Dielectric constant of the materials, k

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Homework Help Overview

The discussion revolves around calculating the dielectric constant of a material inserted between the plates of a capacitor. The problem involves two capacitors, C1 and C2, with given capacitances and voltages, and a change in charge on C2 when the dielectric is introduced to C1.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the implications of whether the voltage remains connected when the dielectric is added. There are attempts to set up equations based on charge redistribution and capacitance changes.

Discussion Status

Participants are actively exploring the relationships between charge, capacitance, and voltage in the context of the problem. Some have proposed equations to find the dielectric constant, while others are clarifying assumptions about the voltage state during the dielectric insertion.

Contextual Notes

There is uncertainty regarding whether the voltage across the capacitors is maintained or disconnected during the dielectric insertion, which affects the interpretation of charge redistribution.

darkeng
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Homework Statement



C1 = 12uF, C2=20uF, and the capacitors are charged so that the voltage across each capacitor is 6V. When a dielectric material is inserted between the plates of C1, the charge on C2 changes by 32 uC. Calculate the dielectric constant of the material, k. (PS. C1 is parallel to C2)

Homework Equations



C=q/v

The Attempt at a Solution



This is my method. 6V = (12x6+20x6+32)/(12k + 20), then solve for k, but it seems like it won't give me the right answer.
 
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darkeng said:

Homework Statement



C1 = 12uF, C2=20uF, and the capacitors are charged so that the voltage across each capacitor is 6V. When a dielectric material is inserted between the plates of C1, the charge on C2 changes by 32 uC. Calculate the dielectric constant of the material, k. (PS. C1 is parallel to C2)

Homework Equations



C=q/v

The Attempt at a Solution



This is my method. 6V = (12x6+20x6+32)/(12k + 20), then solve for k, but it seems like it won't give me the right answer.

Welcome to PF.

Is the voltage disconnected before the dielectric is added?
 
Hi LowlyPion,

The question didn't mention that but I think it's not disconnected.

Thanks
 
darkeng said:
Hi LowlyPion,

The question didn't mention that but I think it's not disconnected.

Thanks

Well if the Voltage is still the same across C2, and the dielectric goes only in C1, C2 is still the same ...

Q2 = C2*V
 
Hi LowlyPion,

oops, my bad. The voltage is disconnected...
 
Before the dielectric was inserted then you had a total charge on the 2 capacitors. It doesn't change that total, but it does redistribute.

(12 μf + 20 μf ) * 6 V = 192 μ C

Now you have a new capacitor that is k*12 μf and you have redistributed the charge between 20 μf and k*12 μf such that there is a transfer of 32 of the 192 μC from C1 to C2.

Before
Q1 = 72 μC
Q2 = 120 μC

after
Q1 = ?
Q2 = ?

When you determine the new C1 (+ dielectric) New/Old = k right?
 
so Q1 becomes 72+32 and Q2 becomes 120-32?

Then how do I find the new voltage?
 
darkeng said:
so Q1 becomes 72+32 and Q2 becomes 120-32?

Then how do I find the new voltage?

They didn't ask that. They want k.
 
Whatever the voltage

Q1/C1 = Q2/C2
 
  • #10
ohhhhhhhhhhhhhhhhhhhhhhhhhhhhhh, i got it now! so Q1 after = 72+32 = 104 uC and Q2 after = 120-32 = 88 uC


104/12*k = 88/20, then k=1.97?


Thanks!
 
  • #11
darkeng said:
ohhhhhhhhhhhhhhhhhhhhhhhhhhhhhh, i got it now! so Q1 after = 72+32 = 104 uC and Q2 after = 120-32 = 88 uC

104/12*k = 88/20, then k=1.97?

Thanks!

Eureka always sounds sweet.

Good luck.
 
  • #12
haha, indeed! Thank you!
 

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