# Capacitor with two dielectrics

1. Aug 9, 2007

### mXCSNT

1. The problem statement, all variables and given/known data
An air-insulated parallel-plate capacitor of capacitance $$C_0$$ is charged to voltage $$V_0$$ and then disconnected from the charging battery. A slab of material with dielectric constant $$\kappa$$, whose thickness is essentially equal to the capacitor spacing, is then inserted halfway into the capacitor. Determine (a) the new capacitance, (b) the stored energy, and (c) the force on the slab in terms of $$C_0$$, $$V_0$$, $$\kappa$$, and the capacitor-plate length L.

2. Relevant equations
$$C = \kappa \frac{\epsilon_0 A}{d}$$ for a parallel-plate capacitor with dielectric
$$\kappa_{Air} = 1.0006$$
$$C = C_1 + C_2$$ for the equivalent capacitance of two capacitors in parallel
$$U = \frac{1}{2}CV^2$$ (energy in a capacitor)
$$V = Q/C$$

3. The attempt at a solution
I took the capacitor to be two capacitors in parallel, one with half the area and therefore half the capacitance of $$C_0$$, the other with dielectric constant $$\kappa$$ and also with half the area.
$$C_1 = \frac{C_0}{2}$$
$$C_2 = \frac{C_0 \kappa}{2 \kappa_{Air}}$$
$$C_{total} = C_1 + C_2 = \frac{1}{2}C_0 \cdot (1 + \frac{\kappa}{\kappa_{Air}})$$
And the energy is just $$\frac{1}{2}C_{total}V_0^2$$

However I know this approach is incorrect. The professor worked this problem in class, and said that such an approach would over-simplify because the potential inside the capacitor would have to jump sharply as you crossed the edge of the inserted slab. He worked through an approach where he took into account that edge effect. Unfortunately I didn't write it down. Also, I am mystified how to get the force on the slab--I guess the slab must pick up some charge, but I don't know how.

2. Aug 9, 2007

### mXCSNT

We also have Gauss' law $$\oint_S \vec{E} \cdot d\vec{A} = \mu_0 Q_{enclosed}$$, and I remember the professor saying something about considering the situation as the slab is gradually moved in from one side a distance x, but I don't have any ideas. I still don't see what causes any force on the slab, unless it means the force on the slab while it is being moved (since the stored energy must increase and the only source for it is the motion of the slab).

3. Aug 9, 2007

### learningphysics

Oops... never mind I was wrong.

Last edited: Aug 9, 2007
4. Aug 9, 2007

### learningphysics

The new capacitance is correct. In this situation charge is fixed. The old and new capacitor have the same charge... not the same voltage. So use that to calculate the energy stored in the new capacitor...

5. Aug 9, 2007

### learningphysics

The force changes as the slab goes into the capacitor... does the question ask for the force at a particular point, or is it asking for a formula for the force at all points?

6. Aug 9, 2007

### learningphysics

I think the question is asking for the average force on the slab... The capacitor loses energy... take the final energy - initial energy... That is the negative of the work done by the capacitor on the slab...

ie : -Fav*(L/2) = final energy - initial energy