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Unformatted text preview: charge Q 1 in the half without the slab, and Q 2 in the half with the slab, in terms of Q and . [±irst Fnd the charge densities σ 1 and 2 .] b. ±ind Δ V in terms of Q , and Fnd the capacitance of the system. c. In which half of the capacitor is the energy density larger? How do you know? a. On the half of the plate above the empty gap the charge density is 1 where E = 1 / ε . On the half above the slab the charge density is 2 , where E = 2 / κε . Thus 2 = κσ 1 and Q 2 = Q 1 . Since Q 1 + Q 2 = Q we Fnd Q 1 = Q 1 + and Q 2 = Q 1 + . b. Since Δ V = Ed and E = Q 1 ⋅ ( A /2) = 2 Q A ⋅ 1 1 + , we have Δ V = 2 d A ⋅ 1 1 + ⋅ Q . Using C = Q / Δ V we Fnd C = 1 + 2 ⋅ A d . c. The energy density is give by u e = 1 2 E 2 . Since E is the same in both halves, u e is larger in the half with the slab. + Q − Q Physics 54 Summer 2011 2...
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This note was uploaded on 10/19/2011 for the course PHYSICS 54L taught by Professor Thomas during the Summer '09 term at Duke.
 Summer '09
 Thomas
 Physics, Force

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