Energy of parallel plate capacitor problem

[meteorologist1]

Hi, I have trouble on the following problem:

Given a parallel plate capacitor, fixed area A, and fixed separation d. Find the energy stored, before and after insertion of a slab of dielectric, which completely fills the space between plates, for each of the two cases:

a) Plates are connected to a battery which maintains constant potential difference

b) Plates are charged with fixed charges +Q and -Q, and battery disconnected.

Explain differences in the two cases.

For part a, I think the answers would be just:
Before insertion of dielectric: U = (1/2)CV^2 where C = A(epsilon)/d and
after insertion of dielectric: U = (1/2)KCV^2 where C = A(epsilon)/d and K is the dielectric constant.

But I'm not sure about part b. Thanks.

 

[dextercioby]

How about $U=\frac{Q^{2}}{2C}$,where C is the electric capacity before and after inserting the diellectric...?

Daniel.

 

[Doc Al]

For part a, I think the answers would be just:
Before insertion of dielectric: U = (1/2)CV^2 where C = A(epsilon)/d and
after insertion of dielectric: U = (1/2)KCV^2 where C = A(epsilon)/d and K is the dielectric constant.

You've got the right idea, but be careful how you express it. The capacitance changes when you insert the dielectric: before insertion, C = A(epsilon)/d; after insertion, C = KA(epsilon)/d.

U = (1/2)CV^2 is always true, but C changes, so U changes from (1/2)[A(epsilon)/d]V^2 to (1/2)K[A(epsilon)/d]V^2. (Don't write U = (1/2)KCV^2; that's not true!)

But I'm not sure about part b.

If the charge is fixed, what happens to the potential difference when the dielectric is inserted. (What happens to the electric field within the plates?)
In this case, both C and V change.

You can also write the stored energy directly in terms of Q and C. (Figure that out.) Then you'd only have to worry about C changing.

 

[meteorologist1]

Ok, I understand now. Thank you very much.