scottdave said:
If the plates are so far apart that they do not interact with each other, then really it is not a capacitor, hence the value of C in your formula ## \frac { CV^2} 4 ## has no meaning.
The following part was not clear. In this part, I am charging the capacitor by keeping the plates very far and still applying V = q/C, which is wrong.
Pushoam said:
The two plates are charged independently of each other, means the charging of one plate does not affect the changing of other plate.
In this way, the work done on the system is ## \frac { CV^2} 4##.
But then , to arrange as a capacitor, I will have to bring the two plates at a distance d.
Charging the capacitor by taking charges from infinity
A) Bringing both positive charges and negtive charges together simultaneously.
This way the charges on both plates kept at distance d will be always equal and opposite.
Potential difference between both plates , V = q/C. So, potential difference between one plate and infinity will always have magnitude V/2 = q/2C.
So, if I charge the capacitor this way, then the work done will be ## \frac 1 4 CV^2.##B) But if I keep the plates very far and then I bring charges upon them separately, then the work done in bringing the charges on both plates will be ## 2\int_{ 0}^{ Q} V dq ## . Here, V = V(q), but unknown to me. V(q) ## \neq q/C ## .
So, I do not know how to calculate work done in this case.
And then some work will be done in bringing the plates at a distance d.
Is this correct?
But, does it mean that when I have brought the plates at a distance d, the total work done on the capacitor is ##½ CV^2##?
The energy stored in the capacitor will always be ##½ CV^2##, irrespective of how I charged it as this energy is potential energy so it should be independent of the process of charging.
But, does the total work done on the capacitor depend on the way we charge it?
And in text-book question of calculating total work done in charging a capacitor,it is assumed that we are charging it by taking electrons from one plate to another plate?