B Charge of isolated parallel plate capacitors

1. Aug 30, 2016

Molar

The other day we were working on a problem which stated :
Q charge is given to the positive plate of an isolated parallel plate capacitor of 10 μF. Calculate the potential difference betweeen the plates.

Our teacher said that as the plate is isolated, Q charge will be divided between the two plates and there would be Q/2 amount of charge on each plate and hence V = Q/2C

I did not understand clearly why the charge will be divided as the capacitor is isolated.
It is clearly stated in the problem that Q charge is given to the +ve plate, so the same amount of -ve charge will be there on the -ve plate. Then V should be just Q/C

This must be some basic concept which I got all wrong. Can someone explain this ?

2. Aug 30, 2016

Staff: Mentor

I'm not sure why your teacher put it that way, but both formulations are true:

Q = CV

Q/2 = CV/2

The first is the more traditional way to think about capacitors, AFAIK.

3. Aug 30, 2016

Molar

Yes it is.
What I did not understand is why the charge will be Q/2 in the capacitor if it is isolated, when Q charge is given to the capacitor.

4. Aug 30, 2016

Staff: Mentor

That's not the traditional way of describing capacitors. You have a current i(t) that pumps electrons from one plate to the other. You integrate that current over the charging time to get the total charge Q that has been displaced. That is the Q in the Q = CV equation.

Was this a university class? What textbook are you using?

5. Aug 30, 2016

jbriggs444

Let me try a slightly different description. Perhaps it will click better.

Suppose that you charge both plates to +Q/2. By symmetry, there is obviously no potential difference between the plates. Now move +Q/2 charge from one plate to the other. You now have +Q on one plate and 0 on the other.

The formula for capacitance tells you how much potential difference results from the second step. As above, you already know that there was no potential difference resulting from the first step.

6. Aug 30, 2016

Molar

Yes, I kind of understand this explanation.
This is undergrad class. We follow Griffith's ELectrodynamics. This was a classwork problem our teacher discussed with us.

Yes, in the second step, there will be a potential difference between the plates. So, does this means that the potential difference,
V = (the charge we need to move to create the potential difference) / capacitance .ie,
V = ( Q/2 ) / C = Q/2C. ??

Also, as we know Q and 0 charges on the two plates does not form a conventional capacitor. For a parallel plate capacitor it is generally +Q and -Q , right (that's what is written in the books)?

7. Aug 30, 2016

Staff: Mentor

Yes. Pulling the electrons off of one plate onto another leaves behind an equivalent amount of positive charge (atoms without electrons).

8. Aug 30, 2016

Molar

Here's another thing I want to know. What would be the difference if the capacitor was not isolated ?

9. Aug 30, 2016

Staff: Mentor

What does that mean? Can you show schematics for "isolated" versus "non-isolated"?

10. Aug 30, 2016

Molar

Even that's what I didn't understand and wanted to know here if that's possible.

When asked, our teacher said : When you connect the two ends of a battery source with the two plates of a capacitor, it is isolated. When you connect one plate to the battery source and make the other plate grounded or 0 V , it is non-isolated.

11. Aug 30, 2016

Staff: Mentor

I don't know what that would mean. You can take a battery-powered circuit and connect a single node to Earth Ground, and it should not change how the circuit operates or is analyzed.

12. Aug 30, 2016

Molar

Because Earth-ground and the battery-negative are at the same potential, right ?

13. Aug 30, 2016

Staff: Mentor

You can Earth ground any single node in a floating circuit and not affect its operation. The traditional telephone power distribution systems in the US actually use -48V and Ground as the rails. The positive terminal of the distribution power supply is Earth grounded.

14. Aug 30, 2016

Molar

That's an interesting information.
Anyway, thank you so much for helping. :)

15. Sep 2, 2016

DarkBabylon

Have you studied electric fields and conductors under their influence by any chance? The question can be solved using fields with those parameters alone, however the physical description is accurate. I am scratching my head to understand why as an undergrad lesson in such a specific problem physicists usually tackle he makes an assumption that would result in no potential difference, instead of there being one.
In physics usually you'd want the student to describe the physical system and understand it rather than arrive at an answer even if it means to use assumptions which go against physics.

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