1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Capacitors in a series circuit W/ one dielectric

  1. Oct 24, 2015 #1
    Problem: When a dielectric slab is inserted between the plates of one of the two identical capacitors in Fig. 25-23, do the following properties of that capacitor increase, decrease, or remain the same: (a) capacitance, (b) charge, (c) potential difference (d) How about the same properties of the other capacitor?


    I said that the potential of the first capacitor decreases and that the charge it stores also increases. For the 2nd capacitor, I said it's capacitance would decrease. I'm not so sure though, I think it may stay the same as well?
    Potential would increase and charge would increase for the 2nd capacitor as well
    . iWruI.png

    Attempt at Solution: The main problem I'm having at solving this is the fact that both charge and voltage for the individual capacitors are variable. Please explain the situation and why the values for potential, capacitance and charge either decrease, increase or stay the same.

    What I think: When a dielectric is added, E between the capacitor decreases by a factor of k so voltage must decrease for the first capacitor and thus the voltage for the 2nd capacitor must increase by the same amount to fulfill Kirchhoff's laws.

    Adding a dielectric also allows for a capacitor to store more charge at the same potential so the first capacitor must store more charge since c = q/v <- direct relationship.

    I'm confused on what happens to the 2nd capacitor. They're in series so there's that inverse relationship and total capacitance decreases.
  2. jcsd
  3. Oct 25, 2015 #2
    Try this: start with the "truths" you know. Eg: 1) kirchoff's rule is always valid, 2) the capacitors are in series so the effective capacitance can be found, 3) because they are in series the charge on both of them at any instant has to be the same thought the value of this charge could vary, 4) again because they are in series we can find a relation between their V1 and v2. These relations should help you come to the conclusions.

    Here is what I got :
    Capacitor 1 (one with dielectric): Q decreases, C increases, V decreases
    Capacitor 2: Q decreases, V increases, C does not change.
    I'm fairly sure I'm correct and they seem to satisfy all the equations too qualitatively.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted