Flux and Gauss' Law

PHY110W ELECTRICITY SUPPLEMENTARY GAUSS’ LAW
Flux and Gauss’ Law
Flux is related to field lines.
Imagine that you can see electric field lines… (Put on the special glasses if you have to!)
Now put a non-metallic object (like a balloon) in the field, so that the field lines pass right through the object.
You’ll notice that some field lines enter the object, some exit it and some just skim the surface.
In the diagram, field lines are seen exiting a small part of a
surface an angle θ to the normal. (A normal is an imaginary line
perpendicular to the surface, usually drawn at the point where a
field line enters or exits the surface.)
The flux through this surface element is just that component of
the field (ie the field lines) which lies parallel to the normal.
Mathematically: flux = (electric field) cosθ
Or, symbolically: Φ = E cosθ
normal
θ
surface
element
field lines
The smaller the angle the field lines make with the normal (ie the
more perpendicularly they exit through the surface), the more flux there is through this part of the surface.
If the lines just skim the surface (ie θ = 90°) the flux is zero (since cos90° = 0).
Now a surface which completely encloses some three-dimensional space (like the surface of an object – eg the
balloon, or a closed rectangular box, or a ball) is known as Gaussian surface. With such a surface you can
clearly see which field lines are entering the surface and which are leaving (exiting) the surface.
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PHY110W ELECTRICITY SUPPLEMENTARY GAUSS’ LAW
We make things mathematically simpler (!) by turning the normal into
the elemental area vector, dA , whose magnitude is equal to the area
of the surface element, and whose direction is always out of the
surface (at right angles to it, like the normal).
So field lines which enter the Gaussian surface (for which θ is larger
than 90°) result in negative flux (because cosθ is now negative).
And, representing the field by the electric field vector, E , we can now
write the equation even more simply (!) as:
Φ = E ⋅dA
area vector,
dA
θ
electric field
vector, E
To find the flux through a whole area, we must sum the flux through each of
the elements which makes up that area, or (as the size of each element → 0)
integrate over the whole area:
total Φ = ∫ E ⋅dA
If the whole area is a complete Gaussian surface, we write: Φ = ∫ E ⋅dA
,
where Φ now represents the net flux through the Gaussian surface.
The importance of this equation is that Gauss’ Law tells us that the net flux through a Gaussian surface is
proportional to the charge enclosed by that surface, or mathematically: ε
0Φ = qenclosed
. Which gives us…
Gauss’ Law:
ε E⋅ dA=
q
∫
0 enclosed
…and from this we derive useful equations for the electric field due to various charge distributions.
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PHY110W ELECTRICITY SUPPLEMENTARY GAUSS’ LAW
Types of questions
Questions on this section usually fall into one of three categories:
1. Calculating the flux through some given surface (maybe part of a bigger, complete, Gaussian surface).
[This type of question is a bit artificial and is really just for practice.]
2. Working with the net flux through an entire Gaussian surface and applying Gauss’ Law to find the
enclosed charge (or vice versa).
3. Using the principle of Gauss’ Law to determine the electric field due to some distribution of charge.
[Which is the main purpose of this section – determining electric fields – so that we can easily determine
the force on a charged particle, and from here the amount of work done in moving it, etc etc.]
Examples:
* * * * *
Type 1. A cubic Gaussian surface is bounded by the planes at x = 0.40 m,
y = 0.60 m and z = 0.40 m as shown in the sketch. The electric field in this
region is non-uniform and is given by E = (2.00 + 3.00x 2 )î N/C. Calculate
the flux through the right hand face (ie the shaded face) of the cube.
Solution: Φ= ∫ E ⋅dA
Over the whole shaded surface x = 0.40 m, so the electric field is a
constant (2.00 + 3.00 × 0.40 2 ) = 2.48 N/C and lies in the positive x
direction, ie parallel to the area vector, so cosθ = 1, and we get
0.6
0.2
y
z 0.4
0.4
Φ
right
= E∫ dA = E A = 2.48 × 0.40 2 = 0,397 N⋅m 2 /C
[Strictly speaking, you should evaluate the dot product more formally using unit vectors.]
x
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PHY110W ELECTRICITY SUPPLEMENTARY GAUSS’ LAW
Type 2. A cubic Gaussian surface is bounded by the planes at x = 0.40 m,
y = 0.60 m and z = 0.40 m as shown in the sketch. The electric field in this
region is non-uniform and is given by E = (2.00 + 3.00x 2 )î N/C. Calculate:
0.6
y
(a)
the flux through each face of the cube;
0.2
(b) the net charge enclosed by the cube.
Solution: (a) This time we must apply Φ = E ⋅dA
over each of the six faces –
∫
z 0.4
0.4
which is actually not as much work as it seems, because the field has no components in the y
or z directions, so there will be no flux through the top, bottom, front and back faces:
Ie Φ top = Φ bottom = Φ front = Φ back = 0 N⋅m 2 /C
We’ve already calculated the flux through the right hand face (in Type 1), so we only have to
follow the same procedure for the left face, where x = 0 m.
Here the electric field is a constant (2.00 + 3.00 × 0 2 ) = 2.00 N/C and again lies in the positive
x direction. This time, however, the area vector points in the negative x direction (because it
always points out of the Gaussian surface), so cosθ = –1, and we get
Φ
left
= E∫ dA =−E A = –2.0 × 0.40 2 = –0,32 N⋅m 2 /C
(b) Using Gauss’ Law, qenclosed
= ε
0Φ (where Φ is the total flux through all six faces)…
∴q enclosed = 8.85 × 10 -12 × (0 + 0 + 0 + 0 + 0.397 + (–0.32)) = +6.80 × 10 -13 C
x
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PHY110W ELECTRICITY SUPPLEMENTARY GAUSS’ LAW
Type 3.
The diagram shows, in cross section, a central metal ball, two
spherical metal shells, and three spherical Gaussian surfaces of
radii R, 2R and 3R, all with the same centre. The charges on the
three objects are: ball, Q; smaller shell, 3Q; larger shell, 5Q.
Show that the magnitude of the electric field at any point on any
one of the surfaces is the same as at any other point on any of
the surfaces.
Shell
R
3R
2R
Gaussian
surface
Solution:
There is no reference to first principles here, so we can simply use the equation for the electric
field due to an enclosed point charge [which was derived earlier (p 17 of Lecture Notes) by
applying Gauss’ Law (in the form ε
0
E⋅ dA = q ) to a point charged surrounded by a
∫
concentric Gaussian sphere of radius r]:
E =
enclosed
q
4πε
r
Q
For the innermost Gaussian surface, r = R, and q (ie enclosed q) = Q, so E =
2
4πε
R
0
2
0
For the next Gaussian surface, r = 2R, and the net enclosed charge is q = (Q + 3 Q), so
4Q
Q
E = 2 2
4 πε (2 R) = 4πε
R
, ie the same value as for points on the innermost surface.
0 0
And so on. (You should be able to finish it on your own now…)
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