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

A cylinder conductor of length ##l##, inner radius ##a## and outer radius ##b## is half-filled with a liquid dielectric as shown in the picture. The cylinder is attached to a battery of constant voltage ##U## and then separated from it. Imagine that the dielectric is leaking and calculate the change of electrical energy when there is no dielectric present. ( meaning the difference between the beginning state and the one were there is no dielectric)

There is a picture below.

## Homework Equations

3. The Attempt at a Solution

I'm gonna guide you through my work and thinking so you can correct me and help me finish the problem.[/B]

##ΔW_e=W_{e2}-W_{e1}## ( The state 2 is when its empty)

##W_{e1}=\frac{1}{2}\frac{Q_1^2}{C_1}## ( ##C_1,Q_1## are the total charge and capacitance of the system 1)

Because there is a dielectric in the first system i have two electric fields, the one above the ##h## and the one below going radial.

##E_1=\frac{Q'_1}{2 \pi ε_0 r}##

##U_1=\frac{Q'_1}{2 \pi ε_0}\ln (b/a)##

##E_2=\frac{Q'_2}{2 \pi ε_0 ε_r r}##

##U_2=\frac{Q'_2}{2 \pi ε_0 ε_r} \ln (b/a)##

Then i express the linear charge densities ##Q'_1,Q'_2##

##Q'_1=\frac{2 \pi ε_0 U_1}{\ln (b/a)}##

##Q'_2=\frac{2 \pi ε_0 ε_r U_2}{\ln (b/a)}##

##Q_{total}=Q'_1*(l/2) + Q'_2*(l/2)##

Something does not seem right here, can you help me with what i got wrong?

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