Finding the charge on a capacitator

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In summary, three capacitors (C1=8μF, C2=4μF, C3=8μF) are connected in a column with a potential difference of 12V. To find the charge of C2, we use the equation ΣC=(Ca*Cb)/(Ca+Cb) and calculate the total capacitance of the system to be 2μF. Using the equation C=Q/V, we find that Qtotal=24μC. However, since the capacitors are in series, the charge on each capacitor is not equal to 1/3 of the total charge. Instead, we use the equation V=V1+V2+V3=Q
  • #1
Dell
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3 capacitators, C1=8μF C2=4μF C3=8μF, are connected in a colum, over a potential difference of 12V, what is the charge of C2??

i know that connecting the capacitators in a colum, ΣC=(Ca*Cb)/(Ca+Cb)

also i know that
C=Q/V, which i think must be the equation i need since i am looking for Q, but i don't think the V i know is the right V, i think its the V for the whole system.

ΣC1,2=(C1*C2)/(C1+C2)=(8*4)/(8+4)=(8/3)μF

ΣC1,2,3=(C3*C1,2)/(C3+C1,2)=(8*(8/3))/(8+(8/3))=2μF

so now i know Vtotal=12V and Ctotal=2μF

C=Q/V
Q=C*V
Qtotal=2*12=24μC

but how do i find Q2 from the Qtotal
 
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  • #2
The V is indeed the potential difference over the equivalent capacitor.

Do you know why capacitors in series add up like 1/C=1/C1+1/C2+...+1/Cn? Imagine at the very start the first plate of the capacitor is getting charged. What will this do to the plate below and the plate below and the plate below etc?
 
  • #3
a sort of chain reaction, 1st will charge 2nd will charge 3rd... so they will all have the same charge?? meaning that
Q2=(1/3)*Qtotal

Q2=Q3=Q1=8μC

is that right?
 
  • #4
You're almost right. The chain reaction you describe is correct and indeed all three capacitors have the same charge. However this charge is not 1/3 of Qtotal.

Lets call the potential difference over the first capacitor V1, over the second V2 and over the third V3. Using your chain reaction argument we know that Q=Q1=Q2=Q3. We also know that in series V=V1+V2+V3 and we know C=Q/V.

Now let's calculate them separately:
V1=Q/C1
V2=Q/C2
V3=Q/C3

V=V1+V2+V3=Q(1/C1+1/C2+1/C3). Looks familiar? Now calculate this Q and see that Q1=Q2=Q3=Q=Qtotal.
 
  • #5
so it would probably be easier to just work it out like that from the beginning instead of adding up all the C's,,
thanks
 

1. How is the charge on a capacitator calculated?

The charge on a capacitator is calculated by multiplying the capacitance of the capacitator (in Farads) by the potential difference (in volts) across the capacitator.

2. What is the unit of measurement for capacitance?

The unit of measurement for capacitance is Farads (F).

3. Can the charge on a capacitator be negative?

Yes, the charge on a capacitator can be negative if the potential difference across the capacitator is negative. This means that the capacitator has accumulated a net negative charge.

4. How does the charge on a capacitator affect its voltage?

The charge on a capacitator is directly proportional to its voltage. This means that as the charge increases, the voltage across the capacitator also increases.

5. Is the charge on a capacitator constant?

No, the charge on a capacitator is not constant. It can change depending on the potential difference across the capacitator and the capacitance of the capacitator. However, the charge on a capacitator remains constant as long as the capacitator is not connected to a circuit.

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