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Unformatted text preview: EE2111 Homework #3 A03 Assigned 11 September 02 Due 17 September 02 Problem 1 . Hambly P3.22 Solution: Using d LW d A C ε ε = = and Kd V = max to substitute into 2 max max 2 1 CV W = , we have WLd K d K d LW W 2 2 2 max 2 1 2 1 ε ε = = . However the volume of the dielectric is WLd Vol = so that ( ) Vol K W 2 max 2 1 ε = Thus we conclude that the maximum energy stored is independent of W, L, and d. To achieve large energy storage per unit volume, we should look for a dielectric having a large value of 2 K ε . Problem 2. Hambly P3.23 Solution: The charge Q remains constant because the terminals of the capacitor are opencircuited. ( ) J V C W C V C Q µ µ 500 2 / 1 1 1000 10 1000 2 1 1 1 12 1 1 = = = × × = = − After the distance between the plates is doubled, the capacitance becomes pF C 500 2 = . The voltage increases to V C Q V 2000 10 500 10 12 6 2 2 = × = = − − and the stored energy is ( ) ( ) J V C W µ 1000 2 / 1 2 2 2 2 = = ....
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 Spring '09
 LABONTE
 Energy, Inductor, Hambly P4

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