Can You Calculate the Charges on Capacitor Plates in a Series Circuit?

[Apteronotus]

Hi I have a question regarding the charge Q that would build up on capacitors in series when there is a potential difference between the plates.

Consider the attached circuit.
1. Knowing the potentials $$\phi_A, \phi_B$$ and the capacitances $$C_A, C_B$$, but not $$\phi_C$$ is there a way of calculating the charges on the plates?

2. Is $$Q_A=Q_{Ca}$$ and $$Q_B=Q_{Cb}$$?

3. Is $$Q_{Ca}=Q_{Cb}$$? If not, what would prevent charge flowing from one plate to the other?

4. The series capacitors are equivalent to a single capacitor having capacitance
$$C_T=(C_AC_B)/(C_A+C_B)$$ and hence charge $$Q_T=C_TV$$. Is $$Q_A=Q_B=Q_T$$?

Sorry this is so long winded, and thanks in advance.

 

[berkeman]

Hi I have a question regarding the charge Q that would build up on capacitors in series when there is a potential difference between the plates.

Consider the attached circuit.
1. Knowing the potentials $$\phi_A, \phi_B$$ and the capacitances $$C_A, C_B$$, but not $$\phi_C$$ is there a way of calculating the charges on the plates?

2. Is $$Q_A=Q_{Ca}$$ and $$Q_B=Q_{Cb}$$?

3. Is $$Q_{Ca}=Q_{Cb}$$? If not, what would prevent charge flowing from one plate to the other?

4. The series capacitors are equivalent to a single capacitor having capacitance
$$C_T=(C_AC_B)/(C_A+C_B)$$ and hence charge $$Q_T=C_TV$$. Is $$Q_A=Q_B=Q_T$$?

Sorry this is so long winded, and thanks in advance.

Capacitors are DC open circuits, so the voltage in the middle is indeterminate, in a sense.

But the way this problem is usually stated, you turn on the voltage source and ramp it up to its final voltage V. During this ramping process, a current flows through the capacitors according to the traditional equation:

$$I(t) = C \frac{dV(t)}{dt}$$

That current results in charge displacement, which gives the final voltage values on the caps. You are correct that the series combination of the two caps results in a lower overall series equivalent capacitance. If the caps are unequal in capacitance value, the larger one will end up with a lower voltage across it.

Now with that, are you able to answer your questions?

 

[berkeman]

Oh, and remember that $$I(t) = \frac{dQ(t)}{dt}$$