Charge on Capacitor plate


by Apteronotus
Tags: capacitor, charge, plate
Apteronotus
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#1
Apr9-10, 05:22 PM
P: 196
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 [tex]\phi_A, \phi_B[/tex] and the capacitances [tex]C_A, C_B[/tex], but not [tex]\phi_C[/tex] is there a way of calculating the charges on the plates?

2. Is [tex]Q_A=Q_{Ca}[/tex] and [tex]Q_B=Q_{Cb}[/tex]?

3. Is [tex]Q_{Ca}=Q_{Cb}[/tex]? 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
[tex]C_T=(C_AC_B)/(C_A+C_B)[/tex] and hence charge [tex]Q_T=C_TV[/tex]. Is [tex]Q_A=Q_B=Q_T[/tex]?

Sorry this is so long winded, and thanks in advance.
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berkeman
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Apr9-10, 05:37 PM
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Quote Quote by Apteronotus View Post
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 [tex]\phi_A, \phi_B[/tex] and the capacitances [tex]C_A, C_B[/tex], but not [tex]\phi_C[/tex] is there a way of calculating the charges on the plates?

2. Is [tex]Q_A=Q_{Ca}[/tex] and [tex]Q_B=Q_{Cb}[/tex]?

3. Is [tex]Q_{Ca}=Q_{Cb}[/tex]? 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
[tex]C_T=(C_AC_B)/(C_A+C_B)[/tex] and hence charge [tex]Q_T=C_TV[/tex]. Is [tex]Q_A=Q_B=Q_T[/tex]?

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:

[tex]I(t) = C \frac{dV(t)}{dt}[/tex]

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
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Apr9-10, 05:39 PM
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Oh, and remember that [tex]I(t) = \frac{dQ(t)}{dt}[/tex]


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