Capacitors in Series: Finding Charge on Each After Reconnection

[funoras]

Homework Statement

Three capacitors, of capacitances C , 2C and 3C respectively are connected in this order in series to a voltage U battery. After they are charged, the middle capacitor (2C) is disconnected and then connected back again with reversed polarity. Find the charges on each of the capacitors after this process.

Homework Equations

The Attempt at a Solution

The equivalent Capacitance
$1/C_a = 1/C + 1/2C + 1/3C$
$C_a = 6C/11$
and the charge on every capacitor is
$q_a=UC_a=6CU/11$

now I'm stuck, i don't know what happens after the disconnection.

 

[gneill]

Homework Statement

Three capacitors, of capacitances C , 2C and 3C respectively are connected in this order in series to a voltage U battery. After they are charged, the middle capacitor (2C) is disconnected and then connected back again with reversed polarity. Find the charges on each of the capacitors after this process.

Homework Equations

The Attempt at a Solution

The equivalent Capacitance
$1/C_a = 1/C + 1/2C + 1/3C$
$C_a = 6C/11$
and the charge on every capacitor is
$q_a=UC_a=6CU/11$

now I'm stuck, i don't know what happens after the disconnection.

The charge q_a that you've found looks good.

Here's an approach that works when you have capacitors that hold an initial charge at the start. From the charge and capacitance you can calculate the potential (voltage) across the capacitor. So for this circuit you can calculate all three potentials. An equivalent circuit model for a charged capacitor is an uncharged capacitor of the same value in series with a fixed voltage supply equal to the initial voltage of the charged capacitor.

Replace all three initially charged capacitors with their equivalent models. To effect the reversal of the 2C capacitor, simply reverse the polarity of its model's voltage supply.

Now, in a series circuit the order of the components does not effect the magnitude or direction of the current (the sum of the terms of the KVL loop equation does not depend upon their order). This means that you can collect all the voltage supplies in the loop together into one supply. You are left with a circuit with a single voltage supply and three uncharged capacitors, and you already know how to find the new charge that will end up on them.

The net charge on each capacitor will be its original charge PLUS this new charge. You'll have to think a bit about how to "add" the new charge to the capacitor that was reversed. Contemplation of the direction in which the current flowed from that capacitor should make it clear.

 

[funoras]

Thank you for the wonderful idea.

So i have calculated each voltage drop
$U_1=q_a/C=6U/11$
$U_2=q_a/2C=3U/11$
$U_3=q_a/3C=2U/11$

now after the reversion, we should have

$U-U_3+U_2-U_1=q_b C_a$
and from here $q_b=36CU/121$
q_b is the additional charge you were talking about. Now for each of the capacitor

$q_1=q_3=q_a+q_b=102CU/121$
and for the reversed one in my opinion
$q_2=-\left|q_a-q_b\right|=-30CU/121$