Dielectric constant of the materials, k (1 Viewer)

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1. The problem statement, all variables and given/known data

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)

2. Relevant equations

C=q/v

3. 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.
 

LowlyPion

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1. The problem statement, all variables and given/known data

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)

2. Relevant equations

C=q/v

3. 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
 

LowlyPion

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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...
 

LowlyPion

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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?
 

LowlyPion

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LowlyPion

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Whatever the voltage

Q1/C1 = Q2/C2
 
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!!!!
 

LowlyPion

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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.
 
haha, indeed! Thank you!
 

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