Recent content by jendead

Ok, that makes sense. Thank you for being patient. :)

Ok, that's starting to make sense.. but why isn't Q also changed to Q/2 since it's being halved at the same time? I apologize for being so dense - I tend to have a really hard time with these concepts sometimes.

Yes, I'm trying to find the potential energy (in joules), not the voltage. I may have been unclear when I said potential.

It's U = q^2/2C.. so if I plug in (q/2)^2/2(2C), which is halving the charge and doubling the capacitance, I'd have q^2/16C.. which still makes no sense. Am I missing something conceptually or algebraically?

I understand that, but I don't see how that relates here. From the look of the answer key, I don't understand where total potential is halved when being spread out over 2 capacitors.

Homework Statement Capacitors A and B are identical. Capacitor A is charged so it stores 4J of energy and capacitor B is uncharged. The capacitors are then connected in parallel. The total stored energy in the capacitors is now ____.Homework Equations U = q^2/2C The Attempt at a Solution Ok...

I think I get it now.. thanks for the help and link :)

Ok, so.. I used V = 4q/C, and got q = 2.5*10^-4C. I then recalculated the voltage drops, and got these.. drop 1 = 2.5*10^-6V drop 2 = 5.0*10^-6V drop 3 = 7.5*10^-6V drop 4 = 1.0*10^-5V These still don't look correct (they are much smaller than anything I would be able to measure in the...

I apologize for bumping, but I still am not sure how to do this correctly.. any ideas?

I have that because when capacitors are connected in series, the voltage is V = q(1/C1 + 1/C2 + 1/C3 + 1/C4), according to my text. So at the first capacitor, V = q/C, then at the second V = q/C + q/C = 2q/C, and so on.. or am I understanding it incorrectly?

Homework Statement This is for a lab I'm doing tomorrow. We will have a 10V circuit with 4 100µF capacitors connected in series. I need to calculate the voltage drop across each capacitor. I also need to create another circuit, but I'll worry about that after I'm sure I'm doing this one...

Phew, got it. I had measured my distances in a weird way, but I fixed it now. :) The answer should have been 8cm, correct? Thanks so much! I have one more question about the etiquette on here. Is it bad form to post more than one question in a day? There's one other problem I'm banging my head...

Ok, I algebra'd it out and got x = -3L/4.. which should give me x = -6cm. I guess that means my original assumption of 0 occurring in the positive side was incorrect? The given answer is very confusing - why would it be in that form? I never actually came across it while finding x.

I'm assuming that the 0 point is somewhere in the positive x region (because the first line has a larger charge - please let me know if my thinking is off). With that assumption, line 1 would be L + x away from the point, and line 2 would be L/2 + x away? Is this on the right track...

Homework Statement Short sections of two very long parallel lines of charge are shown, fixed in place, separated by L = 8.0cm. The uniform linear charge densities are +6.0\muC/m for line 1 and -2.0\muC/m for line 2. Where along the x-axis shown is the net electric field from the two lines...