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Chapter 2 GAUSS’ LAW Recommended Problems: 1,4,5,6,7,9,11,13,15,18,19,21,27,29,31,35,37,39,41,43,45,47,49 ,51,55,57,61,62,69.
Chapter 2 GAUSS’ LAW
Recommended Problems:
1,4,5,6,7,9,11,13,15,18,19,21,27,29,31,35,37,39,41,43,45,47,49
,51,55,57,61,62,69.
ELECTRIC FLUX
Electric flux is a measure of the number of electric filed lines
penetrating some surface in a direction perpendicular to that
surface.
with is the angle between the E and A.
The direction of A, the area of a surface, is always normal to that
surface and points outward for closed surfaces,
cosAE=AE
(by closed surface we mean that surface which divides space into
inside and outside regions).
It is obvious from the last equation that is a scalar quantity with
the SI unit of N.m2/C.
If the electric field E is not constant over the surface in question,
surface
AE d
Note that if E is constant over the surface it should be taken out of
the integration and we recover the first equation as expected.
Consider now the closed surface in
the figure,
If E is outward and 90, hence the
flux through this element is positive.
If E is entering the surface, 90,
and so the flux for such an element
is negative.
From this argument, we can expect that if a field line entering and
leaving the same closed surface the net electric flux through the
closed surface from that line is zero.
E
dA dA
Example 24.1 What is the e.flux through a
sphere that has a radius of R=1.0 m and carries
a charge of q=1.0 C at its center.
Solution: The e.field due to the point charge at
any point on the surface of the sphere is
This field is constant over the surface so we write
q
E
2r
qkE
The field points radially outward and so E & A are parallel, i.e.,
cosAE=AE
12
62
2 1085.8
1010cos4
4
oo
qR
R
q
/CN.m1013.1 25
1cos0
0cosAE
2R
qkE
Example 24.2 A cube of edge l is
oriented in a uniform e.field, as shown.
Find the net e.flux through the surface
of the cube.
The 4-faces named in the figure and the other two unnumbered
faces (the forth and the back faces), that is
dA1 dA2
dA3
dA4
x
y
z
E
Solution: The net flux through the
cube is the sum of the fluxes through the 6-faces of the cube:
forthf
backb AdEAdEAdEAdEAdEAdE
44
33
22
11
The flux through the faces 3, 4, the back, and the forth is zero
because E and dA are perpendicular.
EA -EA
Now the angle between E and dA1 is zero, while the angle
between E and dA2 is 180o 0 EAEA
• Test Your Understanding (1)
A charge Q is placed at the center of a spherical
shell (the red one). If the radius of the shell is
increased (the black one), what happens to the
flux through the shell and the magnitude of the
electric field at the surface of the shell?
a) The flux remains the same and the field decreases.
b) The flux remains the same and the field increases.
c) The flux decreases and the field remains the same.
d) The flux and field both decrease.
Q
Gauss’ Law
the electric flux through any closed surface is equal to the net charge inside that surface divided by o, that is
o
in
q
d AE
The integral is over a
closed surface, called the
Gaussian surface.
The e. field due to the
whole charge distribution.
The surface of the
Gaussian surface. The net charge inside the
Gaussian surface.
q
S1 S2
Coulomb’s law tells us that the magnitude of
the electric field is constant everywhere on
the spherical surface and given as
2rqkE
Since E & A are parallel (=0), then the flux through S1 as
2
24 r
r
qkEAd AE
Let us verify Gauss’ law by considering a positive point charge q
surrounded by two closed surfaces: S1 is spherical, whereas S2 is
irregular.
The figure shows that the number of field lines crossing S1 is the
same as that lines crossing S2 , that is, the flux through the two
surfaces are equal and independent of their shapes.
1
o4But k
o
q
If the charge exists outside a closed surface, the electric field
lines entering the surface must leave that surface. Hence, the
electric flux through that surface is zero.
q
The practical utility of Gauss’ law lies largely in providing a smart
way to evaluate the electric filed for a charge distribution.
For this way to be as easy as possible we must be able to choose
a hypothetical closed surface (Gaussian surface) such that the
electric filed over its surface is constant.
