{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

20115ee1_1_Homework5solutions

# 20115ee1_1_Homework5solutions - Homework#5 solutions 6.4...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online