# Help with capacitors

1. Apr 3, 2007

### spudvr

Hi everyone,
This is my first post on the forum, but I have used the site many times for previous help. My professor started us on capacitors today and just had time to give us a bunch of equations without much explanation of how to solve problems. A problem he gave us to solve so as so:

A 20.0 microFarad capacitor is charged to a potential difference of 900V. the terminals of the charged capacitor are then connected to those of an uncharged 10.0 microFarad capacitor. Find:

a) The original charge of the system.
b) The final potential difference across each capacitor.
c) The final energy of the system.
d) The decrease in energy when the capacitors are connected.
e) Where did the "lost" energy go?

I think i managed a) Q=CV so (20uF)(900V)=18000uC
But thats all I have so far. If anyone could possible help me by setting me off in the right direction with any suggestions it would be greatly appreciated.
Thanks so much
Spudvr

2. Apr 3, 2007

### Staff: Mentor

Thread moved to Homework Help -- Intro Physics. Please remember to post homework and coursework questions in the appropriate Homework Help forum, and not in the main forums.

On your question -- the charge will be conserved, since there is no where for it to go. So you have the initial charge across the first cap, and when you connect the 10uF and 20uF caps in parallel, they will then share that total charge, and since they are in parallel, their final voltages have to match. That should give you enouth hints to answer b-d.

As for e, that is a tricky question. Post your answers for a-d, and we can all offer some hints for you to figure out e.

3. Apr 3, 2007

### spudvr

Ok, this is what i have so far
b)V=Q/(c1=c2)
18000/30=600V
c)Uf=.5(Q1)(V)+.5(Q2)(V)=.5(Q)(V)
.5(18000x10^-6)(600)=5.4J
d) .5(Q)(V)=(18000x10^-6)(900)=8.4J
8.4J-5.4J=2.7J so a decrease in 2.7J

these looking ok?

Last edited: Apr 3, 2007
4. Apr 3, 2007

### Staff: Mentor

Looks good so far. Good work.

Now the tricky part of the question -- I'll only give you one hint for now, because this is the kind of question that your prof is using to see how you think things through.

What happened when the uncharged cap was connected in parallel with the charged cap? You already know that the charge re-distributed itself between the two caps, which means a current flowed briefly in the wire connections between the caps. Now, the hint is to think about the resistance of real wires -- start by assuming there was a 100 Ohm resistor between the two caps when they were connected -- write the equation for the current through that resistor as a function of time as the charge re-distributes. How much energy is lost to that resistor? Now make the resistor 1 Ohm, and do the calculation again. And try again for 1 milliOhm, and so on. What can you say about where the energy goes, even in the limit of nearly-zero resistance wires?

Please show your work on this part e. You can't use my hint as an answer directly without working through the numbers. Your prof is going to want to see your work on this one.

5. Apr 3, 2007

### spudvr

I somewhat understand what you are talking about, however, he has not said a single thing about resistors. I guess ill look ahead in the text and see if i can find it. thanks so much for your help.
spudvr