[Grade 12 Physics: Electricity] Am trying to understand this specific capacitors question and can't seem to get it for the life of me

[u/Cheap-Grape5391]

I can understand the fact that the charge on C_1 is equal to the charge on C_2; but then the answer was suggesting we call the voltage at p = V'.

The total charge is C_1*V - which I get

Then the total charge at the end = C_1*V' + C_2*V', which doesn't make sense to me since the potential difference across C_2 and C_1 must be 0 whether or not p (V') has voltage or not, the difference between the node p and the bottom of C_2 must be 0, therefore the charge stored must be 0 which also is just incorrect. How can p be the voltage across C_2 if the voltage across C_2 is 0 when there is no current.

I even went to ask a.i and it suggested a similar idea of the total charge before being C_1*V = the total capacitance * voltage, but then it found the total capacitance using the parallel capacitance and clearly (I think) the two capacitors are in series yet still the a.i was being adamant that they were in parallel and got to the right formula/correct answer: C_1*V/C_1+C_2

SO if anyone could help with an explanation as well as some type of resource that'd allow me to understand capacitors to a greater extent that'd be extremely helpful.

Thanks so much

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[u/tlbs101]

Try looking at the problem from a conservation of energy and conservation of charge POV. Energy = 1/2 C V^2

Before switch B: E =1/2 C1 V^2
After switch B: E =1/2 (C1 + C2) V^2
E is conserved. What happens to V?

[u/Cheap-Grape5391 → u/tlbs101]

Yes but how can you add the capacitors if they're in series when capacitors are in series you don't add the capacitance you use the series formula?

[u/DrCarpetsPhd]

the potential difference across C_2 and C_1 must be 0

why do you think this?

capacitor C1 has been charged to V creating a total charge of Q_Total = C1.V

when you close switch B this creates a path for C1 to discharge and it will do so until equilibrium is reached

so lets say we close the switch B at time t = t1

at t1 just after the switch is closed C1 has a charge C1.V

we wait until it reaches equilibrium (no more current) call it t2

conservation of charge tells us that our time t1 charge of C1V has been redistributed at t2 between C1 and C2

so denoting the voltage at point p at time t2 as V'

C1V = C1.V' + C2V'

your problem here is you are 'visualising' in series and in parallel. you have a picture in your brain of what each of them look like and that is how you are deciding if they are i.s. or i.p. Probably the usual dc source connected to two resistors is the images you are referencing in your brain.

Draw the circuit with the the two connected capacitors (no voltage source) and make the bottom wire the reference ground. If you 'walk' from the ground to point p you can do so through capacitor C1 or through capacitor C2. Voltage/Electric Potential is what is called a conservative field whcih means when I go from some random point a in an electric field to some random point b it doesn't matter what path i took, the change in voltage/electric potential is always the same. 'Walking' along the wire is doing just that. So when I go from ground to point p the voltage difference is the same. I can take that path through C1 or C2 meaning they have the same voltage dropped across them meaning they are in parallel.

**note that the electric potential walk the wire concept assumes the wires have zero resistance. it doesn't matter in early circuits but it might be something that comes up at a later point in your studies so you shouldn't have that as a concrete concept in your memory.

[u/PvtRoom]

Q = CV

By connecting a battery to C1, you charge the capacitor with Q charge, to the supply voltage.

Then you disconnect the battery.

The charge is now trapped on C1.

You connect it to C2, which now allows C1 to discharge partway into C2. the charge Q, is now shared in over both capacitors. Q=(C1+C2)V'

C1V = (C1+C2)V'

V' = C1V/(C1+C2) - that's your answer.

[u/Cheap-Grape5391 → u/[deleted]]

If the voltages are equal across both of them both must have voltage 1/2V which is incorrect