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http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng2.html#c3

So I have some doubts in this:

The energy stored on a capacitor can be expressed in terms of the work done by the battery.

How does Battery play a role here....is it pushing the charges?

Voltage represents energy per unit charge, so the work to move a charge element dq from the negative plate to the positive plate is equal to V dq, where V is the voltage on the capacitor.

Okay so Voltage is defined as potential energy per unit charge so that's how we get the expression U=qV.

What work done is being talked here upon and from where is this it being transported from...This one seems to be confusing to me.

The voltage V is proportional to the amount of charge which is already on the capacitor.

Element of energy stored: $dU=Vdq=\frac{q}{C} dq$

And finally how did this expression come...can somebody give me an explanation on this concept because I am quite a lot struggling on this one and the further integeration ...why did we have to integrate?

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What does the battery do?
The battery moves electrons from the plate which is already has a net positive charge to the plate which is already has a net negative charge noting that the potential of the positive plate is at a higher potential than that of the negative plate.
To move the electrons in such a way requires work to be done and the battery does that work at the expense of using some chemical energy.

How much work needs to be done to move a charge $\delta q$ through a potential difference $V$?
It is $V\, \delta q$.

An analogy which might help is you moving masses placed on a lower shelf onto masses on a higher shelf such that the masses on the higher shelf form a tower.
So as you move the masses you have to do more work (use more chemical energy) to move successive masses and the increase in potential energy of the masses is larger.

In the case of your example the chemical energy of the battery is being used to increase the electric potential energy as charges are being moved from one plate to the other.

By definition $C=\frac qV$ which gives the work done by the battery (change in electric potential energy stored) is equal to $\frac qC\delta q$ and as more charge is stored in the capacitor (the potential difference between the plates is larger) so more work needs to be done by the battery to transfer charge $\delta q$ from one plate to the other.

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  • $\begingroup$ So work done by battery is getting converted into energy of the capacitor?? $\endgroup$ – user184271 Feb 24 '18 at 12:55
  • $\begingroup$ @harambe That is correct. The capacitor stores electric potential energy. $\endgroup$ – Farcher Feb 24 '18 at 13:07
  • $\begingroup$ @Farcher So work done against electric field in transporting the electron to the other plate gets stored as energy...right? $\endgroup$ – user184272 Feb 24 '18 at 22:51
  • $\begingroup$ @HASEGAWA The battery “pumps” the electron from one plate to the other thus increasing the potential energy whilst at the same time heat is produced in the circuit because it has resistance and there is a current flowing. $\endgroup$ – Farcher Feb 24 '18 at 23:37
  • $\begingroup$ @Farcher So work done by batterygets stored as energy and since it works against electric field it is positive and hence we have to integrate it Okay got that....Is it why energy stored in capacitor is in the form of electric field $\endgroup$ – user184272 Feb 24 '18 at 23:54
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The energy stored on a capacitor can be expressed in terms of the work done by the battery.

Charges need a force to move in your circuit, in this case the battery provides it.

Voltage represents energy per unit charge, so the work to move a charge element ${\rm d}q$ from the negative plate to the positive plate is equal to $V {\rm d}q$, where $V$ is the voltage on the capacitor.

Is the work done upon the charges to transport them in the capacitor. An intuitive way of thinking this is by considering what happens to one single charge once the circuit starts operating. Any charge ${\rm d}q$ will move freely since there aren't other charges in the capacitor. The next charge ${\rm d}q$, however, will feel the electric field generated by the presence of the first one. The third will will the electric field generated by the presence of the previous two, ...

At any time you can calculate this work just and the result is $V {\rm d}q$, where $V$ is the potential difference between the plates

The voltage $V$ is proportional to the amount of charge which is already on the capacitor. Element of energy stored: ${\rm d}U=V{\rm d} q= q {\rm d}q/ C$

This is just a definition of the quantity $C$. You can think of this as how much work do I need to do to charge my capacitor

$$ C = \frac{q}{V} $$

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  • $\begingroup$ So what's the reason for build up of energy in the capacitor?According to my knowledge plus some facts from you the battery is pushing some charges by doing some work which then the charges come to the capacitor and start accumulating....So is this charge build up related to increase in energy of capacitor $\endgroup$ – user184272 Feb 24 '18 at 11:27
  • $\begingroup$ @HASEGAWA Correct, you can calculate the total amount of work and the result is $U\sim q^2$, the more charge you put in the capacitor, the more energy it stores $\endgroup$ – caverac Feb 24 '18 at 12:17

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