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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 Fall '11
 Joshi,Chand
 Electric charge, Prime number, @, Fundamental physics concepts

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