# Why is the voltage the same?

1. Mar 12, 2014

### oneplusone

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

C1 = 6.00 uF
C2 = 3.00 uF
$\Delta V = 20.0$ V

Capacitor c1 is first charged by the closing of s_1. Switch S_1 is then opened and the carved capacitor is connected to the uncharged capacitor b the closing of S_2. Calculate the initial charge acquired by C_1 and the final charge on each capacitor.

2. Relevant equations

C=Q/V

3. The attempt at a solution

See attached.

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I get everything in this solution besides when they set Q_1/C_1 = Q_2/C_2.
Is it ALWAYS true that the voltage across two points are the same?

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2. Mar 12, 2014

### Chronum

If you connected 2 capacitors, remember that their plates are connected by a conductor.

Now what happens when you connect two points with a conductor?

3. Mar 12, 2014

### oneplusone

Current is formed, so there is a voltage?
And since there is no loss of electrons (charge) and the intensity (current) is the same, the voltage is the same?
Is that correct?

4. Mar 12, 2014

### Chronum

Well, as long as there is current flowing, you can't really make a problem out of it, at least not here, because it's not in a steady state.

But yes, when current does stop flowing in the circuit, the voltages are the same on both capacitors.

Now, there are formulae for the final charge/voltage in such an arrangement. I'd suggest you go through your textbook or look through the forums. I've forgotten the formulae myself, but I do remember that they exist.

5. Mar 12, 2014

### Staff: Mentor

You could include the resistance of the wire as a resistor in your crcuit, making it 3 elements. Current will flow through the resistor until the voltage across the resistor falls to zero. At the start, there are different voltages on each side.

6. Mar 13, 2014

### rude man

When S2 is closed the two capacitors are in parallel so by definition their voltage drops must be the same.