Download Gauss`s law and examples with insulators

Gauss’s Law
Gauss’s law is a fundamental relationship about electric fields. Let us motivate it with a
short calculation. Suppose there is a point charge q at the origin. Let’s calculate the electric flux
through a closed spherical surface of radius r centered on the point charge.
1 q
1 q
1 q
q
ˆ =
rˆirdA
dA =
(4π r 2 ) = .
Φ E = ∫∫ E idA = ∫∫
2 ∫∫S
2
S
S 4πε r 2
4πε o r
4πε o r
εo
o
Note that r = constant on the surface of the sphere and that is why it comes out of the integral;
also the surface area of a sphere of radius r is 4πr2.
Now, at first thought it might appear that this derivation hinges on the fact that the charge
lies at the center of the sphere, but this turns out not to be the case. A point charge located
anywhere inside the sphere will produce the same net flux through the surface as evidenced by
the same number of electric lines crossing that surface. Furthermore, the surface need not be a
sphere at all. Any closed surface enclosing the point charge will have the same flux through it,
since the projected area of any closed surface on the radial direction out from the point charge
will be a spherical surface.
On the other hand if the closed surface has no enclosed charge – perhaps a charge lies
outside the closed surface – then the electric flux is equal to zero, since the net number of electric
field lines crossing the closed surface is zero.
Note that the surface chosen is called a “Gaussian surface” and is not a real surface of the
problem, but is arbitrarily chosen according to the whim of the person calculating the flux. We
will see some good choices in the examples below.
Generalizing this, using the notion of superposition of the electric field vectors due to
individual enclosed charges, we can write Gauss’s law as:
q
Φ E = ∫∫ E idA = net ,enclosed ,
S
εo
where E is the net field on the surface – due to all charges, those inside as well as those outside
of S.
This is a very general law – in fact it is one of 4 fundamental equations of
electromagnetism that we will summarize at the very end of this course.
Now let’s see a number of applications of Gauss’s law, first to charged insulators.
Example 1: Spherically symmetric charge distributions
A. A uniformly charged sphere of radius a.
We want to find the electric field everywhere due to this charged sphere. First for r>a,
we choose a Gaussian surface to be a sphere of radius r centered at the sphere center
(which we put at the origin). Then
Q
Φ E = ∫∫ E idA = net ,inside , where Qnet,inside = the total charge Q on the sphere.
S
εo
a
Now, because of the symmetry of the problem, we know first of all that E is radially
r
directed (along the outward normal – and therefore along dA , so that the dot product
E idA = EdA ), and second that E can only depend on r, the distance from the origin and
not on θ and φ, the spherical coordinate angles. Because E only depends on r, it will
necessarily be a constant on the surface S of the sphere of radius r and can therefore come
out of the integral. We therefore have that
Q
S
εo
and solving for E, we find that for r>a, or outside the sphere, but no matter how close to
it, the electric field outside is the same as that of a point charge Q at the origin:
1 Q
E=
(r > a)
4πε o r 2
Now, to find the E field at an interior point, at distance r < a from the origin, we choose
our Gaussian surface to be again a sphere of radius r centered at the origin. The
calculation of the flux proceeds just as above so that, as before,
Φ E = E (4π r 2 ) ,
but now we must calculate the net charge inside our Gaussian surface. But this is given


 4π r 3   Q   4π r 3 
by using the charge density ρ = Q/Volume, as Qnet,inside = ρ 
= 
,
3 
π
4
a
3
3








