Dielectric and energy. I probably did it wrong, no I DID IT WRONG.

[flyingpig]

Homework Statement

[PLAIN]http://img97.imageshack.us/img97/6749/kappam.jpg

Homework Equations

U = (1/2)CV²

V = V₀/k

The Attempt at a Solution

(1) \Delta V_0= \Delta V <==== Electrical potential difference is constant since the battery was never removed.

(2) \kappa Q_0 = Q <=== Since the battery is NOT removed and therefore the potential difference must rema

Divide (2) by (1) and I get

\kappa C_0 = C

U_0 = \frac{1}{2}C_0\Delta V_0^2

Now then the only change is the capacitance and so we have

U = \frac{1}{2}C\Delta V_0^2

U = \frac{1}{2}(\kappa C_0){\Delta V_0^2}

Pull out kappa and we get

U = \kappa \frac{1}{2}C_0\Delta V_0^2

Which should be recognized as U = κU₀For part b) is the question already been answered? That is κQ₀ = Q

 

[flyingpig]

No I take it back, I did it right. Except for part b), I feel like it's asking for something else

 

[SammyS]

Both (a) & (b) are fine.

Perhaps the author of the question had a different method of solution in mind; one in which (a) would be answered before (b).