Try it on paper first. A hint, the key idea and the worked solution are here when you need them.
Two parallel metal plates, each of area , stand apart. The gap is filled with a layered plastic whose dielectric constant grows steadily across the gap:
where is the distance from plate and is the gap, as in the figure. So is next to and next to plate . The plastic touches both plates, with no air left between. Free charge is the charge on the metal; bound charge is the charge that shows on the plastic when its molecules are polarised. A battery holds positive and negative. Ignore the effects at the edges of the plates. Take .
Have you tried it?
Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Plan. Gauss's law, used with the free charge, gives the field at each . Integrate it for the potential difference. The bound charges come from how much the plastic weakens the field at each place.
Step 1. Capacitance (a). With free charge per unit area,
Step 2. Free charge (b). .
Step 3. Field at (c). . At it is half as big, because is twice as big there.
Step 4. Bound charge on the faces (d, e). At the field falls from to , so the face carries . At , where , the face carries .
Step 5. Bound charge in the volume (f). The plastic is neutral, so the rest is spread through it: . It appears because changes from place to place, so the field changes too.
Answer: (a) . (b) . (c) . (d) . (e) . (f) .
Check. The energy is also the sum of over the gap.
Watch out. Using the middle value in gives . Layers across a gap add like capacitors in series, not by their average .
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