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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?

CAPACITOR 1 = CAPACITOR WITH DIELECTRIC CAPACITOR 2 = CAPACITOR WITHOUT DIELECTRIC(ABOVE CAPACITOR 1 IN THE DIAGRAM)

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.

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.

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  • $\begingroup$ You might make answerers' lives easier by clearly labeling the two capacitors (e.g. which one is the "first"?) $\endgroup$ – Daniel Griscom Oct 25 '15 at 1:52
  • $\begingroup$ fixed. C1 = w/ dielectric $\endgroup$ – Nemo_Sol Oct 25 '15 at 1:56
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You have to deal with this in stages:

  1. Adding a dielectric to $C1$ will increase its capacitance. With the same charge present, that means its voltage $V_{C1}$ will go down.
  2. Since $V_{C1}$ goes down, the total of $V_{C1}$ and $V_{C2}$ will now be less than the battery voltage $V_B$. This will start current flowing through the circuit to compensate.
  3. The current will add equal amounts of charge to both capacitors, and increase their voltages, until $V_{C1}+V_{C2}=V_B$.
  4. Once stability returns, the following changes will have been made:

    • The capacitance of $C1$ will have increased, but that of $C2$ will stay the same
    • The charge on both capacitors will have increased, and by the same amount
    • The voltage on $C1$ will have decreased, and that of $C2$ will have increased
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  • $\begingroup$ is VB in '2' the total potential from the battery? Also why would equal amounts of charge be added to both capacitors? $\endgroup$ – Nemo_Sol Oct 25 '15 at 2:11
  • $\begingroup$ Yes, $V_B$ is the battery voltage. And, the same current flows through both capacitors, so the same amount of charge is deposited on each. $\endgroup$ – Daniel Griscom Oct 25 '15 at 2:20
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a) Capacitance C increases b) Charge Q remains unchanged c) if charge Q is constant while C increases, that means voltage V decreases (C=Q/V). d) U decreases ( U=Q^2/2C)

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