# B Energy of a capacitor

#### Hydrous Caperilla

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng2.html#c1
So I have some doubts in this. All I could understand from the derivation is that the battery is doing external work in pushing the electron from the positive plate to the negative plate and this is what is converted into energy according to my teacher.

Now I have some questions

1)Does potential between the plates is created rapidly in a short time and that's why we take the potential constant during the integration? In the integeration U=Vdq so they have taken the potential to be constant and what does dq mean here...Isn't the capacitor charged already

2) How can battery push electrons across the capacitor when capacitor blocks any path of current

30Why is energy of the capacitor channging so that we have to integrate it?

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#### cnh1995

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Gold Member
)Does potential between the plates is created rapidly in a short time and that's why we take the potential constant during the integration? In the integeration U=Vdq so they have taken the potential to be constant and what does dq mean here...Isn't the capacitor charged already
dU=(V+dV)(dq)=Vdq+dV.dq
As both dV and dq are infinitesimally small, their product dV.dq is ignored.
what does dq mean here
The infinitesimal amount of charge that gets added to the capacitor plates in an infinitesimal amount of time (dt)
) How can battery push electrons across the capacitor when capacitor blocks any path of current
It blocks dc only when it is completely charged.

#### Hydrous Caperilla

So the point is that since it is not fully charged the battery can push electrons from one plate to other...looks awesome to me.

Also dq is the charge added to the capacitor by the battery...kay.

But I still didn't understand this one......

dU=(V+dV)(dq)=Vdq+dV.dq
As both dV and dq are infinitesimally small, their product dV.dq is ignored.

Can you tell me two things specifically

1)Does capacitor voltage equal battery voltage rapidly and that's why we used (V+dv) for a charge dq

2)What does dv represent there....By KVL the voltage of the capacitor should be equal to that of battery,shouln't it?

#### Let'sthink

OP is raising a very fundamental question about definition of potential energy. Fds is supposed to give us the potential energy change of a body, presumably a point particle. Let us take two simple cases F along ds and F opposite to ds. If is along ds the body has to gain kinetic energy. But we neglect this or 'wish" it away and say that Fds is the difference potential energy between the two points. Taking the body opposite to F requires some more imagination because given to it self the body will start moving in the direction of F; so first we need to apply at least that much force in opposite direction to keep it there and then apply a little more to make it move in the opposite direction, or instead imagine it has got some velocity (or energy) in opposite direction which is just sufficient to take it though ds and bring it to rest. For potential energy difference for points which are not infinitesimal we take the help of the concept of integration which makes the ds tend to zero. So mathematically it removes the difficulty of getting kinetic energy at all. By saying all this what I am trying to put across is that the integration should not be equated to actual process of taking the body from one point to to other but it should be understood mathematically. If you still insist for the mention of actual process then we have to add infinitely slowly or quasi-statistically to use the term of thermodynamics. In other words the change in potential energy is due to the reversible work done and not the actual process which is always irreversible because something else happens during the process.

#### cnh1995

Homework Helper
Gold Member
What does dv represent there....By KVL the voltage of the capacitor should be equal to that of battery,shouln't it?
Yes, but in steady state. You should consider the resistance in the circuit. The charging time is greatly influenced by the RC time constant (it appears in the exponential charging equation).
When you connect a battery directly across a capacitor, it charges very quickly (with a sharp current spike)because the wire resistance is very small.

#### Hydrous Caperilla

Okay so KVL works at steady state of capacitor but I still didn't understand the dv part tho.....

#### Hydrous Caperilla

But I get the idea how the energy came so I am clear with the derivation.Thank you

#### davenn

Gold Member
So I have some doubts in this. All I could understand from the derivation is that the battery is doing external work in pushing the electron from the positive plate to the negative plate and this is what is converted into energy according to my teacher
Not surprised you have doubts when you are being taught incorrectly
for a start .... electrons move from negative terminal of the power supply ( battery etc) to the positive terminal

2) How can battery push electrons across the capacitor when capacitor blocks any path of current
it cant ( doesn't)

It blocks dc only when it is completely charged.
no it blocks DC (and AC) all the times, NO current flows between the plates ... There is only an APPEARANCE of charges (electrons flowing through the capacitor

As a capacitor energises, for every electron that gets added to the negative plate of the capacitor an electron leaves the positive plate and moves towards the positive terminal of the battery etc. it produces a flow of charge (electrons) in the circuit BUT NOT between the plates of the capacitor

NOTE: because of the above, the net charge on the capacitor is zero, even whilst the capacitor is energising

So the point is that since it is not fully charged the battery can push electrons from one plate to other...looks awesome to me.

Again ... electrons ( charges) move onto one plate and off the other plate

regards
Dave

#### cnh1995

Homework Helper
Gold Member
no it blocks DC (and AC) all the times, NO current flows between the plates ... There is only an APPEARANCE of charges (electrons flowing through the capacitor
It seems that I didn't understand OP's confusion. I thought he was asking 'how capacitors block dc and pass ac' while he was actually asking how current can flow throgh the dielectric.
(If that is so @Hydrous Caperilla , look up displacement current.)

#### jbriggs444

Homework Helper
Again ... electrons ( charges) move onto one plate and off the other plate
Depending on exactly what is meant, this statement may be either true or false.

Electrons are not pushed off one plate, through the dielectric gap and onto the other plate.
Electrons can be pushed the long way around, off one plate, through its terminal, through circuit and battery and back through the opposite terminal onto the other plate.

[But there are a lot of electrons in the wire and many more charge carriers in the battery. It may not be the same electrons flowing all the way from the one capacitor terminal to the other].

#### davenn

Gold Member
Depending on exactly what is meant, this statement may be either true or false.

Electrons are not pushed off one plate, through the dielectric gap and onto the other plate.
Electrons can be pushed the long way around, off one plate, through its terminal, through circuit and battery and back through the opposite terminal onto the other plate.

[But there are a lot of electrons in the wire and many more charge carriers in the battery. It may not be the same electrons flowing all the way from the one capacitor terminal to the other].

absolutely spot on
Tho, I'm not sure they traverse the battery internals. from what I have seen in threads on here and in other places
The battery chemical reaction of charge separation wouldn't allow that to happen ?

Dave

#### jbriggs444

Homework Helper
Tho, I'm not sure they traverse the battery internals. from what I have seen in threads on here and in other places
The battery chemical reaction of charge separation wouldn't allow that to happen ?
Agreed 100%

#### Let'sthink

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng2.html#c1
So I have some doubts in this. All I could understand from the derivation is that the battery is doing external work in pushing the electron from the positive plate to the negative plate and this is what is converted into energy according to my teacher.

Now I have some questions

1)Does potential between the plates is created rapidly in a short time and that's why we take the potential constant during the integration? In the integeration U=Vdq so they have taken the potential to be constant and what does dq mean here...Isn't the capacitor charged already

2) How can battery push electrons across the capacitor when capacitor blocks any path of current

30Why is energy of the capacitor channging so that we have to integrate it?
Regarding 1 and 3, I would like to say that we should not mix integration with process. Integration is mathematical and at best we can imagine an idealized process. Yes we can talk about the process separately. Regarding 2, I would say Chemical reaction in the battery has a tendency to move electrons from the negative end of the battery to the the connected plate of the capacitor through connecting wire. At the same time the reaction has a tendency to move electrons to the positive plate of the battery from the connected plate of capacitor through wire. In fact the very potential difference which you find between two terminals of the battery is due to this tendency of the chemicals contained in battery which contain two different metals to form terminals. As in a water pump the electric motor has a tendency to move water to a higher gravitational potential by giving water velocity similarly the chemical potential energy of the batter y constituents has a tendency to create electrical potential difference between two terminals. This cannot go for long because this build up will start the flow of the ions in the opposite direction as a result of electrical potential difference(pd) build up. So when battery is on open circuit. the chemical process and electrical current through chemicals maintain a dynamic equilibrium. If there is a tendency to reduce this external pd the chemical reaction will make it up and thus the chemical energy will be continuously converted into electrical energy. As in the case of lifting water by pump electrical energy gets converted into gravitational energy of water.

