How does the addition of a dielectric affect the energy stored in a capacitor?

[EmmaK]

Homework Statement

A parallel plate capacitor has a capacitance C when there is a vacuum between the 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.

Homework Equations

C=C₁+C₂
C=$$\stackrel{\epsilon_0 A}{d}$$

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??

 

[ideasrule]

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

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

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??

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

 

[EmmaK]

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

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

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

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

 

[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 <br /> <br /> [itex]U_1 =1/2 *1/2 C_0 \epsilon[/tex]_r V^2 [itex]=1/2 \epsilon[/tex]_r *U_0 ?[/itex][/itex][/itex]