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Capacitors

# 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.

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• • 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.
• What does David mean when he says the "charges dropped"? • Should'nt calculas be involved in some form in order to derive the energy stored ? • 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?
• 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 ? • wouldn't it be wrong to say that current will flow at ? Isn't current already a flow of charge • 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? • 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? • 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
• I guess cell phone "batteries" are technically capacitors? • 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.
• if energy is lost when charging an ideal capacitor, why does capacitor inductor tank circuit exist? • 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)
• I don't get what he means by 'charge gets dropped through ---- of the initial voltage'. 