Capacitor charge in series and parallel circuit

[Joshb60796]

Homework Statement

Three capacitors, with capacitances C1 = 4.0 μF, C2 = 3.0 μF, and C3 = 2.0 μF, are
connected to a 12 -V voltage source, as shown in the figure. What is the charge on
capacitor C2 ?

Homework Equations

Q=VC
Q = charge
V = voltage
C = capacitance
Ceq = equivalent capacitance

The Attempt at a Solution

C1 and C2 are in series and therefore should be summed as in Ceq=1/((1/C1)+(1/C2)) because the voltage across them will be shared proportionally. If I multiply Ceq by the voltage I get a charge of Q12 which is for both capacitors. I need to find the charge on C2 so I am multiplying the charge on both, Q12, by the proportion of C2 to C1 which is 3/4. I am getting 15.4 microCoulombs but my answer should come to 16 microCoulombs. I need help on this questions. Am I doing this correctly? Is my thought process correct?

 

[tiny-tim]

(try using the X₂ button just above the Reply bos

Hi Joshb60796!

(try using the X₂ button just above the Reply bos )

C1 and C2 are in series …

Nooo …

C₂ and C₃ are in parallel, and then C₁ is in series with their resultant.

Try again. ​

 

[Joshb60796]

Hi Joshb60796!

(try using the X₂ button just above the Reply bos )


Nooo …

C₂ and C₃ are in parallel, and then C₁ is in series with their resultant.

Try again. ​

I was aware that C₂ and C₃ were parallel which is why I didn't include C₃. Aren't C₂ and C₃ seeing the same voltage so if the voltage is connected for a long time C₃ doesn't factor into the charge that is held on C₂? I thought it was just Capacitance and Voltage that has to do with Charge, Q.

 

[tiny-tim]

Aren't C₂ and C₃ seeing the same voltage …

yes

so if the voltage is connected for a long time C₃ doesn't factor into the charge that is held on C₂?

yes it does, C₂ has to share its charge with C₃ (and btw, not equally)

I thought it was just Capacitance and Voltage that has to do with Charge, Q.

yes, but you have to use the total capacitance of C₂ and C₃ combined to find how that combines with C₁

find the capacitance of C₂ and C₃,

then find the total capacitance …

show us what you get ​

 

[Nickie]

Your thought process is correct, but there may be a small error in your calculation. When finding the equivalent capacitance for capacitors in series, the formula is Ceq = 1/((1/C1)+(1/C2)). However, in your calculation for the charge on C2, you are using the formula Ceq=1/((1/C1)+(1/C2)) instead of Ceq = (C1 * C2)/(C1 + C2). Using the correct formula, the charge on C2 should be 16 microCoulombs.