Hi,

i had some general questions about capacitors. First of all, when a capacitor charges up, do actual charges travel through the plates? like does it charge up to a threshold (the capacitance value) and then the charges travel from one plate to the other through the air/dielectric? Also, whether arranged in series or parallel, i understand that voltage varies with the connection. But im confused as to whether the charges remain the same in both connections. for example, if you had a 2 capacitors of different values, arranged them in series, would they contain the same charge in each capacitor as they would if they were arranged in parallel?

 

First of all, when a capacitor charges up, do actual charges travel through the plates?

Not at all - charge is driven onto one plate and off the other.

Our Ebook has some good material on capacitors (other stuff as well). Why not look through it and see if they make more sense? - http://www.allaboutcircuits.com/vol_1/chpt_13/1.html

 

Any two capacitors connected in series have same charge stored in them but different voltage because the capacitors are different. When they are in parallel, that means they have same voltage across them but since they have different capacitance values they store different amounts of charge.

- Raj
http://pic16f628a.blogspot.com/
http://picboard.blogspot.com/

 

A single capacitor:

q = CV

With a 1F capacitor and 1V source you have:

q = (1F)(1V) = 1C

Two 1F capacitors in parallel. Since they are in parallel they must have the same voltage. Charge simply adds:

q = (1F)(1V) + (1F)(1V) = 2C

This is equivalent to a single 2F charged to 1V:
Ceq = q/V
Ceq = 2C / 1V
Ceq = 2F

Capacitors in parallel simply add in value.

Two 1F capacitors in series. They are in series so must have the same current and hence they must store the same charge. Pushing a charge into the first equivalent must a charge onto the second.

This gives you:
C1 * V1 = C2 * V2
1F * V1 = 1F * V2
V1 = V2

Or V1 = V2 = 0.5 V

This means equal capacitors equally split voltage.

How about equivalent capacitance?
From the point of view of the battery you only pushed q = 1F*0.5V = 0.5C yet you managed to get V1 + V2 = 1 V.
This means you have:
Ceq = q/V
Ceq = 0.5C / 1V
Ceq = 0.5F

Capacitors in series add like resistors in parallel, where two equal ones give you half the value.

 

Back in the dark ages (1960's), I was taught that the charge retained in a capacitor is the result of the electron orbit in the dielectric being sent into an elliptical shape as the result of an excess of electrons on one conductive plate and a shortage of electrons on the other plate.. As the cap has a path to discharge applied, the return of the electrons to a normal circular orbit is what supplies the energy. Sort of like putting a rubber band into a stretch to store energy.

 

While the analogy works for practical purposes Bill that's not how capacitors actually work.

kiloman, except for leakage current electrons don't pass from one plate to another, the plates potential difference provides energy to move electrons on the other plate through the electric field, no actual electrons are directly transferred from one plate to the other, it's the electric field interaction between the two plates that causes energy to be transferred. Energy moving and charges being transfered does NOT mean electrons are shooting between the plates =)

 

hi i am new here .. I am glad to join this forum community . I am really interested in physics .. I am A student .. In class I discussed every topic of physics with my all friends and teachers .. I am agree with rajbex because it is the two different capacitor which have different capacitance does not have the same charge ... But have same potential across ..!!!