# Capacitors Help

Watch
Announcements

Page 1 of 1

Go to first unread

Skip to page:

Circuits and electrical components have always been my weak point in physics, so I just want to check I have the correct basic understanding of capacitors before I go on to learn more about them.

If a charged capacitor discharges in a circuit containing only that capacitor and another uncharged capacitor with a capacitance of half the charged capacitor, the charge held in each capacitor in the end will be in the ratio of their capacitance, but the sum of the charge will be equal to the initial amount that was stored in the first capacitor.

Is this because Q=CV, voltage=energy per coulomb and because the first capacitor is acting as an emf source, and the second capacitor is the only other component in the circuit, the potential difference over both is equal (as energy supplied per coulomb by the first is the amount supplied by each coulomb of charge in the second). So Q1/C1 = Q2/C2. Charge is conserved, so the initial charge held in the first capacitor is divided up according to the ratio of their capacitance.

This is where it gets a bit messy....

Although I haven't learned that much about it yet, I know that the rate of charge discharge is exponential.

So this means the current in the circuit varies, but the potential difference over the capacitors doesn't (?).

V=IR, so the capacitor's have to overcome their own resistance, varying with the rate of discharge/charge, in delivering the charge, and the work done against this resistance will account for a loss of energy in dividing the charge between the two resistors.

I did a question regarding calculating the energy stored when the initial capacitor was charged, and then calculating the total energy stored in both capacitors after the charge was shared amongst them both. I got the right answer, but there was a significant energy drop. I'm trying to figure out what could account for such a big loss when we're assuming zero wire resistance.

Thanks in advance (and I hope I've been clear enough in explaining my ideas)

If a charged capacitor discharges in a circuit containing only that capacitor and another uncharged capacitor with a capacitance of half the charged capacitor, the charge held in each capacitor in the end will be in the ratio of their capacitance, but the sum of the charge will be equal to the initial amount that was stored in the first capacitor.

Is this because Q=CV, voltage=energy per coulomb and because the first capacitor is acting as an emf source, and the second capacitor is the only other component in the circuit, the potential difference over both is equal (as energy supplied per coulomb by the first is the amount supplied by each coulomb of charge in the second). So Q1/C1 = Q2/C2. Charge is conserved, so the initial charge held in the first capacitor is divided up according to the ratio of their capacitance.

This is where it gets a bit messy....

Although I haven't learned that much about it yet, I know that the rate of charge discharge is exponential.

So this means the current in the circuit varies, but the potential difference over the capacitors doesn't (?).

V=IR, so the capacitor's have to overcome their own resistance, varying with the rate of discharge/charge, in delivering the charge, and the work done against this resistance will account for a loss of energy in dividing the charge between the two resistors.

I did a question regarding calculating the energy stored when the initial capacitor was charged, and then calculating the total energy stored in both capacitors after the charge was shared amongst them both. I got the right answer, but there was a significant energy drop. I'm trying to figure out what could account for such a big loss when we're assuming zero wire resistance.

Thanks in advance (and I hope I've been clear enough in explaining my ideas)

0

reply

Report

#2

(Original post by

Is this because Q=CV, voltage=energy per coulomb and because the first capacitor is acting as an emf source, and the second capacitor is the only other component in the circuit, the potential difference over both is equal (as energy supplied per coulomb by the first is the amount supplied by each coulomb of charge in the second). So Q1/C1 = Q2/C2. Charge is conserved, so the initial charge held in the first capacitor is divided up according to the ratio of their capacitance.

This is where it gets a bit messy....

Although I haven't learned that much about it yet, I know that the rate of charge discharge is exponential.

So this means the current in the circuit varies, but the potential difference over the capacitors doesn't (?).

V=IR, so the capacitor's have to overcome their own resistance, varying with the rate of discharge/charge, in delivering the charge, and the work done against this resistance will account for a loss of energy in dividing the charge between the two resistors.

I did a question regarding calculating the energy stored when the initial capacitor was charged, and then calculating the total energy stored in both capacitors after the charge was shared amongst them both. I got the right answer, but there was a significant energy drop. I'm trying to figure out what could account for such a big loss when we're assuming zero wire resistance.

Thanks in advance (and I hope I've been clear enough in explaining my ideas)

**hhattiecc**)Is this because Q=CV, voltage=energy per coulomb and because the first capacitor is acting as an emf source, and the second capacitor is the only other component in the circuit, the potential difference over both is equal (as energy supplied per coulomb by the first is the amount supplied by each coulomb of charge in the second). So Q1/C1 = Q2/C2. Charge is conserved, so the initial charge held in the first capacitor is divided up according to the ratio of their capacitance.

This is where it gets a bit messy....

Although I haven't learned that much about it yet, I know that the rate of charge discharge is exponential.

So this means the current in the circuit varies, but the potential difference over the capacitors doesn't (?).

V=IR, so the capacitor's have to overcome their own resistance, varying with the rate of discharge/charge, in delivering the charge, and the work done against this resistance will account for a loss of energy in dividing the charge between the two resistors.

I did a question regarding calculating the energy stored when the initial capacitor was charged, and then calculating the total energy stored in both capacitors after the charge was shared amongst them both. I got the right answer, but there was a significant energy drop. I'm trying to figure out what could account for such a big loss when we're assuming zero wire resistance.

Thanks in advance (and I hope I've been clear enough in explaining my ideas)

When thinking about voltage and charge, it's quite intuitive to use the water pressure analogy where there is a force per unit area. Filling up a water tank raises the pressure throughout the tank dependent on the depth of the water at any given place. Raising a tank of water above ground gives the water potential to do work under the force of gravity.

Capacitance is analogous to the volume of water in the water tank. Filling the capacitor tank up with charge, increases the voltage pressure in the tank dependent on how much charge is stored for a give surface area. In this case, the voltage pressure is created in the electric field between the surface areas of the two capacitor plates.

The difference is that in the water analogy, gravity provides the acceleration force whereby raising the tank height above ground stores potential energy (P.E. = mgh.)

In the capacitor, the acceleration force is provided by other charges. i.e. the imbalance of charges at the receiving end of the current (charge) conduction path.

Emptying one capacitor (discharge) reduces the voltage pressure in that tank, but must simultaneously increase it in the receiving capacitor. As the first capacitor discharges, it's potential must fall and with less voltage, that means the current (charge per second) must also fall.

The key thing to remember is that charge is a conserved property. The voltage pressure is a measure of the potential (Joules per Coulomb) as a function of the surface area of the capacitor plates and the separation distance between them. As the charge is transferred, the potential Joules per Coulomb also falls.

In other words, the charge transferred from the first capacitor must be the same as the sum of the charges in the two parallel capacitors, and since capacitor voltage is a function of surface area, the charge force is conserved and simply now distributed between the two thus reducing the combined voltage pressure and hence potential to do work. {The individual electron-charges are all still present but are now neutralised (recombined with +ve charge atoms in the receiving capacitor) or spread more thinly within the conduction energy band and therefore the force they exert per unit volume also must reduce.}

The difference in potential energy available per coulomb of charge between the two cases is simply a measure of the charge/unit area change. Putting a resistance in the current path converts the potential to heat during the transfer but the final outcome in potential energy stored between the capacitors ends up the same.

The difference in potential energy is given up during the transfer process.

A surprising number of misconceptions regarding this abound throughout the internet including some otherwise good websites.

0

reply

X

Page 1 of 1

Go to first unread

Skip to page:

### Quick Reply

Back

to top

to top