Electric Flux
Recall that when electric field lines are carefully drawn that the density of lines (the
number of lines crossing a unit area perpendicular to the lines) is proportional to the electric field
magnitude.
Therefore if we have a flat rectangular area A perpendicular to the electric field, the
number of electric field lines crossing the area is proportional to
Φ E = EA , where this is called the flux of E where E is a constant and is
perpendicular to the area A.
The concept of the flux of a vector is not limited to electric fields but can be applied to
any vector field, e.g. the velocity field for fluid flow.
Here, we will gradually generalize this definition to cases
a. when E is not perpendicular to the area
b. when E is not a constant
c. to curved surfaces
d. and finally to closed, rather than open, surfaces.
a. With flux associated with the number of field lines crossing the surface, it is
straightforward to generalize to the case when E is not perpendicular to the area.
Imagine a bull’s eye target. First when arrows are aimed at the target perpendicular to
the surface, you have the greatest probability to hit the bull’s eye. If, by mishap, the
bull’s eye were oriented so that the normal to its surface were not horizontal, but were up
at some angle, then the area presented to you as you shoot your arrows would decrease –
the projected area you would see would be smaller. By how much? Well, if A’ is the
projected area perpendicular to E , A’ = A cosθ, where the angle is that between E and
the normal to the surface. So in this case we have that the electric flux is given by
Φ E = EA cos θ
having a maximum value of EA when the normal is along the E and a minimum of 0
when it is perpendicular to E and presents zero projected area.
b. Now suppose that E ≠ constant. How shall we define the electric flux over some area A?
We divide A up into small areas ∆Ai. Let’s make these into vectors with the direction
given by the direction of the normal vector (perpendicular to the surface), so that
∆Ai = ∆Ai nˆ
where n̂ is a unit vector perpendicular to the area ∆Ai.
n
Then we can define the electric flux over the small
E
area ∆Ai to be
∆Φ E = Ei i∆Ai = Ei ∆Ai cos θ i ,
where the angle is (from the usual dot product)
that between the two vectors.
To get the total electric flux over the total area A, we first imagine that the small areas
∆A become differential areas dA and then we integrate over the total area A to find
Φ E = ∫ d Φ E = ∫∫ E idA ,
R
where the integral is the usual double integral over the plane area A. Note that HRW
write this integral with a single integral sign in a short-hand notation. Don’t get confused
– it is really a surface integral.
c. What happens, next, if the surface is curved in space?
You will see a lot more of this from Julius, but here we simply quote that the result will
look almost the same, but with the plane region R replaced by the curved surface S, so
that
Φ E = ∫∫ E idA
E
S
n
d. Lastly, what happens if the surface is closed?
This means that the surface encloses a volume
so that there is an inside and an outside volume
bounded by the closed surface. Then we simply write
that the electric flux is given by
Φ E = ∫∫ E idA
dA
S
where the circle on the integrals indicates that S is a closed surface and where now, to be
specific, the normal to the surface is always taken as the outward normal – that is
pointing toward the external volume, pointing outward from the surface.
Example 1: Find the flux through a cube of side L oriented with its faces parallel to the
Cartesian axes and with a uniform electric field along the + x direction.
Solution: Note that there are 6 faces to the cube.
y
For the two faces with normals along ± k (front and back):
E
E ikˆ = 0 , so ΦE = 0 for these.
Similarly the flux through the the two faces with normals
along ± j are zero
x
But the two faces with normal along ± I each contribute to
ˆ + E i(−iˆ) A = 0 .
the total flux: Φ E = E iiA
z
So, we conclude that the total flux through the cube is zero.
Example 2: Find the flux through a hemisphere of radius a due to a uniform electric field along
its axis as shown.
z
We want to calculate
E idA where
∫∫
S
y
x
E = Eo kˆ and dA = dA rˆ since the outward normal points
along the radial direction. Remembering that in spherical
coordinates dA = a2sinφ dφ dθ , we have
Φ E = Eo ∫∫ kˆirˆ(a 2 sin φ ) dφ dθ
where the limits of integration on φ are 0 to π/2 and on θ
are 0 to 2π. Noting that the dot product involves two unit
vectors and that the angle between them is φ, we have
E
Φ E = Eo a 2
π /2
∫
0
2π
cos φ sin φ dφ ∫ dθ
0
But the second integral is just 2π and the first integral can be done directly to give
π /2
sin 2 φ
1
= , so that the total electric flux is given by Φ E = Eoπ a 2 .
2 0
2
Notice that, if we had been clever, we might have guessed this result since it represents the
product of the uniform electric field and the projected area perpendicular to the electric field –
namely the area of the circle in the x-y plane that bounds the hemisphere.
Example 3: Find the electric flux through a closed sphere of radius a due to a uniform electric
field.
Using the previous result, this is fairly straightforward. Divide the sphere into two
hemispheres and use the previous result for each – remembering that when you take kˆirˆ in the
integral and use the limits of π/2 to π that this contribution is negative for the bottom
hemisphere. Therefore the total electric flux through the sphere is zero.