I'm having a bit of trouble with this problem, you can neglect any effect at the boundaries.

A parallel plate capacitor, whose plates have a distance 2d, is charged and then isolated. Keeping the charge on each plate constant, we introduce, parallel to the plates, a slab of dielectric of thickness d, and relative dielectric constant k. We then extract the insulating slab, and introduce a conducting slab, also of thickness d. Knowing that, in order to extract the insulating slab, we need to do a work W1, and to extract the conducting slab a work W2 = 3W1, compute k.

So since Q is constant, initially the capacitor has capacitance;

C0 = Q/V0 = A(ε0)/2d

after the insulating slab of dielectric is added we compute;

C = kC0=εA/2d where ε = k(ε0)

and when the conducting slab is introduced, the initial capacitor essentially becomes two such that the inverse of the total capacitance:

1/Ct = 1/C1 + 1/C2, where C1 = C2 = A(ε0)/(d/2), therefore Ct = A(ε0)/d

I know this isn't enough to solve the problem, and that the electrostatic energy stored in the capacitor U0 = C0(V0^2)/2 = QV0/2 = Q^2/2C0.

What is W1? Is it the sum of the changes in energy when the slabs are inserted and removed? and how can I use this to find k?


closed as off-topic by John Rennie, Bill N, Kyle Kanos, Jon Custer, Qmechanic Nov 14 '17 at 16:27

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You can compute the total energy of the system with the slab inserted and not inserted. The difference between those states is the work done to move from one state to the other. Reading the question, the work $W_1$ is the work done to extract the dielectric slab - so it's the difference between the energy stored in the capacitor with the slab inserted, and when there is vacuum (air) in the space between the plates.

The energy density in a dielectric medium is $\frac12 \mathbf{E \cdot D}$ - see for example this. You can integrate that expression over the volume to get the total energy, given the electric field in the space between the capacitor plates is uniformly distributed. Note that the dielectric medium is thinner than the space between the plates. Also note that there is no electric field inside a conductor - so all the energy will be in the air gap. This should be enough to solve the problem.


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