Ok so the relationship between the charge Q1 when S1 is closed and the two resulting charges when S2 is closed is given by the Q1' = 72μC - Q2'
Sooo...the potential difference across the 2 capacitors are equal therefore leading to the equations that wasn't making sense to me (ΔV').
With...
Yes I understand that it is one or the other and they will not be both "charged" at the same time but I just don't know how to solve the algebra equation above...this will give me what I need.
(72μC) - Q2'/(4μF) = Q2'/(7μF)
Consider the circuit shown in the figure below, where C1 = 4.00 µF, C2 = 7.00 µF, and ΔV = 18.0 V. Capacitor C1 is first charged by closing switch S1. Switch S1 is then opened, and the charged capacitor is connected to the uncharged capacitor by closing S2.
(a) Calculate the initial charge...
Maybe E=KEo...but again I don't know how to make that work with this problem. I am not really looking for an answer for I have already turned in the wrong one. I am just trying to figure out how I got it wrong and how to do it right. :)
I am sorry. I don't really know how to find a PE of an item without any numbers. I am used to a length or something to help me out. That is why I am having such a hard time with this problem. :(
I was thinking that because q is the same distance that A and C/B and D would equal each other because they have the same potnetial energy. If this is not the case due to the shape of each I would say that A>C>D>B due to the triangle being smaller and having a greater potential energy? Thank you...
Rank the electric potential energies of the systems of charges shown in the figure below from largest to smallest. Indicate equalities if appropriate. (Use only ">" or "=" symbols. Do not include any parentheses around the letters or symbols.)
I was thinking A=C>B=D
Please help me if I am wrong!