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# Energy of a capacitor

This video explains the potential of a capacitor and how they function in a circuit. By David Santo Pietro. . Created by David SantoPietro.

## Want to join the conversation?

- Where are capacitors used?(15 votes)
- A capacitor can store electric energy when it is connected to its charging circuit. And when it is disconnected from its charging circuit, it can dissipate that stored energy, so it can be used like a temporary battery. Capacitors are commonly used in electronic devices to maintain power supply while batteries are being changed.(23 votes)

- What does David mean when he says the "charges dropped"?(14 votes)
- Should'nt calculas be involved in some form in order to derive the energy stored ?(5 votes)
- Yes, I think calculus should be involved in, I have tried to do it, bu still missing something :

I plot V in function of Q for the capacitor, imagine you have only 4 positives charges on one side of the capacitor, and 4 negatives charges on the other side. Because V=∆U/Q (where ∆U is electrical potential energy of the charges, Q the charges), And ∆U=F·distance=Q·E·distance (where E is the Electrical Field) thus V=E·distance.

E=KQ/r^2=K·(number of charges/Q_total)·Q_total/r^2

V=Vmax for Vmax=Emax·distance, V decreases as the capacitor discharges because charges are less and less separated from each other over time, thus E between the plates of the capacitor decreases as time goes on.

For all the negatives charges (4)on the left side:

V=((K((4/4)Q_total))/r^2)·distance=(4/4)KQtotal·distance/r^2=(4/4)·some constant=Vmax

For 3 negatives charges on the left side :

V=(3/4)·some constant=(3/4)Vmax

For 2 negative charges on the left side :

V=(2/4)·some constant=(1/2)Vmax

For 1 negative charge on the left side :

V=(1/4)·some constant=(1/4)Vmax

For 0 negative charge on the left side :

V=(0/4)·some constant=0·Vmax

When you Plot V in function of Q : V=Q/Qtotal·Vmax , the curve is linear.

When you do the integral the curve from 0 to 4 :

∫V(Q)dQ=∫Vmax·Q/Qtotal·dQ=[Vmax·(Q/Qtotal)^2/2]from 0 to 4=(Vmax·(4/4)^2/2)-(Vmax·(0/4)^2/2)=Vmax·8 J

(Since [∫V(Q)dQ]=Energie/Charges·charges, ∫V(Q)dQ is in Joules.)

Since ∫V(Q)dQ=∫Vmax·Q/Qtotal·dQ=Vmax·1^2/2 we have : Total Electrical potential Energy=1/2·Vmax

So there is no Q in the final equation :

Electrical potential energy = 1/2Vmax, someone could help me please?(2 votes)

- How can be measure the charge in a capacitor, empirically? For example. We have an ammeter to measure current, a voltmeter to measure voltage. So is there a particular instrument to measure the charge? Maybe even instantaneously measure the capacitance ?(3 votes)
- However you can indeed measure capacitance with a multimeter!(1 vote)

- wouldn't it be wrong to say that current will flow at0:48? Isn't current already a flow of charge(3 votes)
- You are right. Saying
*current flows*is a habit mostly common among engineers. In actuality, charge is the only thing that flows, so it is more correct to say that charge flows. It is such a common habit that most people will accept it if you say that current flows, though.(3 votes)

- how is a capacitor discharged? i understand that the electrons build up on the negative plate and a voltage is induced between the positive plate and the dielectric, but what actually causes the charge to be released? does the dielectric have to break?(2 votes)
- The capacitor has to be connected to a circuit that allows it to discharge.

Dielectric breakdown is of course an alternative, undesirable method of discharge.(5 votes)

- Consider a capacitor is connected to a battery and charged to its maximum charge holding capacity. Now, if I connect the same capacitor to another battery of a different emf, how do I calculate the heat lost in the process?(2 votes)
- Hello Aanchit,

You would need to know more about the circuit including:

* internal resistance of the original battery

* equivalent series resistance of the capacitor

* internal resistance of the 2nd battery

* resistance of the wire(s)

If we know all of these values we could perform a calculation where the power lost is equal to current squared times resistance.

Note that all of this work would only get us in the ballpark since batteries are non-linear devices...

Regards,

APD(3 votes)

- I guess cell phone "batteries" are technically capacitors?(3 votes)
- It is, sort of, if you are considering only their uses; both batteries and capacitors store charge to be used in various circuits. However, capacitors simply store charge on the metal pieces on either side of a barrier. There is no other change in the actual capacitor.

On the other hand, batteries use chemical changes to store charge. (In electrochemical terms, batteries have anodes, cathodes, and electrolytes in them. The anode in the battery will react with the electrolyte, resulting in a compound and releasing electrons as a byproduct. Simultaneously, the cathode in the battery will react with the electrolyte to form a compound and absorb the released electrons. This is, in essence, electricity.)

A single-use battery lasts as long as the anode and/or the cathode supplies last. Phone batteries, which are rechargeable, are different from single-use batteries in that they use chemical reactions that can be reversed. In other words, they can again form the original anodes and cathodes with an external inflow of electrons (current).

Hope this helped! Knowing electrochemistry (or just redox reactions) will help immensely in understanding batteries.(2 votes)

- if energy is lost when charging an ideal capacitor, why does capacitor inductor tank circuit exist?(2 votes)
- Hello Ekeh,

If we had ideal capacitors and ideal inductors the energy would never be lost. The tank circuit would "ring" forever. In the real world we always have resistance and so some of the energy is lost and we end up with a decaying waveform.

