Electrical Field of a Parallel Plate Capacitor

In summary, the conversation covers a review of Gauss' Law and the calculation of electric fields, capacitance, and voltage for different capacitor configurations. The main problem discussed is calculating the voltage and time for a spherical shell capacitor with a steady current and a resistor connected to the ground. The solution involves using the formula dq/dt=i and substituting expressions to obtain the final answer in terms of R, I, and C.
  • #1
meister
56
0
Just going through some review questions and wanted to clear up some questions I had. I figure you guys like showing off how smart you are.

First Question:
Consider the parallel plate capacitor classroom demonstration.
A. Write a mathematical expression for Gauss' Law. Describe the meaning of Gauss' Law in words.
I've already got this part.


B. Using Gauss' Law: draw a picture of one plate, draw a suitable Gaussian surface, and calculate the electric field due to the charge density, [sig]C/m^2, on that plate.
Pretty sure I've got this part but I just wanted to check. It specifies one plate, and I wound up with E = [sig]/(2e), where e is the permittivity of free space.


C.What is the total electric field: external to the plates and inside the plates?
Well, there is no electric field outside of the plates, correct? And inside the plates I would assume would be 2 * [sig]/(2e), but that's sort of an educated guess...


D. Estimate the capacitance for the plates, based on your observations of the plate geometry and a calculation of the theoretical capacitance.
This part throws me because on the practice sheet there is so much room to write. The answer I got was C=e*A/d, with d being the distance between the plates and A being the surface area of the plates.


The second question is easier but the last section is giving me fits. I won't even bother to type the full question. The problem involves a single spherical shell capacitor, of radius a, connected with a wire to a resistor, R ohms, which is connected to the ground. The shell has a steady current i amps that is delivered to the sphere not using the wire. The section giving me problems is:

If the capacitor has no charge on it to begin with, to what voltage Vf will it charge and roughly how long, T, will it take to reach that voltage. Answers should be in terms of i, C, and R.
I got [del]V=iR, but I don't think that's what he's looking for...and as for the time, I assume it has to do with i=[del]Q/[del]T, but I can't seem to put it all together.



Any help at all on these problems is greatly appreciated. Thanks.
 
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  • #2
The main problem I'm having with that last problem is that it says the charge doesn't flow through the resistor. Would this matter? I think I managed to solve for [del]t anyway. [del]t=CR.
 
  • #3
Originally posted by meister
B. Using Gauss' Law: draw a picture of one plate, draw a suitable Gaussian surface, and calculate the electric field due to the charge density, [sig]C/m^2, on that plate.
Pretty sure I've got this part but I just wanted to check. It specifies one plate, and I wound up with E = [sig]/(2e), where e is the permittivity of free space.

Yes.

C.What is the total electric field: external to the plates and inside the plates?
Well, there is no electric field outside of the plates, correct? And inside the plates I would assume would be 2 * [sig]/(2e), but that's sort of an educated guess...

That's right too. There should be a derivation in your book, so that will take the guesswork out of it.

D. Estimate the capacitance for the plates, based on your observations of the plate geometry and a calculation of the theoretical capacitance.
This part throws me because on the practice sheet there is so much room to write. The answer I got was C=e*A/d, with d being the distance between the plates and A being the surface area of the plates.

That is the correct expression for a capacitor when fringe effects can be neglected. Also, when they say "estimate the capcitance", I think they are looking for a number.

The second question is easier but the last section is giving me fits. I won't even bother to type the full question. The problem involves a single spherical shell capacitor, of radius a, connected with a wire to a resistor, R ohms, which is connected to the ground. The shell has a steady current i amps that is delivered to the sphere not using the wire. The section giving me problems is:

If the capacitor has no charge on it to begin with, to what voltage Vf will it charge and roughly how long, T, will it take to reach that voltage. Answers should be in terms of i, C, and R.
I got [del]V=iR, but I don't think that's what he's looking for...and as for the time, I assume it has to do with i=[del]Q/[del]T, but I can't seem to put it all together.

OK

The main problem I'm having with that last problem is that it says the charge doesn't flow through the resistor. Would this matter? I think I managed to solve for [del]t anyway. [del]t=CR.

How did you get that? To me, when it says the charge does not flow through the resistor, it means that all the charge is stored in the capacitor. That means that dq/dt=i, where q=the charge on the capacitor.
 
  • #4


Originally posted by Tom
How did you get that? To me, when it says the charge does not flow through the resistor, it means that all the charge is stored in the capacitor. That means that dq/dt=i, where q=the charge on the capacitor.
I substituted different expressions into one another to get that result. I forgot to post that the question wanted the answer in terms of R, I, and C.

My work is as follows:
[del]V=iR; this is just a definition which is why I think that it's wrong.

For the second part:
i=[del]Q/[del]t ergo [del]t=[del]Q/i=C[del]V/i=CiR/i=CR
 

1. What is an electrical field?

An electrical field is a region in space around an electrically charged object where other charged particles will experience a force.

2. How is the electrical field of a parallel plate capacitor calculated?

The electrical field of a parallel plate capacitor is calculated by dividing the voltage between the plates by the distance between the plates. This can be expressed as E = V/d, where E is the electrical field, V is the voltage, and d is the distance between the plates.

3. What factors affect the strength of the electrical field in a parallel plate capacitor?

The strength of the electrical field in a parallel plate capacitor is affected by the distance between the plates, the voltage applied, and the type of material used for the plates.

4. How does the electrical field change if the distance between the plates is increased?

If the distance between the plates of a parallel plate capacitor is increased, the electrical field will decrease. This is because the voltage remains constant, but the distance increases in the denominator of the equation for electrical field (E = V/d).

5. What is the relationship between the electrical field and the capacitance of a parallel plate capacitor?

The electrical field is directly proportional to the capacitance of a parallel plate capacitor. This means that as the electrical field increases, the capacitance also increases. This relationship can be expressed as E = Q/C, where E is the electrical field, Q is the charge on the plates, and C is the capacitance.

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