20115ee1_1_Homework5solutions

20115ee1_1_Homework5solutions - Homework #5 solutions 6.4...

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Unformatted text preview: Homework #5 solutions 6.4 a)Vo is held constant due to battery. b) = so capacitance changes by a factor of 0.1. c) = −∇ = d) = e) = D changes by a factor of 0.1 || Q decreases by a factor of 0.1 f) = g) az E changes by a factor of 0.1. . = . changes by a factor of 0.1 decreases by a factor of 0.1. 6.5 E will be a function of x so to find Vo we integrate =− . ! == − (2 + 2 ∗ 10( ) . = = ! = 4.43 ∗ 10, || ∗ (0.02) = 451 ./ 6.6 a) now charge remains the same on both plates. Assuming the distance between plates is much less than the dimensions of the plates b) = || = 1 = az V therefore must increase by a factor of 10. Q stays the same so capacitance changes by a factor of 0.1 c)E remains the same d)D remains the same e) Q stays the same f) 0 stays the same g) = . increases by a factor of 10. 6.7 6.8 a) Using conductor boundary conditions definition =− = 2 . So ∙ 97 67 =- Therefore = − b) =− = 34 (3) 5(3) 34 (3) 6. 5(3) 7 ? = ( )67 , here charge density may be a function of ρ. By Now we need to relate this to given variables Vo, d using where d is the distance between the plates. 67 . 3; :1 + <; = 67 . c) = 3 E >− A1 + @ B C D = −2F 1 A 2 d) = = 6.9 2F H A + G 4B || 1 2 2 + 4 4B2 C C 6.11 ...
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This note was uploaded on 12/05/2011 for the course EL ENGR 1 taught by Professor Joshi,chand during the Fall '11 term at UCLA.

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