Prove to me how capacitors in parallel can have the same V across plates

[mhrob24]

*If at any point I say something incorrect or its clear I don't have the right understanding of something, please point it out and correct me. I need to be sure I'm understanding it all correctly. So please don't answer unless you're willing to read this entire post*

So I know capacitors in series have same the charge (Q) stored on each of their plates because of conservation of charge (the magnitude of the charge moved through the circuit across the battery terminals must be conserved so as the charge travels, each conducting capacitor plate with have the same magnitude charge Q). I also know that its possible that the potential between each capacitor is different. My book doesn't directly explain why after stating this, but I can see how that's possible :

V = Q/C

When doing the calculation for capacitance between 2 parallel plates, we get: C = ε0*A/D. Thus, for parallel plate capacitors (or any other type of capacitor for that matter), the capacitance only depends on the geometry (area) of the conducting plates and the distance between them. So, you can effectively change the voltage of a capacitor without messing with the magnitude of charge on the plates because capacitance is independent of Q (and also of V). So that makes sense to me. So not all the capacitor plates in a series will have the same area or distance between them, but they will all have the same magnitude Q stored.

What DOESN'T make sense to me is the fact that each capacitor connected in parallel will have the same V across their plates, but at the same time, its also possible that each capacitor stores a different charge.

So the voltage in our case is basically the work done in moving a charge in an electric field divided by the charge's magnitude. So, the voltage depends on the electric field, not the charge moving in it. So when we look at a capacitor connected in parallel, we know that the charge Q stored on its set of plates could possibly be different from the charge stored on another capacitor within the parallel connection. With that said, the electric field DOES depend on the charge stored on the capacitor (that's what causes the uniform field between the plates in the first place). So, this means that the electric field between each capacitor in the parallel connection can't be the same if we know that its possible for the charge on one capacitor can be different from another capacitor. Thus, if the electric field on one capacitor can be different from another, and the voltage depends on the electric field, how can each capacitor have the same V?

 

[mhrob24]

The only way I can kind of see how it makes sense is if you use V = Q/C and plug in Q*d/ε0*a for V. Then you get:

Q*D/ε0*a = Q/C. Then, the charge Q cancels so V won't depend on the electric field or the charge Q on the capacitor plates...but this just goes against everything I learned previously about potential difference between a uniform electric field (V = E*D, where E depends on magnitude of charge Q stored on each plate)...

 

[phinds]

Thus, if the electric field on one capacitor can be different from another, and the voltage depends on the electric field, how can each capacitor have the same V?

You are looking at that backwards. The field depends on the voltage.

Looking at it another way, how could you possibly have parallel caps (or anything else) with different voltages? That would imply that there was a zero-resistance wire with different voltages at different point. Not possible.

 

[Dale]

Summary: n/a

So, this means that the electric field between each capacitor in the parallel connection can't be the same if we know that its possible for the charge on one capacitor can be different from another capacitor.

This part is where you are going wrong (I think).

For each capacitor you have a relationship between V and Q. In a series circuit all of the capacities have the same Q, so we solve for each V. In a parallel circuit all of the capacitors have the same V, so we solve for each Q.

Either way there is a unique solution.

 

[mhrob24]

Ok, let me see if I've got this correct.

So, E ∝ Q. However, Q doesn't reach the plates without the V from the battery moving it. So I can see how the poster before you is correct by saying that the E field depends on the V, not the other way around. So, basically, this is saying E ∝ V. If you rearrange C = Q/V to solve for Q, you get:

Q = C * V.

Thus, this proves that each capacitor in a parallel circuit CAN store a different charge on its plates while still having the same V as the other capacitors in the circuit because you can simply change the capacitance (which only depends on geometry of the plates and distance between them) without changing the V (Since C doesn't depend on Q or V)...right?

Also, the poster before you stated that its impossible to have a different voltages for each capacitor in a circuit. However, my understanding is that it is possible in a series circuit, because V = Q/C, so by changing the capacitance on one capacitor in a series circuit, you are changing the potential difference between the plates of that capacitor. Thus, you add up the potential drops between each capacitor to get your total supply voltage (Unless he meant that its impossible to change the SUPPLY voltage of the circuit, which is correct. You can't do that.)

 

[phinds]

Also, the poster before you stated that its impossible to have a different voltages for each capacitor in a circuit. However, my understanding is that it is possible in a series circuit,

Yes, of course it's possible in a series circuit. I specifically, and clearly, spoke about a parallel circuit. They are not even remotely the same thing.

 

[hutchphd]

In parallel all components must have the same potential difference across their terminals because each terminal is interconnected by "perfect" conductors. That is what @phinds said. Please don't misrepresent a comment and then disagree!

 

[anorlunda]

Temporarily closed for moderation.

Edit: No, I changed my mind. Closed permanently. The OP question has been answered. I'm going to clean up the off topic stuff about interpretation of words.