Energy Density with a Dielectric

[racecar_]

Homework Statement

Two parallel plates, having a vacuum in between them, are separated by d=0.01 m apart and are connected to a battery maintaining 9V between the plates and a charge of Q0 magnitude on each plate. The separation between the plates is then raised to 2d, the battery voltage is increased to 18V, and a dielectric is added with dielectric constant 2. What is the energy density difference after the addition of the dielectric?

Relevant Equations

energy density = (1/2)εE^2

I am confused about how the electric field changes in this problem - is E' = E/K_e=E/2? Is E = V/d a correct usage?

When I solve it this way, the answer is incorrect:
change in energy density = (1/2)ε(E'²- E²) = (1/2)ε(E²/4 - E²) = (1/2)ε(-3/4)(V/2d)².

What am I doing wrong? Thanks.

 

[TSny]

I am confused about how the electric field changes in this problem - is E' = E/K_e=E/2?

This equation does not apply in your situation. This equation would be valid if, after charging the capacitor, you disconnect the capacitor from the battery and then add the dielectric.

Is E = V/d a correct usage?

Yes. This is generally valid for any parallel plate capacitor.

 

[kuruman]

The obvious approach is to calculate the stored energy in each case and divide by the volume.

 

[racecar_]

This equation does not apply in your situation. This equation would be valid if, after charging the capacitor, you disconnect the capacitor from the battery and then add the dielectric.

So the electric field will stay the same? For the equation energy density = (1/2)εE^2, what changes then?

The obvious approach is to calculate the stored energy in each case and divide by the volume.

Yeah, I think it's a better approach. In what situations should I use finding the stored energy and dividing the volume rather than (1/2)εE^2, and vice versa? Thanks.

 

[kuruman]

Either approach should work if you know what you’re doing. Here you made a key mistake by using ε for both vacuum and dielectric between the plates instead of for only the dielectric.

 

[TSny]

So the electric field will stay the same?

Yes.

For the equation energy density = (1/2)εE^2, what changes then?

Following up a little on @kuruman's post. The energy density of an electric field in a vacuum is $\frac1 2 \varepsilon_0 E^2$. This expression is modified for an electric field inside a dielectric. Have you covered this in your course?

If not, then use the approach of finding the stored energy divided by the volume.