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A parallel-plate capacitor has plate area A and plate separation d. Two dielectric slabs of dielectric constants K1 = 2 and K2 = 3 completely fill the space between the plates (slab K1 adjacent to the positive plate, slab K2 adjacent to the negative plate, each of thickness d/2). If sigma is the free-charge surface density on the positive plate, find the bound-charge surface density at the interface between the two dielectrics.
- sigma/3
- sigma/6
- -sigma/6
- -sigma/3
Correct answer: -sigma/6
Solution
Since D is continuous, D = sigma everywhere between the plates. The electric field in slab K1 is E1 = sigma/(K1*epsilon0) and in slab K2 is E2 = sigma/(K2*epsilon0). The bound charge at the interface is sigma_b = epsilon0*(E2 - E1) = sigma*(1/K2 - 1/K1) = sigma*(1/3 - 1/2) = -sigma/6.
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