What Does V Represent in the Capacitor Equation q=C*V?

[Learnphysics]

Capacitor concept problem

Capacitors.

I have a general idea of how they work:

They don't allow charges to pass through, so the charges build up on one plate and cause a voltage across the plates.. until the Capacitor has the same voltage as the greatest voltage in the circuit, then when that voltage is turned off the capacitor now has the highest voltage in the circuit and so the charge flows out of it, around the circuit and back to it's other plate.

The equations associated with this are q=C*V, and i=C*(dv/dt).

My question is this, what does the V stand for in the first equation? Voltage across what? voltage where?

I had an idea that the dv/dt refers to the difference in voltage across the plates at anyone point in time. ANd that made a lot of sense to me. (eg, when there's a high difference across the plates current flows, when there's no difference, no current flows).

But I'm beginning to suspect thinking of the dv/dt as a simple DIFFERENCE in voltage is going to cause problems, and i need to begin considering it as a infinitesimal difference in voltage, over an infinitesimal difference in time?

Or do i even need to do this, can i continue thinking about it as just a difference in voltage.

How does dv/dt differ from the "V" in q=C*V.

Also: wikipedia says that charge cannot flow through a capacitor, it simple behaves like a membrane in a pipe. I know that while the capacitor is charging, it can be thought of as electrons are accumulating on one of the plates. But if the electrons are accumulating on one of the plates, how can current be flowing through the entire circuit? Current must only be flowing from the voltage source to one plate of the capacitor.

Is this correct?

 

[willem2]

Capacitors.

I have a general idea of how they work:

They don't allow charges to pass through, so the charges build up on one plate and cause a voltage across the plates.. until the Capacitor has the same voltage as the greatest voltage in the circuit, then when that voltage is turned off the capacitor now has the highest voltage in the circuit and so the charge flows out of it, around the circuit and back to it's other plate.

This isn't always true. Suppose you have two resistances in series with a battery (a voltage divider) and you connect the capacitor across one of the resistances.
The final voltage across the capacitor will be the same voltage that there would be across the resistance if the capacitor wasn't there.

The equations associated with this are q=C*V, and i=C*(dv/dt).

My question is this, what does the V stand for in the first equation? Voltage across what? voltage where?

voltage across the capacitor.

I had an idea that the dv/dt refers to the difference in voltage across the plates at anyone point in time. ANd that made a lot of sense to me. (eg, when there's a high difference across the plates current flows, when there's no difference, no current flows).

But I'm beginning to suspect thinking of the dv/dt as a simple DIFFERENCE in voltage is going to cause problems, and i need to begin considering it as a infinitesimal difference in voltage, over an infinitesimal difference in time?

yes.

Or do i even need to do this, can i continue thinking about it as just a difference in voltage.

How does dv/dt differ from the "V" in q=C*V.

dv/dt is the derivative with respect to time. You should have studied differentation in
calculus here.

Also: wikipedia says that charge cannot flow through a capacitor, it simple behaves like a membrane in a pipe. I know that while the capacitor is charging, it can be thought of as electrons are accumulating on one of the plates. But if the electrons are accumulating on one of the plates, how can current be flowing through the entire circuit? Current must only be flowing from the voltage source to one plate of the capacitor.

The charge on one plate of the capacitor will attract or repel electrons on the opposite plate, so it gets an opposite charge. This opposite charge is the only reason you can put so much charge on the plates.

 

[Learnphysics]

dv/dt is the derivative with respect to time. You should have studied differentation in
calculus here.

I have a general understanding of how calculus and differentiation works. But what I'm confused about is why dv/dt.

Why an infinitesimal difference in voltage on TIME? I mean it kinda sounds like it's suggesting only with a change in voltage over TIME will we see a current. (rather than a change/difference in voltage outright).

Know what i mean?

The dv/dt kinda reminds me of faraday's law, eg only when there's a CHANGING magnetic field does current get induced.

It kinda sounds like i = C*dv/dt is saying only if i "CHANGE the voltage" will i have a current. (and this isn't generally true, any difference in voltage will cause a current, not a change in voltage/time).

It's almost like saying if i had a controllable voltage source, and i took that knob and turned it really fast I'd have more current, than if the knob was already automatically stuck at a very high voltage.

(If the knob was stuck at 100V, and the capacitor was not charged at all.. according to i = C*dv/dt, there'd be no current, as there is no change in voltage)

Or does dv/dt refer to the change in the voltage AS the thing is charging up?

 

[Dale]

I mean it kinda sounds like it's suggesting only with a change in voltage over TIME will we see a current.

That is correct for a capacitor. If the voltage across a capacitor is constant wrt time then there is no current.

It kinda sounds like i = C*dv/dt is saying only if i "CHANGE the voltage" will i have a current. (and this isn't generally true, any difference in voltage will cause a current, not a change in voltage/time).

It is generally true for a capacitor. You are thinking about a resistor when you say that any difference in voltage will cause a current.

It's almost like saying if i had a controllable voltage source, and i took that knob and turned it really fast I'd have more current, than if the knob was already automatically stuck at a very high voltage.

Yes, that is exactly what it is saying if you place your voltage source across the capacitor.

(If the knob was stuck at 100V, and the capacitor was not charged at all

This is a contradiction. If the capacitor is stuck at some voltage then the capacitor is charged to Q=CV.

Actually, these two equations are related:
Q = C V
dQ/dt = d(CV)/dt
I = C dV/dt

 

[Learnphysics]

That is correct for a capacitor. If the voltage across a capacitor is constant wrt time then there is no current.


It is generally true for a capacitor. You are thinking about a resistor when you say that any difference in voltage will cause a current.


Yes, that is exactly what it is saying if you place your voltage source across the capacitor.


This is a contradiction. If the capacitor is stuck at some voltage then the capacitor is charged to Q=CV.

Actually, these two equations are related:
Q = C V
dQ/dt = d(CV)/dt
I = C dV/dt

Ah this is starting to make more sense, these formulas and relationships here don't consider charging and discharging TIME do they? They assume if i connect a 100v power source to a capacitor, that the capacitor will immediately be charged to 100v. (rather than charged up as described by V=Vi(1-e^-t/rc)?

 

[Dale]

Ah this is starting to make more sense, these formulas and relationships here don't consider charging and discharging TIME do they? They assume if i connect a 100v power source to a capacitor, that the capacitor will immediately be charged to 100v. (rather than charged up as described by V=Vi(1-e^-t/rc)?

Actually, I = C dV/dt is the equation that governs the charging and discharging time. In fact, V = Vi (1-e^-t/rc) is a solution to the equation I = C dV/dt for a step input on an RC circuit.

 

[madmike159]

I found a simulation here which might help you understand it a bit better.

http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=31