# Homework Help: Two capacitors in parallel rearranged

1. Oct 8, 2015

### haikuberryfin

• Thread moved from the technical math forums, so no Homework Help Template is shown
I just had a test, and I'm wondering how to do one of the free responses on it. The first couple parts, which I'll omit, were pretty simple, but I got kind of lost toward the end.

Two Capacitors (C1=10microF C2=40microF) are connected in parallel to a 10V battery. The capacitors are then disconnected from the battery and rewired so that their positive sides are connected to the other capacitor's negative side.

a) What is the voltage drop across each capacitor?
b) Does the potential energy increase, decrease, or remain the same. If it changes, what is the percentage increase or decrease of the system?
c) Why did the capacitor system gain or lose energy if it did? Where did that energy come from or go?

For part "a," I found original Qeq. Qo = (10microF + 40microF) * 10V = 500microC;
Q1 = Q2 = 250microC;
V1 = 250microC / 10microF = 25V; V2 = 250microC / 40 microF = 6.25V

Moving onto part "b," I thought conceptually that the system would have to lose energy since charges would have to be redistributed across the plates to even out.
I then used U = Q2 / (2 * C) = 500microC2 / (2 * 8microF) = .015625J
This doesn't make sense since that is a big increase from the .0025J of the original configuration.

I treated part "c" like I did conceptually for part "b" and went with it decreasing.

If anyone could help me with this, I'd greatly appreciate it.

2. Oct 8, 2015

### ehild

That is wrong. What is equal if two elements are connected in parallel?

3. Oct 8, 2015

### haikuberryfin

Connected in parallel, Voltage drops are equivalent. Aren't the capacitors in series though after the switch occurs?

4. Oct 8, 2015

### ehild

Questin a) asks the charges before
No. The charge can not leave the capacitors, it only can recombine. There will be net charge on the plates in contact. When you disconnect the capacitors they do not have equal charges. There is Q1 on one and Q2 on the other. Connecting + to -, the magnitude of the net charge on the connected plates is |Q1-Q2|. If the capacitors are connected in series, that net charge is zero.

5. Oct 9, 2015

### rude man

First, I assume the problem meant to read "The capacitors are then disconnected from the battery and rewired so that each cap's positive side is connected to the other's negative side." So they are still in parallel but one is flipped in polarity.
Assuming the above, how can there be two voltages? The caps are connected to each other directly.
You need to solve part (a) correctly before moving on to b and c.
Suggestion: rather than connecting the caps to each other directly, put a resistor between them and solve for v1(t) and v2(t), the two ends of the resistor. Then let the resistance → 0.