(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

An air-insulated parallel-plate capacitor of capacitance [tex]C_0[/tex] is charged to voltage [tex]V_0[/tex] and then disconnected from the charging battery. A slab of material with dielectric constant [tex]\kappa[/tex], 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 [tex]C_0[/tex], [tex]V_0[/tex], [tex]\kappa[/tex], and the capacitor-plate length L.

2. Relevant equations

[tex]C = \kappa \frac{\epsilon_0 A}{d}[/tex] for a parallel-plate capacitor with dielectric

[tex]\kappa_{Air} = 1.0006[/tex]

[tex]C = C_1 + C_2[/tex] for the equivalent capacitance of two capacitors in parallel

[tex]U = \frac{1}{2}CV^2[/tex] (energy in a capacitor)

[tex]V = Q/C[/tex]

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 [tex]C_0[/tex], the other with dielectric constant [tex]\kappa[/tex] and also with half the area.

[tex]C_1 = \frac{C_0}{2}[/tex]

[tex]C_2 = \frac{C_0 \kappa}{2 \kappa_{Air}}[/tex]

[tex]C_{total} = C_1 + C_2 = \frac{1}{2}C_0 \cdot (1 + \frac{\kappa}{\kappa_{Air}})[/tex]

And the energy is just [tex]\frac{1}{2}C_{total}V_0^2[/tex]

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.

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# Homework Help: Capacitor with two dielectrics

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