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Why is E=1/2QV?

Okay so I appreciate that as a capacitor is being charged it begins with zero charge stored on it, and slowly fills up. But I don't understand why the voltage varies.

The book says 'Each time we add a little extra charge (Q) this has to be done by increasing the voltage... [as] we are doing work against the repulsion between like charges'. I don't understand this because why is the battery now all of a sudden choosing how much energy it, gives each unit of charge? Surely the voltage should be constant, if units of charge have excess energy surely they can just use it up as heat like they do in normal battery light circuits? I don't understand why the battery is now selective about how much energy each unit of charge receives?
(edited 10 years ago)
Reply 1
Bump, I'm interested.
Original post by FarmerMan
Okay so I appreciate that as a capacitor is being charged it begins with zero charge stored on it, and slowly fills up. But I don't understand why the voltage varies.

What does voltage (pd) measure?
The pd on the plates is a measure of the energy required to move a charge from zero potential onto the plate. The more charge on the plates, the bigger the repulsive (Coulomb) force stopping further charge being moved there. So as charge builds up on the capacitor, it needs more work to put further charge there. More work means the potential is greater on the plates. That is the definition of potential.


Which is why the book says...

The book says 'Each time we add a little extra charge (Q) this has to be done by increasing the voltage... [as] we are doing work against the repulsion between like charges'.


I don't understand this because why is the battery now all of a sudden choosing how much energy it gives each unit of charge?

It isn't. Why do you think this?
Each unit of charge moving through the battery gets the same amount of energy, qV.


Surely the voltage should be constant, if units of charge have excess energy surely they can just use it up as heat like they do in normal battery light circuits? I don't understand why the battery is now selective about how much energy each unit of charge receives?


In order to move a charge q onto a capacitor at pd V you would need qV of energy.
However, the capacitor is not permanently at a pd V while it is being charged.
It starts at 0 and rises to V
So initially it needs zero work to put the 1st charge on and qV to put the last one on.
If you take the whole process from 0V to V and zero charge to Q, the energy required was 0.5qV.
0.5V can be thought of as the mean value of the pd during charging.

And here we have the tricky point.
The battery has delivered energy to the charges = QV but the capacitor has only stored 0.5QV of energy as a result.
The interesting fact is that it doesn't matter what the resistyance of the circuit is that is used to charge the capacitor from the battery, the amount of energy lost in the process is always 0.5QV
Reply 3

I don't understand this because why is the battery now all of a sudden choosing how much energy it gives each unit of charge?
It isn't. Why do you think this?

The battery has delivered energy to the charges = QV but the capacitor has only stored 0.5QV of energy as a result.

I think I get it now. The reason I thought the battery was choosing how much energy it gives each unit of charge is because; I was assuming that all the energy supplied by the battery was being stored by the capacitor. But as you said the capacitor will only receive (when it is fully charged) 1/2 the energy supplied by the battery. So am I right in saying, that the battery is providing every unit of charge with QV amount of energy, it's just that the first electrons to go on the capacitor will need less energy, so will use up there left over energy as heat. But the electrons which come on the capacitor later will need more energy, so will use up less energy as heat, until eventually the last electron will use up no energy as heat, as it needs all of its energy to get on the capacitor.

Thank you very much for your reply, I hope what I said above was right.
(edited 10 years ago)
Original post by FarmerMan
I think I get it now. The reason I thought the battery was choosing how much energy it gives each unit of charge is because; I was assuming that all the energy supplied by the battery was being stored by the capacitor. But as you said the capacitor will only receive (when it is fully charged) 1/2 the energy supplied by the battery. So am I right in saying, that the battery is providing every unit of charge with QV amount of energy, it's just that the first electrons to go on the capacitor will need less energy, so will use up there left over energy as heat. But the electrons which come on the capacitor later will need more energy, so will use up less energy as heat, until eventually the last electron will use up no energy as heat, as it needs all of its energy to get on the capacitor.

Thank you very much for your reply, I hope what I said above was right.


You can certainly think of it that way.

You just have to remember that only half the energy supplied by the battery gets stored as electrical energy in the capacitor, irrespective of the value of the resistance in the connecting circuit.
Reply 5
In capacitor there is an insulator between the plates so no current flowes through it...and electrons given by the battery stores in dielectric due to induction....when storage charge increases it's voltage also increases.. current due to these electrons is called displacement current......But in case of battery there is an electrolyte between the anode and Cathod..and this electrolyte is a good conductor of electricity......so in this there is no charge storage ..so....it's voltage remains constant.......

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