Register to reply

Charge on Capacitor plate

by Apteronotus
Tags: capacitor, charge, plate
Share this thread:
Apteronotus
#1
Apr9-10, 05:22 PM
P: 203
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.
Attached Thumbnails
Series CapacitorColor.JPG  
Phys.Org News Partner Engineering news on Phys.org
Future phones to use blood and speech to monitor HIV, stress, nutrition
Neuron circuit may enable pitch perception applications
Quasi-distributed temperature sensors from draw-tower fabrication technology
berkeman
#2
Apr9-10, 05:37 PM
Mentor
berkeman's Avatar
P: 40,926
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
#3
Apr9-10, 05:39 PM
Mentor
berkeman's Avatar
P: 40,926
Oh, and remember that [tex]I(t) = \frac{dQ(t)}{dt}[/tex]


Register to reply

Related Discussions
Surface charge density on the outer surface of a plate in a parallel plate capacitor General Physics 9
Parallel plate capacitor (voltage and charge relative to distance) General Physics 2
Charge on Parallel Plate Capacitor Introductory Physics Homework 4
A proton is released from rest at the positive plate of a parallel-plate capacitor. Introductory Physics Homework 16
Determine the charge density on either plate (capacitor) Introductory Physics Homework 4