Connected Capacitors: Charge & Potential Diffrences

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When two capacitors, C1 and C2, are connected, the charge from the initially charged capacitor C1 (Q0) redistributes between the two capacitors. The charge on each capacitor becomes Q1 = Q0/2 and Q2 = Q0/2, leading to equal charge distribution. However, the potential difference across each capacitor is not zero, as this would violate Kirchhoff's loop rule (KVL). Instead, the voltage across each capacitor must satisfy KVL, ensuring the total voltage in the circuit remains consistent. The correct approach confirms that charge distribution must adhere to KVL principles for accurate results.
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Homework Statement


An isolated capacitor C1 carries a charge Q0. Its wires are then connected to those of a second capacitor C2, previously uncharged. What charge will each carry now? What will be the potential difference across each?

Homework Equations


C=Q/V

The Attempt at a Solution


I figured the charge would eventually become evenly distributed (no measure of time was given). So, I figured...

Q1=Q2=1/2 Q0

The potential difference (voltage) would be zero because the charges are equal. ?
 
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Are you sure?

Lets see what happens if we take your first answer as correct. You said (from a very obvious perspective) that charge gets equally distributed, but then if you apply Kirchhoff's loop rule (KVL) across the closed circuit, doesn't it get violated?? The total voltage difference across the loop turns out to be:

Q0/2C1-Q0/2C2

which is not equal to zero... which means that you answer isn't correct. Any any case you can't violate KVL!

Indeed you try to distribute the charge in such a way that KVL is satisfied, and since the sum of the charges present in the plates has to be Q0... you will get the correct answer!
 
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