This can be attained if the following remarks are satisfied:
(i) The charge distribution must have a high degree of symmetry
(spherical, cylindrical with infinite length, plane with infinite
extends).
(ii) The Gaussian surface should have the same symmetry as that
of the charge distribution.
(iii) The point at which E is to be evaluated should lie on the
Gaussian surface.
(iv) If E is parallel to the surface or zero at every point, then
0 AE d
(v) If E is perpendicular to the surface at every point, and since E
is constant, then
EAd AE
• Test Your Understanding (2)
Consider the charge distribution shown in
the figure. The charges contributing to the
electric flux, , through S ' and to the
electric filed, E, at a point on S ' is:
a) All the four charges contributing to both and E.
b) Only q2 and q3 contributing to both and E.
c) Only q2 and q3 contributing to while the four charges
contributing to E.
d) Non of the four charges contributing to both and E.
• Test Your Understanding (3)
Four closed surfaces, S1 through S4,
together with the charges Q , -Q , and -2Q are sketched in the figure shown. (The
colored lines are the intersections of the
surfaces with the page.) The surface that
has the largest electric flux is:
a) S1 b) S2 c) S1 and S2 d) S3
Referring to the same figure, the surface that has the smallest
electric flux is:
a) S4 b) S2 c) S2 And S4 d) S1
Example 24.4 Find the e.f a distance r from a point charge q.
Solution Since the charge distribution is spherical, we
choose the Gaussian surface as a sphere of
radius r. Now
q E
dA
r
Gaussian surface
It is clear that E and dA are parallel, E is constant over the
surface.
o
qdAE
o
qrE
24 24 r
qE
o
o
in
qdAE
q is the total charge enclosed by Gaussian surface
a) We choose an arbitrary point outside the shell.
Example 24.6 A thin spherical shell
of radius a has a total charge Q uniformly
distributed over its surface.
Solution
Find the magnitude of the electric field at a point
a) outside the shell a distance ra
b) inside the shell a distance ra r
Gaussian surface Q
a
It is clear that E is normal to the surface at every point in the
Gaussian surface, that is E and dA are parallel, so we write
o
in
qdAE
oinq
EA o
Q
Since the charge distribution is spherical, we choose a spherical
Gaussian surface concentric with the shell.
But A, the area of the Gaussian surface is 24 rA
2o4 r
QE
b) Now we choose a point inside the shell.
In this case the Gaussian surface is inside
the shell.
It is clear that there is no charge inside the
Gaussian surface, that is
0inq 0E
r
Q
a
Example 24.5 An insulating sphere
of radius a has a total charge Q uniformly
distributed through its volume.
r
Gaussian surface
Q a
Calculate the electric filed E
a) outside the sphere a distance ra
b) inside the sphere a distance ra
Solution a) Again, and because the spherical symmetry of the charge
distribution, we select a spherical Gaussian surface of radius r, concentric with the sphere.
As for the case in the previous example we write
o
in
qdAE
o
QEA
Substituting for A by 24 rA
2o4 r
QE
b) In this case we choose a spherical
Gaussian surface of radius ra.
To find the charge qin within the Gaussian
surface of volume Vin, we use the fact that
inin Vq
where is the volume charge density. Knowing that
r
Gaussian surface
Q
a
334 rVin
334
anda
Q
3
3
a
Qrqin
Now, applying Gauss’ law we obtain
o
in
qdAE
3o
324
a
QrrE
3o4 a
QrE
Note that at the center of the sphere, r=0, E=0. Is it reasonable?
Example 24.7 Find the electric field
E at a distance r from an infinite line
charge of uniform density .
+
+
+
+
+
+
+
+
+
+
+
+
dAc
dAb
dAb
h r Solution
As a Gaussian surface we select a circular
cylinder of radius r with height h and
coaxial with the line charge.
Since the cylinder has three surfaces, the integral
in Gauss's law has to be split into three parts:
the curved surface, and the two bases.
o
in
qddd
cbb
AEAEAE
From the symmetry of the system, E is parallel to both bases.
Furthermore, it has a constant magnitude and directed radially
outward at every point on the curved surface of the cylinder.
