Homework Help: Capacitor and dielectriscs

1. Jan 10, 2010

EmmaK

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

A parallel plate capacitor has a capacitance C when there is a vacuum between teh plates. How does this change if the gap between the plates is half filled with a dielectric with dielectri constant $$\epsilon$$r

Calculate the change in energy stored in the capacitor if the dielectric is inserted while the capacitor is connected to a battery of fixed voltage V.

2. Relevant equations

C=C1+C2
C=$$\stackrel{\epsilon_0 A}{d}$$
3. The attempt at a solution

For the first part, i assumed it was equivalent to 2 capacitors (one with the dielectric and one without in parallel)

C=$$\stackrel{\epsilon_0 0.5A}{d}$$ =0.5C

C2= $$\epsilon_r$$*$$\stackrel{\epsilon_0 0.5A}{d}$$ - 0.5$$\epsilon_$$C

C(total) = 0.5C+0.5$$\epsilon_r$$C.

Second part, not really sure how to start...

V is fixed, and U=0.5CV2.. So if C increases by (1+$$\epsilon$$r) U increases by the same factor??

2. Jan 10, 2010

ideasrule

Why in parallel? The battery's voltage is being applied across the 2 capacitors, not across each capacitor individually.

Yes, except C doesn't increase by (1+$$\epsilon$$r).

3. Jan 10, 2010

EmmaK

Not entirely sure, just seen other examples where they said this

C(total)=C(0.5+0.5$$\epsilon$$r) so it increases by a factor of (0.5+0.5$$\epsilon$$r)?

Last edited: Jan 10, 2010
4. Jan 10, 2010

EmmaK

tried it again...

$U_0 =1/2 C_0 V^2$ and $U_1 =1/2 C_1 V^2$

[itex] C_1 = 1/2 \epsilon[/tex]r so

[itex] U_1 =1/2 *1/2 C_0 \epsilon[/tex]r V^2 [itex] =1/2 \epsilon[/tex]r *U_0 ?