How does capacitance affect voltage in a capacitor?

[MitsuShai]

Ok, I'm just trying to get this straight in my head. We are going over capacitance in voltage in lab today.

1. The electric field will decrease if charges are taken off the capacitor plates because the electric field depends on the movement of charge in the capacitor. Thus the voltage in the plates will also decrease because voltage is proportional to the electric field. Right?

2. If the capacitance of the circuit fixed then how can the time constant be changed?
Well if you can't change the time constant if the capacitance is fixed.


Formula: V(t)=V(0)e^(-(delta)t/time constant)

3. At t=0 V(t)/V(0)= 1. While decaying, V(t)/V(0) falls and after a while it will be 1/20. If the time constant increased, will the time for V(t)/V(0)=1/20 change and why?
Yes because the formula V(t)=V(0)e^(-(delta)t/time constant) will become delta t= time constant ln(20) and as shown, the time constant and time is proportional. Thus the time will increase when the time constant increases.


Diagram: http://i324.photobucket.com/albums/k327/ProtoGirlEXE/100_0696.jpg

4. What is the source of emf of the circuit when switch is open?
When the switch is open the capacitor will be discharging and the battery cannot provide a continuous current when the switch is open, so the capacitor provides the source of emf.

5. Express the current through the voltmeter in terms of total current I and the resistances R and Rmeter.
I thought current doesn't flow in a voltmeter but the picture says otherwise and also the next question, so I'm not sure about this, I don't think I should use the junction rule because the current I doesn't have a resistance. The voltage of the meter is the same as the voltage across PQ because its in parallel, so I got this so far:
I meter= (I resistor * R)/R meter I don't know how to find it in terms of just I and instead of I resistor

6. What is current through meter when R meter=R.
I meter=I, since the resistances are the same, they cancel.

7. Did the current get smaller?
No, it got bigger, since the current is free from resistances.


Diagram: http://i324.photobucket.com/albums/k327/ProtoGirlEXE/100_0698.jpg

8. Find current though meter, using resistance of meter, R small, and current I.
Again like the 5th question, I'm not sure but I got this far: I meter= (I small* R small)/R meter, don't know how to get it in I instead of I small

9. Is the current smaller than in circuit 2?
yes I figured this out mathematically, if a small R is on the numerator and a bigger R in the denominator then I would be smaller than before.


I have no idea what this paragraph is sayaing:
Diagram: http://i324.photobucket.com/albums/k327/ProtoGirlEXE/100_0701.jpg
This is a pretty amateur explanation but there's still an e there so it should be exponential.
If the that equation that is presented in the paragraph is constant then the time constant should not change, because...I don't know how to explain this using physics

 

[cepheid]

Ok, I'm just trying to get this straight in my head. We are going over capacitance in voltage in lab today.

1. The electric field will decrease if charges are taken off the capacitor plates because the electric field depends on the movement of charge in the capacitor.

Not the movement of charge in the capacitor. I would say that the electric field depends on the amount of charge in the capacitor.

Thus the voltage in the plates will also decrease because voltage is proportional to the electric field. Right?

In this situation, yeah.

2. If the capacitance of the circuit fixed then how can the time constant be changed?
Well if you can't change the time constant if the capacitance is fixed.

I am sorry, but this is incorrect. What two quantities does the time constant depend on?

Formula: V(t)=V(0)e^(-(delta)t/time constant)

3. At t=0 V(t)/V(0)= 1. While decaying, V(t)/V(0) falls and after a while it will be 1/20. If the time constant increased, will the time for V(t)/V(0)=1/20 change and why?
Yes because the formula V(t)=V(0)e^(-(delta)t/time constant) will become delta t= time constant ln(20) and as shown, the time constant and time is proportional. Thus the time will increase when the time constant increases.

Right. The longer the time constant is, the more time it takes for the voltage to decay by any given factor. (The time constant is just the time required to decay by a factor of e).

 

[MitsuShai]

Not the movement of charge in the capacitor. I would say that the electric field depends on the amount of charge in the capacitor.

Really? The reason I thought it was the movement of charge was because an electric field is generated by the movement of charge or maybe I'm getting this confused with the magnetic field which is generated by the movement of particles.

I am sorry, but this is incorrect. What two quantities does the time constant depend on?

It depends on the resistance and capacitance, so the time constant can change if the resistance changes. But I was thinking, in the equation C=Q/V, if C is fixed shouldn't the Q & V be fixed too and R=V/I (current in a circuit can't change either), so R should be fixed as well.

 

[cepheid]

Really? The reason I thought it was the movement of charge was because an electric field is generated by the movement of charge or maybe I'm getting this confused with the magnetic field which is generated by the movement of particles.

It's magnetic fields that are generated by moving charges. There is an electric field associated with any electric charge, even in a static situation. The electric field of a charge at a point in space is just the electric force per unit charge that it exerts at that point (in other words, it is the force that would be experienced by a positive test charge of 1 coulomb if you were to place it at that point in space near the charge in question).

It depends on the resistance and capacitance, so the time constant can change if the resistance changes. But I was thinking, in the equation C=Q/V, if C is fixed shouldn't the Q & V be fixed too

No, the Q and V should not be fixed. Let me explain.

A physical quantity like resistance or capacitance is an inherent property of the device/component in question. Any chunk of material is going to have some electrical resistance. That resistance depends only on the material that the chunk is made out of, and its geometry. A resistor is just a component whose material and geometry have been designed in such a way that the component has a very precisely-determined resistance. If it has a certain resistance value "R", then R is always its resistance*, regardless of what circuit you put it into and what voltages and currents you expose it to. What Ohm's law (V = IR) means is that if you put a voltage V across the resistor (when it is in a complete circuit) then the current that flows through it must be equal to V/R, where R is the resistance of that resistor. You can put any voltage across it you like, and the resulting current through it will be in accordance with Ohm's law. But changing the voltage doesn't change the resistance, since that is determined by the properties of the component itself.

The same thing is true for a capacitor -- its capacitance depends only on its geometry and the material it's made out of. What C = Q/V means is that since C is a constant for a given capacitor, increasing the voltage V across the capacitor means increasing the amount of charge Q that is stored on it. In other words, if V goes up, then Q must also go up, in order to keep their ratio the same. Another consequence of this is that the larger a capacitor is, the more charge you can store on it at the same voltage.

So, in this example, even if C is fixed (because you are using a specific capacitor), you can still change R by swapping out one resistor for a different one, thereby changing the time constant.

*Note that resistance changes with temperature, since the conducting properties of any given material change with temperature.

 

[MitsuShai]

Thank you for that explanation.