So, while discussing problems with friends, I came across a capacitors problem which looked something like this:
So, my questions are:
Can this be called a capacitor even though it same polarity of charge on the plates?
Charge – After closing the switch, will there be equal chares on both plates (each ($Q_1 + Q_2$)/2) in order to minimise repulsion? — Update : I’ve figured this out using Guass’s law. You can ignore this part.
How do I find the energy in the first configurations? Can I find the electric field due to one plate, integrate it to get potential and then multiply the charge on the other plate to get the energy? Or is there any other method? — Update : I have found a method a for this. Find the voltage (charge on inner surface)/C and then use $(1/2)(C)(V^2)$. But, I am not sure whether this is correct because it only takes into account the charges on the inner surfaces.
In the second case, will the energy stored be zero because the potential difference is zero? Then again, there are some positive near each other, repelling each other and held together by by the metal plates only. Shouldn’t these have a potential energy just like two point positive charges kept some distance apart? This is why I think that the method in question 3 (i.e, the formula $(1/2)(C)(V^2)$) could be wrong when the plates have unequal charges.
If energies found in part 3 and 4 are not equal, where would the difference go? I think it can't be heat since there is no resistance (assume ideal wires) — Update : I have learned through a comment on my answer to a similar question that this energy is lost in the form of EM radiations.
Your help would be appreciated.
PS: Sorry if there are too many questions. I felt that they are all related to each other, and so I put them in a single post.