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Conservation of charge

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Flyback

Well-Known Member
Hello,
Our contractor tells us that if you have two 1F capacitors, each containing 1 Coulomb of charge, then if they are placed in series, he thinks that you then have two coulombs of charge in the series stack.
However, the charge of the series stack is still 1 coulomb.
He quotes conservation of charge at us...saying if you only had 1C of charge in the series stack then 1C of charge would have been destroyed, and that he says, cannot happen.
How can we tell him? We showed Q=CV etc but to no avail
 

JLNY

Active Member
You are correct. The charge remains at 1C, but the electrical potential (i.e. the voltage) of the series stack is doubled, hence conservation of energy is conserved. if we think about the "middle" plates, the charge on them is isolated from the circuit, so any current that flows will cause the charge to cancel out internally.

Would something like this really poor MS paint diagram help?
 

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ci139

Active Member
((((sh¡t&sh¡t)))) . . . so - Δq=ΔU·C , P=UI ,U=ΔA/Δq , I=Δq/Δt (Attempting the energy conservation showcase) ΔE=P·Δt = UIΔt = U(Δq/Δt)Δt = (Δq/ΔC)Δq=Δ²q/ΔC . . .

. . . 1F||1F=1/(1/1F+1/1F) =1/2·1F , incase ΔE=Const. the Δq=²√(̅C̅o̅n̅s̅t̅.̅·̅Δ̅C̅)̅'=±i√(̅C̅o̅n̅s̅t̅.̅·̅½̅)̅' :wideyed: -- but this would be valid for variable capacitor (with external force applied to pull appart the plates) not tiled/combined one . . .

computing . . .
must beeEcsv.png but i don't want to FW►► here

there's something but i've not verified it yet . . . http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capac.html#c3 . . . -- but this is not your case coz you tile pre-charged caps -- where it must be easy to show P=UI must preserve so Pdt=dtUI also e.g. E.stored=no change only the I against t graph would vary ... eEcsv-xls.png
 
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ronsimpson

Well-Known Member
Most Helpful Member
if you have two 1F capacitors, each containing 1 Coulomb of charge, then if they are placed in series, he thinks that you then have two coulombs of charge in the series stack.
Question: "you have two 1F capacitors, each containing 1 Coulomb of charge," How much charge do you have now? 2?
 

MikeMl

Well-Known Member
Most Helpful Member
The effective capacitance is halved. The voltage is doubled. Energy and charge are both preserved.
 

Ratchit

Well-Known Member
Hello,
Our contractor tells us that if you have two 1F capacitors, each containing 1 Coulomb of charge, then if they are placed in series, he thinks that you then have two coulombs of charge in the series stack.
However, the charge of the series stack is still 1 coulomb.
He quotes conservation of charge at us...saying if you only had 1C of charge in the series stack then 1C of charge would have been destroyed, and that he says, cannot happen.
How can we tell him? We showed Q=CV etc but to no avail
Both you and the contractor are wrong. A cap contains the same amount of of net charge when energized to 0 , 10, 100, or 1000 volts. That is because for every amount of charge stored on one plate, the same amount of charge is depleted from the opposite plate, resulting in a net total difference of zero. The amount of energy stored at the different voltages varies according to E = 1/2 C V^2? A cap is a energy storage device. It does not store net charge. Energy is charge times volts. The two caps referenced above will each have a total of 1 coulomb of charge separation. Putting them in series will give 2 volts at first and zero volts when their 1 coulomb separation drops to zero. Putting them in parallel will give 1 volts at first and zero volts when their 2 coulomb separation drops to zero. Either way, their energy storage is the same.

This is a good example why I do not advocate the word "charge" when you really mean "to energize" a battery, cap, or anything else.

Ratch
 
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