TPR SWB Physics Passage 36 #6 (parallel capacitors)

[C5b6789]

I attached the question and answer explanation below.

I think there's an error in the answer explanation so it should be Q1' + Q2' = Q1

Conceptually, why is the sum of the new charge magnitudes for both capacitors C1 and C2 equal to the original charge magnitude Q1 for just C1?

I understand their "quicker" way of thinking about this question, but I'm not understanding their first method of solving it mathematically through Q1' + Q2' = Q1

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[C5b6789]

is anyone able to provide help? thank you

 

[DrknoSDN]

I attached the question and answer explanation below.

I think there's an error in the answer explanation so it should be Q1' + Q2' = Q1

Conceptually, why is the sum of the new charge magnitudes for both capacitors C1 and C2 equal to the original charge magnitude Q1 for just C1?

I understand their "quicker" way of thinking about this question, but I'm not understanding their first method of solving it mathematically through Q1' + Q2' = Q1

1) Yes, it appears to be a typographical error and should read Q1' + Q2' = Q1.

2) Conceptually you establish a charge on C1 with the battery connected. Once you disconnect the battery, whatever charge is on the plates is a set value. When you close the switch you are distributing that charge (lets call it 1 charge), between both capacitors. So the original charge (Q1), is distributed out and ends up being equal to the new charge on capacitor 1 (Q1') and capacitor 2 (Q2')..
Net eqn is: Q1 => Q1' + Q2'

3) The conceptual solution is the same as the mathematical one but put into words. You have some charge to start, and when you close the switch the total capacitance is tripled so without any battery to add more charge the voltage is cut to 1/3.

Good catch on that typo.

 

[C5b6789]

1) Yes, it appears to be a typographical error and should read Q1' + Q2' = Q1.

2) Conceptually you establish a charge on C1 with the battery connected. Once you disconnect the battery, whatever charge is on the plates is a set value. When you close the switch you are distributing that charge (lets call it 1 charge), between both capacitors. So the original charge (Q1), is distributed out and ends up being equal to the new charge on capacitor 1 (Q1') and capacitor 2 (Q2')..
Net eqn is: Q1 => Q1' + Q2'

3) The conceptual solution is the same as the mathematical one but put into words. You have some charge to start, and when you close the switch the total capacitance is tripled so without any battery to add more charge the voltage is cut to 1/3.

Good catch on that typo.

Great thank you for the explanation, I need to work on a stronger understanding of how capacitors relate to charge.