To answer your question, the tank circuit acts as a filter. It is especially useful in radio frequency circuits where among other things it can be used to "tune" the radio and select a particular radio station.

Regards,

APD(1 vote)

- 3:14I don't get what he means by 'charge gets dropped through ---- of the initial voltage'.(2 votes)

## Video transcript

SPEAKER 1: Check
out this capacitor. Look at what happens if I
hook it up to this light bulb. CHILDREN (IN UNISON): Boo! SPEAKER 1: Yeah,
nothing happened because the capacitor
is not charged up. But if we look at up
to a battery first, to charge up the capacitor,
and then hook it up to the light bulb, the
light bulb lights up. CHILDREN (IN UNISON): Ooh! SPEAKER 1: The
reason this happens is because when a capacitor is
charged up, it not only stores charge, but it stores
energy as well. When we hooked up the
capacitor to the battery, the charges got separated. These separated charges
want to come back together when given the chance,
because opposites attract. So if you complete the
circuit with some wires and a light bulb,
currents going to flow. And the energy that was
stored in the capacitor turns into light and heat that
comes out of the light bulb. Once the capacitor
discharges itself, and there's no more
charges left to transfer, the process stops and
the light goes out. The type of energy that's
stored in capacitors is electrical potential energy. So if we want to figure
out how much energy is stored in a
capacitor, we need to remind ourselves
what the formula is for electrical potential energy. If a charge, Q, moves
through a voltage, V, the change in electrical
potential energy of that charge is just Q times V. Looking
at this formula, what do you think the energy would
be of a capacitor that's been charged up to a
charge Q, and a voltage V? CHILDREN (IN UNISON): Q times V! SPEAKER 1: Yeah, and that's what
I thought it would have been, too. But it turns out the energy of
a capacitor is 1/2 Q times V. CHILDREN (IN UNISON): Boo! SPEAKER 1: Where does
this 1/2 come from? How come the energy
is not just Q times V? Well, the energy of
a capacitor would be Q times V if during
discharge, all of the charges were to drop through the
total initial voltage, V. But during discharge,
all of the charges won't drop through
the total voltage, V. In fact, only the first
charge that gets transferred is going to drop through the
total initial voltage, V. All of the charges that get
transferred after that are going to drop through
less and less voltage. The reason for this is that
each time a charge gets transferred it decreases
the total amount of charge stored on the capacitor. And as the charge on the
capacitor keeps decreasing, the voltage of the
capacitor keeps decreasing. Remember that the
capacitance is defined to be the charge stored
on a capacitor divided by the voltage across
that capacitor. So as the charge goes down,
the voltage goes down. As more and more charge
gets transferred, there'll be a point
where a charge only drops through 3/4 of
the initial voltage. Wait longer, and
there'll come a time when a charge gets transferred
through only a half of the initial voltage. Wait even longer, and
a charge will only get transferred through a
fourth of the initial voltage. And the last charge to
get transferred drops through almost no
voltage at all, because there's
basically no charge left that's stored
on the capacitor. If you were to add
up all of these drops in electrical
potential energy, you'd find that the total drop
in energy of the capacitor is just Q, the total
charge that was initially on the capacitor, times
1/2 the initial voltage of the capacitor. So basically that 1/2 is there
because not all the charge dropped through the total
initial voltage, V. On average, the charges dropped through
only a half the initial voltage. So if you take the charge stored
on a capacitor at any moment, and multiply by the voltage
across the capacitor at that same
moment, divide by 2, you'll have the energy
stored on the capacitor at that particular moment. There's another form of this
equation that can be useful. Since capacitance is defined
to be charge over voltage, we can rewrite this
as charge equals capacitance times voltage. If we substitute the
capacitance times voltage in for the charge, we see
that the energy of a capacitor can also be written as 1/2
times the capacitance times the voltage across
the capacitor squared. But now we have a problem. In one of these formulas
the V is squared, and in one of these formulas
the V a not squared. I used to have trouble
remembering which is which. But here's how I remember now. If you use the formula
with the C in it, then you can see the V squared. And if you use the formula
that doesn't have the C in it, then you can't
see the V squared. So these are the two
formulas for the energy stored in a capacitor. But you have to be careful. The voltage, V,
in these formulas refers to the voltage
across the capacitor. It's not necessarily the voltage
of the battery in the problem. If you're just looking
at the simplest case of one battery that
has fully charged up a single capacitor,
then in that case, the voltage across
the capacitor will be the same as the
voltage of the battery. So if a 9-volt battery
has charges up a capacitor to a maximum charge
of four coulombs, then the energy stored
by the capacitor is going to be 18 joules. Because the voltage
across the capacitor is going to be the same as
the voltage of the battery. But if you're looking at a case
where multiple batteries are hooked up to
multiple capacitors, then in order to find the
energy of a single capacitor, you've got to use
the voltage across that particular capacitor. In other words, if you
were given this circuit with these values, you
could determine the energy stored in the middle
capacitor by using 1/2 Q-V. You would just have
to be careful to use the voltage of that
capacitor, and not the voltage of the battery. Plugging in five
coulombs for the charge lets you figure out that
the energy is 7.5 joules. CHILDREN (IN UNISON): Ooh! [MUSIC PLAYING]