Download Discussion Question 3A

Discussion Question 3A
P212, Week 3
Electric Flux
Electric Flux refers to the net amount of electric field flowing through a surface. A way to write this is:
G G
G G Q
flux, Φ ≡ ∫ E ⋅ dA . Why do we care about flux? Because Gauss’ law states that flux = ∫ E ⋅ dA = enc ,
ε0
which gives us an easy way to calculate electric fields for charge distributions. So, as first step, let’s
G G
see how to find ∫ E ⋅ dA , or the net flux, through some surfaces … [Remember that you care about net
flow through a surface, which is the difference between flow in and flow out]
(a) An infinitely thin imaginary plane (a “Gaussian surface”) is placed in
a uniform electric field as shown. Is there a net flux through the plane?
There is a net flux.
K
E0
Would there be a net flux if the plane were rotated 90º through an axis
perpendicular to the plane of paper?
If the plane were rotated 90º, there would be no net flux through the plane.
(b) An imaginary sphere is now placed in the uniform field. Is there a
net flux through the sphere? Is there a net field in the sphere?
There is no net flux through the sphere K(the same number of lines enter
as leave). The net field in the sphere is E .
K
E0
Is Gauss’ Law relevant for finding the field in this case? Why?
Gauss’ Law is not relevant because net flux is zero.
(c) A Gaussian sphere encloses charges Q1 and Q2 at its center, where
Q1 + Q2 < 0. Draw the electric field lines due to these charges.
Field lines, shown in red, point toward negative charges.
G
Now use arrows to show the direction of dA .
G
dA , shown in green, always points out normal to surface.
Is the net flux through the sphere positive, negative, or zero?
G
G
Because dA and E point in different directions, the net flux is negative.
(d) What is the surface area, ∫ dA , of the sphere in part (c) if it has a radius R?
Q 1 ,Q 2
G
2
dA
∫ = 4π R
Is Gauss’ Law relevant here? Gauss’ Law is relevant since there is a net flux and symmetry.
G
Find E at the surface of the sphere.
G G G
G (Q + Q )
(Q + Q )
2
∫ E ⋅ dA = E 4π R = 1 ε 0 2 → E = ε 014π R22
1
Discussion Question 3A
P212, Week 3
Electric Flux
G
(e) The sphere is now placed in a uniform electric field, E0 . Draw
an arrow showing the direction of the net field at point A. Recall
Q1 + Q2 < 0 .
How does the magnitude of the field at point A compare to the
field at point B?
Equal in magnitude (but different in direction)
How does the magnitude of the field at point C compare to the
field at point D?
Field at D > field at C
G
Write an expression for E at point C.
G ⎛
Q + Q2 ⎞
By superposition, E = ⎜ E0 − 1
⎟ xˆ
4πε 0 R 2 ⎠
⎝
2
K
E0
•A
•
D
Q 1, Q 2
•B
•C