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 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.

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_rC.

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

C_1 = 1/2 \epsilon[/tex]_r so <br /> <br /> U_1 =1/2 *1/2 C_0 \epsilon[/tex]&lt;sub&gt;r&lt;/sub&gt; V^2 =1/2 \epsilon[/tex]&amp;lt;sub&amp;gt;r&amp;lt;/sub&amp;gt; *U_0 ?