Trouble w/ setting up the equations for capacitive circuits

[Blockade]

Homework Statement

Homework Equations

.Q = CV_ab

The Attempt at a Solution

I am having a hard time understanding when to put the capacitance in the denominator. I get that Voltage = Charge/Capacitance, but for the equation circled in red "q₁/c + q₂/c = q₃c" ... I don't get why each of those charges have their corresponding value of capacitance under it. If you look at two equations up "q₂+q₃ = q₄" does need the capacitance in it.

 

[gneill]

I am having a hard time understanding when to put the capacitance in the denominator. I get that Voltage = Charge/Capacitance, but for the equation circled in red "q1/c + q2/c = q3c" ... I don't get why each of those charges have their corresponding value of capacitance under it. If you look at two equations up "q2+q3 = q4" does need the capacitance in it.

If you're summing or comparing voltages (potential differences) then you want to divide the charge by its associated capacitance to obtain the PD across the given capacitor.. V = Q/C. If you're just summing and comparing charges, then you just sum or compare them directly.

 

[J Hann]

Wouldn't it be simpler to reduce the circuit in a manner similar to reducing resistive circuits?
For capacitors in parallel:
C = C1 + C2 + ... + Cn
For capacitors in series:
1 / C = 1 / C1 + 1 / C2 + ... + 1 / Cn
After you have the appropriate equivalent capacitances then you can begin analyzing charges and voltage drops.

 

[gneill]

Wouldn't it be simpler to reduce the circuit in a manner similar to reducing resistive circuits?
For capacitors in parallel:
C = C1 + C2 + ... + Cn
For capacitors in series:
1 / C = 1 / C1 + 1 / C2 + ... + 1 / Cn
After you have the appropriate equivalent capacitances then you can begin analyzing charges and voltage drops.

Sure. But then you "lose" the individual capacitors with that reduction. Since you eventually want to find their individual charges and voltages that could be problematical.

On the other hand it can be used as a method towards the solution. If you first reduce down to a single equivalent capacitance then you can find the charge on it: $Q = V_{ab}*C_{eq}$. That will also be the charge on the lone series capacitor at the bottom of the original circuit, so you can determine its voltage and by extension, the voltage across the other group of capacitors. It's a short bit of work in a similar fashion to deal out the voltages and charges to the rest.