lec 5 physics 72
DESCRIPTION
Physics 72TRANSCRIPT
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Lecture 5Electric flux
August 14, 2015
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Application of dipoles: microwave heating
Lecture 5 August 14, 2015 2 / 17
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Application of dipoles: microwave heating
Lecture 5 August 14, 2015 3 / 17
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Application of dipoles: microwave heating
Lecture 5 August 14, 2015 4 / 17
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Flux in fluid flow
Volume flow rate through a loop of area A:
dV
dt= Av
Lecture 5 August 14, 2015 5 / 17
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Flux in fluid flow
dV
dt= Av = (A cos)v
Lecture 5 August 14, 2015 6 / 17
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Electric flux
E = EA
Define the area vector :~A = A n
where n points perpendicular to the area , so that the electric flux can bewritten as
E = ~E ~A = EA cos
Lecture 5 August 14, 2015 7 / 17
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Electric flux
Lecture 5 August 14, 2015 8 / 17
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Electric flux
Lecture 5 August 14, 2015 9 / 17
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Electric flux
Lecture 5 August 14, 2015 10 / 17
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Electric flux
Example
A disk of radius 0.10m is oriented with its normal unit vector at to auniform electric field of magnitude 2.0 103N/C. What is the electricflux through the disk?
Lecture 5 August 14, 2015 11 / 17
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Electric flux
Solution:
E = EA cos = (2.0 103N/C)(pi(0.10m)2) cos 30 = 54N m2/C .
Lecture 5 August 14, 2015 12 / 17
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For curved surfaces and nonuniformfields,
E =
~E d ~A
Lecture 5 August 14, 2015 13 / 17
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Electric flux
Example
Consider a cylinder of radius with its axis parallel to a uniform field ofmagnitude E . What is the total electric flux passing through the cylinder?
Lecture 5 August 14, 2015 14 / 17
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Electric flux
Solution:Write the flux for surfaces labeled a, b and c . Note that Aa = Ac .
E = a + b + c
= EAacos 180 + EAbcos 90 + EAccos 0
= EAa + 0 + EAc= 0 .
Lecture 5 August 14, 2015 15 / 17
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Quiz!
What is the total electric flux passing through the cube?
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Quiz!
Solution:
E ,1 = EA cos 135 = EA
(22
)E ,2 = EA cos 45
= EA(
+22
)E ,3 = EA cos 135
= EA(22
)E ,4 = EA cos 45
= EA(
+22
)E ,5 = EA cos 90
= 0
E ,6 = EA cos 90 = 0
tot = E ,1 + . . .+ E ,6 = 0 .
Lecture 5 August 14, 2015 17 / 17