HW5_solutions

HW5_solutions - P.4-1 The upper and lower conducting plates...

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P.4-1 The upper and lower conducting plates of a large parallel-plate capacitor are separated by a distance d and maintained at potentials V 0 and 0, respectively. A dielectric slab of dielectric constant 6 . 0 and uniform thickness 0 . 8 d is placed over the lower plate. Assuming negligible fringing eFect, determine (a) the potential and electric ±eld distribution in the dielectric slab, (b) the potential and electric ±eld distribution in the air space between the dielectric slab and the upper plate, (c) the surface charge densities on the upper and lower plates. (d) Compare the results in part (b) with those without the dielectric slab. d 0.8d V=V0 V=0 Solution: Neglecting fringing, the potential in each region is the solution Laplace’s equation in one dimension: 2 V = d 2 V dx 2 = 0 Hence, we have V ( x ) = b c 1 x + c 2 , 0 x 0 . 8 d c 3 x + c 4 , 0 . 8 d < x d The resulting electric ±eld in each region is given by E = b ˆ a x c 1 , 0 x 0 . 8 d ˆ a x c 3 , 0 . 8 d < x d and the electric ²ux density by D = b ˆ a x ǫ r ǫ 0 c 1 , 0 x 0 . 8 d ˆ a x ǫ 0 c 3 , 0 . 8 d < x d The boundary conditions to be applied to obtain c 1 , c 2 , c 3 , and c 4 are (1) V (0) = 0 (2) V ( d ) = V 0 (3) V (0 . 8 d - ) = V (0 . 8 d + ) (4) D x (0 . 8 d - ) = D x (0 . 8 d + ) 2
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The resulting system of equations can be written as 0 1 0 0 0 0 d 1 0 . 8 d 1 0 . 8 d 1 ǫ r 0 1 0 c 1 c 2 c 3 c 4 = 0 V 0 0 0 Solving the linear system, we obtain c 1 = V 0 d (0 . 8 + 0 . 2 ǫ r ) c 2 = 0 c 3 = ǫ r V 0 d (0 . 8 + 0 . 2 ǫ r ) c 4 = 0 . 8 ( ǫ r 1) V 0 (0 . 8 + 0 . 2 ǫ r )
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HW5_solutions - P.4-1 The upper and lower conducting plates...

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