# TPR SWB Physics Passage 36 #6 (parallel capacitors)

#### C5b6789

##### Full Member
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

#### Attachments

• Screen Shot 2014-07-24 at 11.18.18 AM.png
36.9 KB · Views: 5
• Screen Shot 2014-07-24 at 11.18.36 AM.png
77.3 KB · Views: 6

#### C5b6789

##### Full Member
is anyone able to provide help? thank you

#### DrknoSDN

##### Full Member
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

##### Full Member
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.

This thread is more than 7 years old.

Your message may be considered spam for the following reasons:

4. It is very likely that it does not need any further discussion and thus bumping it serves no purpose.
5. Your message is mostly quotes or spoilers.