Capacitors & Circuits in General

[MedPR]

---

Members don't see this ad.

A few questions if you guys don't mind.

1. How does the distance between the plates of a capacitor affect the capacitance? Based on the equation C=Q/V, it seems like the distance separating the plates is irrelevant, but that doesn't seem logical to me.

2. When doing circuit problems, I know that conventionally current (protons) flow from the big line to the small line, but I also know that protons don't really flow and that in reality the electrons flow the opposite direction (small line to big line). I've only encountered a few examples where this actually matters, but I'm still not very sure on how to approach the problems. For example, when the switch is closed in the following circuit, conventional current goes through the capacitor before it goes through the resistor, but electrons go through the resistor before going through the capacitor. My questions are

2a. Does it matter what order capacitors and resistors are in if they are in series to each other anyway?

2b. If asked "If the switch is not closed, will any current flow through the resistor or capacitor?" would I answer yes, or would I answer no.. Electrons can flow, but conventional current cannot.

http://media.photobucket.com/image/RC circuit/snoweangel27/figure-25-70.gif

 

[milski]

1. For a plate capacitor, it's inversely proportional to d - C=ε0 * A/d. The capacitance in C=Q/V is the capacitance of a specific capacitor - it will not tell you anything about what capacitance depends on.

2a. No.

2b. It's an open circuit - there is no current in any sense. No sort of charge will flow through that circuit until you close the switch.

The only time where convention vs real charges would matter is if you're trying to do something to the charges with an electric or magnetic field. In that case the velocity and mass of the moving charges will matter.

 

[MedPR]

1. For a plate capacitor, it's inversely proportional to d - C=ε0 * A/d. The capacitance in C=Q/V is the capacitance of a specific capacitor - it will not tell you anything about what capacitance depends on.

2a. No.

2b. It's an open circuit - there is no current in any sense. No sort of charge will flow through that circuit until you close the switch.

The only time where convention vs real charges would matter is if you're trying to do something to the charges with an electric or magnetic field. In that case the velocity and mass of the moving charges will matter.

What about in a situation where they give you a circuit with only a capacitor then they ask "In which scenario will the capacitor reach maximal charge the fastest" If the resistor is in front of the capacitor wouldn't it take longer for the capacitor to charge?

 

[mcloaf]

What about in a situation where they give you a circuit with only a capacitor then they ask "In which scenario will the capacitor reach maximal charge the fastest" If the resistor is in front of the capacitor wouldn't it take longer for the capacitor to charge?

Yes.

 

[MedPR]

Yes.

Ok so which side would "in front" be? In between the big line and the capacitor, since that's the way conventional current flows? Or in between the small line and the capacitor, since that's the way that electrons actually flow?

 

[milski]

What?! The order should not matter. It will be slower if there is a resistor vs there is no resistor. Is that what you meant, mcloaf?

 

[Class1P]

intuition takes a back seat here a little.

What he is confused about is that he is thinking electrons flowing through a resistor then hitting the capacitor plate make it take longer to charge than when electrons never go through a resistor ( a case where there is a resistor on the + side of the capacitor)

You have to remember that the (+) side of the battery is sucking in electrons and that will happen slower with a resistor on that side. So the (+) side of the capacitor is gaining a (+) charge at a slower rate and the electrons will not be pulled to the (-) side of the capacitor as fast. At least that is how I think of it. So as long as your resistor is the same, the position of it should not matter.

Edit: It might be helpful to think of a capacitor as just an electro-magnetic field. The positive side is becoming more positive and an induced EM field between the plates. The negative plate ""feels" that EM field and electrons want to be on the positive side, but they can't be so they just collect at the negative plate. The electrons never actually flow between the capacitor's plates unless your dielectric breaks down at some really high voltage.

 

[mcloaf]

What?! The order should not matter. It will be slower if there is a resistor vs there is no resistor. Is that what you meant, mcloaf?

Yeah, sorry for being vague--I should've addressed the order of circuit elements.

😳

 

[MedPR]

Ok so the order of circuit elements makes no difference. The only thing to consider is how to add them when they are in series vs in parallel.

But if you add more resistors in series, the capacitor will charge less quickly. If you add resistors in parallel, you reduce the Req though, so the capacitor will charge more quickly?

 

[milski]

Yes, you got it. Order does not matter. Higher Req leads to longer time to charge. Final charge is a constant for a given voltage.

 

[MedPR]

Yes, you got it. Order does not matter. Higher Req leads to longer time to charge. Final charge is a constant for a given voltage.

Ok and since charge is quantized, does that mean the Q in C=Q/V increases/decreases in specific increments?

 

[milski]

Ok and since charge is quantized, does that mean the Q in C=Q/V increases/decreases in specific increments?

In theory - yes. In practice the discrete quantums are so tiny that you can ignore them. I have not seen a case where it would matter and I've seen calculus applied on C=Q/V as if C,Q,V were smooth functions and not discrete ones.

 

[chiddler]

so the capacitor will charge more quickly?

Do you mean more quickly relative to no resistors in circuit? Or more quickly relative to a resistor in series with a capacitor?

 

[MedPR]

In theory - yes. In practice the discrete quantums are so tiny that you can ignore them. I have not seen a case where it would matter and I've seen calculus applied on C=Q/V as if C,Q,V were smooth functions and not discrete ones.

Ok thank you!

Do you mean more quickly relative to no resistors in circuit? Or more quickly relative to a resistor in series with a capacitor?

More quickly relative to having a resistor in series or less resistors in parallel. I was just making sure that if you decrease Req, you increase how quickly the capacitor will charge/discharge.