Dielectric Capacitor: Calculation and Force Exertion

[sheepcountme]

Homework Statement

You have 2 square metal plates of side length L, separated by a very small distance d. The two plates are held at fixed potential difference ΔV by a battery. A thin slab with dielectric constant κ and thickness d is inserted a distance x into the gap between the plates. (a) What is the equivalent capacitance? (b) How much energy is stored in the capacitor? (c) What is the magnitude of the electric force exerted on the dielectric slab? (d) Does this force tend to pull the slab into the gap, or to repel it from the gap? Hint: If you place two capacitors side by side and wire them so that they always have the same electric potential across them, the total capacitance is equal to the sum of the individual capacitances.

Homework Equations

C_k=kC
k=E_vacuum/E_dielectric
C=q/deltaV

The Attempt at a Solution

I'm a bit confused by what they are asking for really, if they expect us to actually get numbers out of this, I don't see how it is possible. As for part a, C_k=kq/deltaV
But q isn't a known constant in this situation so I don't know if this is an acceptable answer.

 

[collinsmark]

For part (a) you can use other equations for a parallel plate capacitor to determine q. If you don't have those equations handy, you can use Guass' law, and the definition of electric potential to re-derive the approximate equations (which I'm pretty confident is where your parallel plate capacitor equations came from to begin with). Break it up into two capacitors, using the hint given in the problem statement.