EA
The total charge inside the Gaussian surface is only
the charge of the part inside the surface, i.e.,
o
2
inqrhEEA
rE
o2
Note that if the wire is not too long its ends will be closed to any
Gaussian surface. Since the electric field at, and closed to the
ends is not uniform it will be impossible to manage the integral of
Gauss’ law.
o
2
hrhE
qin = h. Now we write
h r
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
Example 24.8 Find the electric
field E due to a nonconducting,
infinite plane with uniform surface
charge density .
Gaussian surface
Solution
To solve this problem we select as a Gaussian surface a small
cylinder whose axis is perpendicular to the plane and whose ends
each has an area A.
o
in
qddd
cbb
AEAEAE
As we do in the previous example we write Gauss’ law as
With qin is given by Aqin
dAb
dAb
dAc
EA
EA
Furthermore, E is directed normally outward and has a constant
magnitude at each point on the two ends of the cylinder.
This means that the third integral vanishes and the first two
integrals each reduce to EA.
o
2
AEA
o2
E
Note that this result agrees with the result of Example 1.9.
It is left as an exercise to show that the problem can be solved
using a Gaussian surface in the shape of parallelepiped.
• Test Your Understanding (4)
A point charge Q is located just above
the center of the flat face of a
hemisphere of radius R as shown. The
electric flux through the flat face of the
hemisphere is:
a) Q/ 4o b) Q/ 2o c) -Q/ 4o d) -Q/ 2o
CONDUCTORS IN ELECTROSTATIC
EQUILIBRIUM
If a conductor is charged, charges will move a way from each
other due to the repulsion force between them.
For the charges to be as far a way from each other as they can,
they will move to the outer surface of the conductor.
Conductors with no motion of charges are said to be in
electrostatic equilibrium. Such conductors have the following
property:
(i) Any excess charge will reside entirely on the outer surface
of an isolated conductor.
Bearing this property in mind, if a
Gaussian surface is constructed inside
such a conductor, it will not enclose
any charge. Using Gauss’s law, we
conclude that
Gaussian surface
(ii) The electric field must be zero inside any conductor in
electrostatic equilibrium.
(iii) The electric field just outside a conductor is always
perpendicular to the surface of the conductor and equal to
o
If this is not the case, the free charges will move along the surface
and this violate the condition of equilibrium.
Let us now use Gauss’s law to calculate the
magnitude of the electric filed just outside a
charged conductor. To do so we draw a
Gaussian surface in the shape of small
cylinder as shown. Gauss’s law then gives
E
o
in
qddd
cbb
AEAEAE
But the base inside the conductor has no flux through it since E=0,
and the flux through the curved surface is zero since E is normal
to the area vector of this surface. Hence, the last two integrals
vanish leaving us with
o
in
qEA
o
E
o
A
Example 24.10 A conducting sphere of
radius a has a net charge 2Q. Concentric with
this sphere is a conducting spherical shell of
inner radius b and outer radius c and has a net
charge of -Q.
a) Find the electric field in the regions inside the sphere, between
the sphere and the shell, inside the shell, and outside the shell.
b) Determine the induced charge on the inner and outer surfaces
of the shell.
Solution
Since the sphere is conducting we conclude that the electric field
in the first region is zero, i.e., E1=0.
a
b c
2Q
-Q
In the second region we select a spherical Gaussian surface with
radius a r b.
o
22 4
inqrE
2o
24
2
r
QE
In the third region the electric filed is again must be zero since this
region is inside the shell which is a conductor, i.e., E3=0.
In the last region outside the shell we construct a Gaussian
surface with radius r c.
a
b c
2Q
-Q
Since E is constant in magnitude over the Gaussian surface and
normal to it, we find from Gauss’ law
This surface enclose a total charge of (-Q+2Q)=Q. Gauss’ law is
then gives
o
22
24
QrE
2o
44 r
QE
If one construct a Gaussian surface in that region with radius
b < r < c the net charge inside that surface must be zero. So the
electric field E3 is zero as expected.
b) The charge on the sphere induces a
charge of -2Q is on the inner sphere.
The induced charge on the outer surface will be 2Q. Therefore,
the net charge on the outer surface will be Q.
2Q
-Q -2Q +2Q
Induced charges
Q
zero
