# Why is the voltage same here ?

1. Mar 12, 2014

### oneplusone

1. The problem statement, all variables and given/known data

C_1 = 5uF
c_2 = 10uF
C_3 = 2uF

Potential difference between a and b is 60 V.
What is the charged stored on C_3

2. Relevant equations

C=Q/V

3. The attempt at a solution

See attached.

My question is, why is the charge distribution from A to C the same as the distribution from C to B?
Sorry if the solution is unclear; I find it unclear as well.

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• ###### Screen Shot 2014-03-12 at 4.21.43 PM.png
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2. Mar 12, 2014

### Chronum

You can start by calculating equivalent capacitance between both a-c and c-b.

3. Mar 12, 2014

### oneplusone

Say the capacitance between a-c is X and between c-b is Y.

Does this mean the voltage is still the same at a-c and c-b even if $X\ne Y$??

Im very confused (sorry!)

4. Mar 12, 2014

### Chronum

No. The two equivalent capacitances are in series. They do not share common terminals, so it's not necessary that they have the same voltage.

5. Mar 12, 2014

### oneplusone

Okay, I'm still not getting it…

I know how to find the total charge by using C = Q/V.
I also know that we can make two effective capacitors (from a-c and b-c). So now it reduces to a known voltage from a-b and two equivalent capacitors, and the total charge. How do i proceed from here to find the charge on each of the two capacitors?

6. Mar 12, 2014

### Staff: Mentor

Capacitors in series receive equal Q (assuming there was no initial charge on any). Being in series, they see the same current, and for the same time/s.

7. Mar 12, 2014

### ehild

Those equivalent capacitors are in series. The charge is the same on the series capacitors, like the current is the same in series resistors.

Imagine you join a pair of terminals of two capacitors and connect the free terminals to a battery.

Electrons will migrate from one plate to the positive terminal to the battery. That plate becomes positively charged, gets q charge. That positive charge attracts the negative charges on the opposite plate, its inner surface becomes negatively charged, by -q. As the plate itself is neutral, the outer surface becomes positive, but it neutralizes with the negative charges arriving from the plate of the other capacitor. That plate becomes positively charged with q. At the end, the other plate of the second capacitor becomes negative, with -q. Both capacitors have the same charge q, and their voltage are U1=q/c1 and U2=q/C2 and U1+U2=U(battery)

ehild

#### Attached Files:

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8. Mar 12, 2014

### oneplusone

So is this correct thinking:

You combine the capacitors into two capacitors in series, and since the charges on both of those are equal, you use C=Q/V so C_1V_1 = C_2V_2. Since C_1 and C_2 and V_1+v_2 is known, you can solve for it.

However, how do you solve if they are parallel? After breaking down the circuits, what do you do? (the top one specifically--how do you solve for it if it's parallel)

9. Mar 12, 2014

### Simon Bridge

You may have series and parallel mixed up.
Series is one-after-the-other, parallel is side-by-side. The diagram attached to post #7 (go look) is for capacitors in series.

When you want to think about combining capacitors, you want to go right back to basics with the physical parallel plate setup.
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/pplate.html

Take two physical caps, same capacitance, so they have the same area plates A and the same separation between the plated d. You understand how the capacitance depends on A and d?

If in parallel, this is the same as a single capacitor with twice the area but same separation.
If in series, this is the same as a single capacitor with the same area but twice the separation.

Repeat for the case that A and d are different so that the capacitances are different.

10. Mar 13, 2014

### ehild

The parallel capacitors have the same voltage across each. So U=q1/C1=q2/C2.

whild