#### Dale

Mentor
Okay so KVL works at steady state of capacitor but I still didn't understand the dv part tho.....
KVL works at all times, not just at steady state.

#### theodoros.mihos

$C$ is an empirical parameter for any device.
Try to add small charges $\delta{q}$ to a device that $V$ depends linearely to $Q$ inside, like a common capacitor.
Adding total energy you can see a Gaus's summation that provide a factor 1/2.

#### Khashishi

2) How can battery push electrons across the capacitor when capacitor blocks any path of current
In an ideal capacitor, electrons don't move across the gap. The electrons take the long way around (through the circuit).

#### Hydrous Caperilla

So are the carriers pumped out through the battery and reach the other plate?

#### Let'sthink

So are the carriers pumped out through the battery and reach the other plate?
That is nicely put. Little correction in place of carriers if we say charges it will be better. Inside battery charge carriers are mobile and outside in wire electrons are mobile. That is what chemical reaction or chemical potential does. Rather that is the tendency. It continues till equilibrium is established between, you may say, electric potential and electro-chemical potential.

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#### Hydrous Caperilla

I was talking about the charge carriers in capacitor being pumped out through battery....Sorry it came out as slightly misunderstood by my side

#### Let'sthink

I was talking about the charge carriers in capacitor being pumped out through battery....Sorry it came out as slightly misunderstood by my side
There are no mobile charge carriers in the capacitor there are only electrons on the surface of each plate of capacitor. Technically current does pass through the dielectric or even vacuum but it is not carried by mobile charges. Change of electric field with time constitutes the current.

#### Khashishi

So are the carriers pumped out through the battery and reach the other plate?
Yeah. If you hook up a battery to a capacitor, one end of the battery pulls electrons from one end of the capacitor, and one end pushes electrons on to the other end of the capacitor. Meanwhile, positive ions flow within the battery. Despite being called a "circuit", electrons don't go all the way around.

#### CWatters

Homework Helper
Gold Member
When you initially connect a capacitor to a battery the capacitor starts off discharged (zero voltage). So one end of the wire connecting the two is at battery voltage and the other is at zero. This causes electrons to flow down the wire from the battery onto one plate of the capacitor. The surplus of electrons on that plate causes electrons to be repelled from the other plate. So it looks like electrons flow into one plate and out of the other but they are not the same electrons. This process causes a voltage to build up that opposes the battery voltage. Eventually the capacitor voltage matches the battery voltage and the flow of electrons stops.

If the source/ battery voltage is constant (DC) things stay as above. No current flows so we say that it blocks DC.

If the source voltage changes (could be AC or any strange waveform you like) there be a temporary voltage difference between the source and capacitor that causes electrons to flow in such a way that the capacitor voltages always tries to keep up with changes to the source voltage. This flow of electrons looks like any other current so we say that capacitors pass AC.

If you put a resistor between an AC source and a capacitor you make it harder for electrons to flow limiting the current so it takes longer for the capacitor to catch up with changes to the source voltage. This means the capacitor voltage tends to lag behind the source voltage.

Useful maths...

Q=VC
Differentiate both sides..
dQ/dt = CdV/DT

dQ/dt is the current (I) so..

I = CdV/DT

In other words the current that flows into a capacitor depends on how fast you change the voltage on the capacitor. If you connect a voltage source to a discharged capacitor you are asking the voltage to change rapidly so the current might be very high.

If you integrate the equation above you get

V = 1/C * the integral of I

Sorry can't do integral symbol on this tablet.

One way to think about this is that the voltage on a capacitor depends not only on the current "now" but on the past history of the current. In effect the capacitor is storing information about the past.

Inductors are just as interesting.

#### Let'sthink

Even while charging or discharging alternating current passes through capacitors. When we say ac passes through capacitor we mean any changing, The current which passes through capacitors is known as dD/dt or displacement current it is equivalent to motion of charges. It will produce all effects of a current caused by motion of charges. Just from outside you will not be able to tell whether the current is caused due to motion of charges or due to changing electric field.