ##. Is there any particular reason the solution would neglect the bound charge as well in such a case? Don't both types of charges affect the total electric field and thus potential in the capacitor?

 

[Ronie Bayron]

I was just looking through a few different solutions in Griffiths EM and I must have not realized it, but do bound and free charges both contribute to the overall electric field?

For example: when dealing with a capacitor with a dielectric between it, one of the solutions wants to find the capacitance through ##C = \frac{Q}{V}##. But looking through the solution, it states ## Q = \sigma_f A ##. Is there any particular reason the solution would neglect the bound charge as well in such a case? Don't both types of charges affect the total electric field and thus potential in the capacitor?

 

[Ronie Bayron]

It is important that you understood ionization and de-ionization concept.
Material, let's say Lithium 7 for example (no. protons)=(no. electrons) , then net charge is zero. Therefore, no q when material is not charged.

However the introduction of excess electron (negative ionization) say (-2) make the material excess negative -2charges/atom-the net charge/atom.
The removal of electrons say 2 electrons in the orbitals makes the material (+2) charges/atom.

Those are the conditions where charge is made available or stored in the dielectric. The net charge of a material.
The ability of material to be ionized or de-ionized dictates permitivity. Dielectric has that capacity.

 

[TheCanadian]

It is important that you understood ionization and de-ionization concept.
Material, let's say Lithium 7 for example (no. protons)=(no. electrons) , then net charge is zero. Therefore, no q when material is not charged.

However the introduction of excess electron (negative ionization) say (-2) make the material excess negative -2charges/atom-the net charge/atom.
The removal of electrons say 2 electrons in the orbitals makes the material (+2) charges/atom.

Those are the conditions where charge is made available or stored in the dielectric. The net charge of a material.
The ability of material to be ionized or de-ionized dictates permitivity. Dielectric has that capacity.

Okay. To pose my particular problem, I have attached the solutions to 2 problems.

In the first image's problem, we are being asked to find the capacitance between a capacitor with one half filled with dielectric, and the other half filled with air. In this case, the separate the two sides and only use free charge when solving for the capacitance in the side that contains the dielectric. Why is that so? Why don't they consider the total charge on the plates in this case since isn't C = Q/V where Q is the total charge?

In the second image's problem, in particular for part c, there is simply 2 parallel plates with a uniform polarization between them, and we are trying to find the capacitance. The solution says the total surface charge density is equal to P (the polarization) which is also equal to the total bound charge density in this instance. But why in this case do we consider the total charge density (i.e. including the bound charge) while in the former example, when finding the capacitance of the parallel plates with a predefined charge density on it, we only consider the free charge density?

Please let me know if you need any further clarification with my problem.

 

[Ronie Bayron]

Okay. To pose my particular problem, I have attached the solutions to 2 problems.

In the first image's problem, we are being asked to find the capacitance between a capacitor with one half filled with dielectric, and the other half filled with air. In this case, the separate the two sides and only use free charge when solving for the capacitance in the side that contains the dielectric. Why is that so? Why don't they consider the total charge on the plates in this case since isn't C = Q/V where Q is the total charge?

O, wow. I would not dare anymore look to the solution equations you posted but I'll try to explain conceptual ideas with you. This problem is just actually having 2 capacitors in series connection (one with air and one with dielectric). The concept of capacitor is like pressure tank set up below. Water molecules are the electrons pushed by the voltage potential. They enter and exit at one terminal. This is exactly what happens in the capacitor. All charges are dumped on the negative terminals and when connected to a lower potential, electrons discharge to the path of least resistance.

In the second image's problem, in particular for part c, there is simply 2 parallel plates with a uniform polarization between them, and we are trying to find the capacitance. The solution says the total surface charge density is equal to P (the polarization) which is also equal to the total bound charge density in this instance. But why in this case do we consider the total charge density (i.e. including the bound charge) while in the former example, when finding the capacitance of the parallel plates with a predefined charge density on it, we only consider the free charge density?

Please let me know if you need any further clarification with my problem.

I am sorry I did not get the figure very well. May be you could post a picture or something.