Calculating Capacitor Charges with Dielectric Insertion

[jehan60188]

Homework Statement

A 24 V battery is used to charge two identical capacitors CL = CR = 6 µF, in parallel
so Q(CL) = Q(CR) = 0.000144 C

the battery is removed, and the energy stored in the system is .003456 J

Next a dielectric with k = 1.4 is inserted between the plates of CR, fully filling the available space. (The dielectric carries zero net charge.)

Calculate the new values of the charges on each capacitor.

Homework Equations

C=Q/V (capacitance = charge/electric potential)
C_new = C*k (capacitance = capacitance in a vacuum times K)
capacitors in parallel add (C = c1 + c2)
capacitors in series add their inverses (C =(1/C1 + 1/C2)^-1)

The Attempt at a Solution

the capacitance of the system is (1/8.4 + 1/6)^-1 = 3.5 uF
since the charge doesn't change, it's still 0.000144*2 C
voltage over either capacitor is the same

V*CL + V*CR = Q

things break down from here. any pointers?


thanks!

 

[SammyS]

Homework Statement

A 24 V battery is used to charge two identical capacitors CL = CR = 6 µF, in parallel
so Q(CL) = Q(CR) = 0.000144 C

the battery is removed, and the energy stored in the system is .003456 J

Next a dielectric with k = 1.4 is inserted between the plates of CR, fully filling the available space. (The dielectric carries zero net charge.)

Calculate the new values of the charges on each capacitor.

Homework Equations

C=Q/V (capacitance = charge/electric potential)
C_new = C*k (capacitance = capacitance in a vacuum times K)
capacitors in parallel add (C = c1 + c2)
capacitors in series add their inverses (C =(1/C1 + 1/C2)^-1)

The Attempt at a Solution

the capacitance of the system is (1/8.4 + 1/6)^-1 = 3.5 uF
since the charge doesn't change, it's still 0.000144*2 C
voltage over either capacitor is the same

V*CL + V*CR = Q

things break down from here. any pointers?

thanks!

I can see two ways two solve this. Neither involves considering that the capacitors are in series.

One is to consider the two capacitors to be in parallel, both while being charged, and also while the dielectric is being inserted.

The effective capacitance changes in this case, but the charge remains the same. You can then find the potential difference across the plates.​

The other is to consider the capacitors separately, realizing that after the dielectric is inserted, the extra charge, q, gained by C_R is equal to the amount of charge lost by C_L. The voltage across one is equal to the voltage across the other.

 

[jehan60188]

still lost, but I'm going to poke at the problem some more.
would love clarification and/or further information!

how much charge is gained by CR?

i figured it out

the charge in the system remains the same
the (now) 8.4 uF capacitor is responsible for 100 * 8.4/(6+8.4) percent of the charge, so i take the total charge (2*.000144) and multiply it by 8.4/(6+8.4)

thanks!