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Determine the charge on each and voltage across each.

  • Thread starter negat1ve
  • Start date
1. Homework Statement

Two identical capacitors are connected in parallel. Each aquires a charge Q0 when connected to source voltage V0. The voltage source is the dissconnected and a dielectric(K=3.2) is inserted to one of the capacitors. Determine the charge on each and voltage across each.

2. Homework Equations
Given from problem:
C10 = C20
Q10 = Q20
V10 = V20 = V0

3. The Attempt at a Solution

I have an example in my txtbook showing a capacitor is that charged by a battery. The battery is then disconnected and a dielectric inserted. The result is that the charge is held constant(because there is nowhere for it to go) and that the voltage drops by a factor of K.
V = V0/K
This is throwing me off because do i want to say the same thing about the voltage of the capacitor in my problem. They are in parallel so that would mean that both capacitors would have their voltages dropped by this factor of K? If this is true i have the problem solved but something doesn't feel right about it. Anyone know how to handle this problem already? Thanks for your time as always!
 

LowlyPion

Homework Helper
3,079
4
Re: dielectrics

Consider then what the new equivalent capacitance is and proceed accordingly.
 
Re: dielectrics

I said that the dielectric is added to C2. So the new capacitance for C2 is
C2 = KC20 and C1 is still
C1 = C10

Using the voltage drop result from my textbook, if i apply this i get that
V2 = V1 = V0/K

I also know that the equivalent charge Qeq of the initial setup with the battery connected will be the same Qeq after beacuse the the charge is conserved.
Qeq0 = Q10 + Q20 = Q1 + Q2 = Qeq

Is this information correct?
 

LowlyPion

Homework Helper
3,079
4
Re: dielectrics

I said that the dielectric is added to C2. So the new capacitance for C2 is
C2 = KC20 and C1 is still
C1 = C10

Using the voltage drop result from my textbook, if i apply this i get that
V2 = V1 = V0/K

I also know that the equivalent charge Qeq of the initial setup with the battery connected will be the same Qeq after beacuse the the charge is conserved.
Qeq0 = Q10 + Q20 = Q1 + Q2 = Qeq

Is this information correct?
Not quite. Certainly the total charge doesn't change. But what about the Voltage?

For instance what is your new equivalent capacitance? And what has that done to the voltage over both capacitors, and hence the charge distribution on each?
 
Re: dielectrics

Not quite. Certainly the total charge doesn't change. But what about the Voltage?

For instance what is your new equivalent capacitance? And what has that done to the voltage over both capacitors, and hence the charge distribution on each?
My new equivalent capacitance is
Ceq = C1 + C2
= C10 + K*C20

So the total capacitance has increased.

And Ceq = Qeq / Veq
where Qeq = Q1 + Q2 stays the same
so Veq = V1 = V2 has to decrease to balance out the equation.

I wanna say that Q1 = CV = C10*V0/k = Q10/k
So for Qeq to stay the same
Q2 would have to equal
Q2 = Q20(3/2)K
yes????
 

LowlyPion

Homework Helper
3,079
4
Re: dielectrics

My new equivalent capacitance is
Ceq = C1 + C2
= C10 + K*C20

So the total capacitance has increased.

And Ceq = Qeq / Veq
where Qeq = Q1 + Q2 stays the same
so Veq = V1 = V2 has to decrease to balance out the equation.

I wanna say that Q1 = CV = C10*V0/k = Q10/k
So for Qeq to stay the same
Q2 would have to equal
Q2 = Q20(3/2)K
yes????
Isn't your new equivalent capacitance 4.2*C?

Previously you had Ceq = C + C
But after dialectric it's Ceq = C + 3.2*C = 4.2*C

Q = V*C

Means that Vo*2*C = V*4.2*C

That means V = 2*Vo/4.2
 
Re: dielectrics

Isn't your new equivalent capacitance 4.2*C?

Previously you had Ceq = C + C
But after dialectric it's Ceq = C + 3.2*C = 4.2*C

Q = V*C

Means that Vo*2*C = V*4.2*C

That means V = 2*Vo/4.2
wow im an idiot haha that makes so much sense and is so much easier than thinking about it the way i was. thanks for the help!!!
 

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