 3 
r3
so that Qnet ,inside = Q 3 . Then substituting this into Gauss’s law, we have that
a
 Q   r3 
1 Qr
E (4π r 2 ) =    3  or E =
(r < a).
3
ε
a
4
πε
a
o
 o  
You can check that at r = a the two results agree, so that E is necessarily continuous.
B. A thin charged spherical shell of radius a
Again we divide space into two regions: r > a and r < a. For r > a, the argument is
identical to the previous one and we get the same point charge result. For r < a, we can
evaluate the flux over a Gaussian sphere at radius r in the same way, but the net enclosed
charge is zero and so we conclude that E = 0 inside the spherical shell. In this case E is
not continuous across the surface of the sphere, but has a discontinuous jump from zero
1
inside to
Q/a2 = σ/εo, just outside the shell.
4πε o
Φ E = E ∫∫ dA = E (4π r 2 ) =
r
a
ρ
Λ
Example 2: Cylindrically symmetric charge distributions
What is the E field around an infinite line of charge with linear charge density λ?
Choose a Gaussian surface to be a cylinder of radius r and length L with its axis on the
line of charge. By symmetry, the E field must be along the cylindrical coordinate radial
direction and can only depend on r and not on θ. Therefore the normal to the Gaussian
surface is along the electric field direction and E = constant on its surface. The flux
through the end caps of the closed cylinder is zero because the electric field is
perpendicular to the normal to the end caps. Then we can find the electric flux to be
Φ E = ∫∫ E idA = EA = E (2π rL) ,
S
where the term in brackets is the surface area of the Gaussian cylinder. According to
Gauss’s law, this is equal to the enclosed charge. The enclosed charge is simply λL so
λL
1 2λ
that we have E =
=
.
ε o (2π rL) 4πε o r
Does the line of charge need to be infinite. In principle, yes, since otherwise we
will not have the required symmetry. On the other hand so long as r is not too far from
the wire and the wire is sufficiently long the result we found will be valid.
Example 3: Planar symmetry
A. What is the electric field from an infinite plane sheet of positive charge with uniform
surface charge density σ?
Symmetry in this problem dictates that the electric field must be perpendicular to
the plan and point away from its positively charged surface. Also, the electric field can
only depend on the perpendicular distance away from the plane. Let’s orient the plane in
the y-z plane so that the x-axis is perpendicular to the surface. Then an appropriate
Α Gaussian surface is a small cylinder of cross-sectional area A that cuts through the plane
and is symmetric in x, as shown, extending a distance x = L to both sides of the plane.
To calculate the flux over the cylinder, we separately consider the flux over the
cylindrical surface and the two end caps. Since the cylindrical walls are perpendicular to
the surface, its outward normal lies parallel to the surface and hence in the flux integral,
this contribution vanishes due to the dot product between two perpendicular vectors. The
integrals over the two end caps are equal and both equal to simply EA. The charge
enclosed by the cylinder is equal to σA, since only area A of the plane is enclosed.
Putting this together, Gauss’s law states that 2EA = σA/εo or that E = (σ/2)/εo, a constant.
We reach the conclusion that the electric field is constant and directed perpendicularly
outward from the surface.
B. What is the electric field everywhere from two parallel infinite plane sheets of charge
separated by a distance d, with equal and opposite uniform surface charge densities?
Since the E field from each plane has just been found, we simply use
superposition to find the net E in each region. To the left or right of both planes E = 0,
since the two fields, oppositely directed, cancel. In between the two planes, these two
fields add so that Ebetween planes = σ/εo. This will be important when we study capacitors!!
d
-σ
+σ
We will now show how to write Gauss’s law as a differential equation. For this we need
the divergence theorem:
∫∫ E idA = ∫∫∫ ∇i E dV .
S
V
We first can write that
q = ∫∫∫ ρ dV , where ρ is the volume charge density. Therefore Gauss’s law can
V
be written as
1
ρ dV .
V
ε o ∫∫∫V
Since the volume is totally arbitrary, we can equate the integrands to get a differential equation:
ρ ( x, y , z )
∇i E ( x, y , z ) =
.
εo
This can be written in terms of the electric potential by substituting
E = −∇V
∫∫∫
∇i E dV =
and re-writing the previous equation in terms of the scalar potential. This is how more difficult
problems are actually computed, first finding V and then the electric field.
We have
ρ ( x, y , z )
∇i( −∇V ( x, y , z ) ) =
,
εo
which can be written in short-hand as
ρ ( x, y , z )
∇i∇V = ∇ 2V = −
.
εo
The left hand side is read as “del-squared of V” and the operator ∇ 2 is a scalar operator and is
known as the Laplacian operator. In Cartersian coordinates the Laplacian of V is written as
∂
∂ ˆ ∂ ˆ  ∂ ˆ ∂ ˆ ∂ ˆ
∂ 2V ( x, y, z ) ∂ 2V ( x, y, z ) ∂ 2V ( x, y, z )
j + k i i +
j + k  V ( x, y , z ) =
∇ 2V =  iˆ +
+
∂y
∂z   ∂x
∂y
∂z 
∂x 2
∂y 2
∂z 2
 ∂x
ρ ( x, y , z )
is usually the best way to solve a complicated problem. Once
εo
V is known the electric field can be calculated by taking the gradient of V.
Solving ∇ 2V ( x, y , z ) =