Download Handout 2: Electric flux and Gauss` Law Electric flux Consider a

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Handout 2: Electric flux and Gauss’ Law
Electric flux
Consider a uniform electric field 𝐄 passing
through a surface area 𝐴 as in Figure 1. The direction of
the vector area 𝐀 is defined to be perpendicular to the
surface. The electric flux through the surface is defined
as
Φ = 𝐄 ∙ 𝐀.
If electric field is not constant, the electric flux is given in
form of integral over a surface 𝑆:
Φ =
𝐄 ∙ 𝑑𝐀.
𝑆
Example 1 A thin circular disc of radius 𝑅 = 10 cm is
placed in a uniform electric field 𝐸 = 500 NC-1. The
normal to the disc is making angle 𝜃 = 30° with the electric field. Determine the flux through the disc.
Example 2 Consider a closed triangular box resting within a horizontal electric field of magnitude 𝐸 = 7.8 ×
104 NC-1 as shown in Figure.
Calculate the electric flux through
a) he vertical rectangular surface,
b) the slanted surface,
c) the entire surface of the box.
Figure 1: Electric flux
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Gauss’ law
Consider Figure 2. An electric charge is inside a
closed surface. The electric field from the charge causes
electric flux through the surface. Gauss’ law states that
the total electric flux through a closed surface is equal to
the net electric charge inside the surface divided by 𝜀0 :
𝐄 ∙ 𝑑𝐀 =
𝑆
𝑄𝑖𝑛
.
𝜀0
It can be deduced from the Gauss’ law that if there is no
net charge inside a closed surface, the electric flux
through the whole surface must be zero. The example of
this can be seen in Figure 3.
Figure 2: Flux from a charge
inside a closed surface
Example 3 Charge 𝑞1 = 1.5 μC is placed inside cube and
charge 𝑞2 = 2.0 μC is outside. Determine the total electric flux through the cube.
Example 4 Given that 𝑄 = 1.6 nC, determine the electric
flux through the surface 𝑆1 , 𝑆2 , 𝑆3 and 𝑆4 .
Example 5 A charge 𝑞 = 1.5 × 10−7 C is located at distance 𝑑/2 from the center of a square plate of length 𝑑.
Calculate the flux through the plate.
Figure 3: The total flux through the
closed surface is zero because 𝑄𝑖𝑛 = 0.
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Applications of Gauss’ law
Gauss’ law is a powerful technique used to find
electric field from a system of charge with certain symmetry (spherical, cylindrical). The method of finding
electric field by using Gauss’ law follows the following
steps:
1. Imagine electric field lines from the charge.
2. Draw Gaussian surface such that 𝐄 is perpendicular to the surface and the magnitude of 𝐄 is constant throughout the surface:
𝐄 ∙ 𝑑𝐀 = 𝐸
𝑆
𝑑𝐴.
Figure 4: Spherical Gaussian surface
𝑆
3. Determine 𝑄𝑖𝑛 .
4. Solve for 𝐸:
𝐸
𝑑𝐴 =
𝑆
𝑄𝑖𝑛
.
𝜀0
Point charge Point charge creates electric field
with spherical symmetry. Therefore, it is sensible to
choose the Gaussian surface to be a sphere of radius 𝑟
around the charge as in Figure 4. From Gauss’ law,
𝑞
𝐄 ∙ 𝑑𝐀 = 𝐸
𝑑𝐴 = 𝐸(4𝜋𝑟 2 ) =
.
𝜀0
𝑆
𝑆
Therefore, the electric field on the Gaussian surface
𝑞
𝐸(𝑟) =
.
4𝜋𝜀0 𝑟 2
Line of charge A line of charge creates electric
field with cylindrical symmetry. Hence, it is helpful to
draw the Gaussian surface to be a cylinder of radius 𝑟
and length 𝑙 around the line of charge as in Figure 5. Because the flat surfaces of the cylinder are parallel to the
field, the flux through these surfaces is zero; the flux
comes from the curved surface of the cylinder only.
From Gauss’ law,
𝐄 ∙ 𝑑𝐀 = 𝐸
𝑆
𝑑𝐴 = 𝐸(2𝜋𝑟𝑙) =
𝑆
𝑄𝑖𝑛
.
𝜀0
The electric field on the Gaussian surface is given by
𝐸 𝑟 =
𝑄𝑖𝑛
𝜆
=
,
2𝜋𝜀0 𝑟𝑙
2𝜋𝜀0 𝑟
Figure 5: Cylindrical Gaussian surface
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where 𝜆 = 𝑄𝑖𝑛 𝑙 is charge per unit length along the line.
It should be noted that the above treatment is valid only when the line is infinitely long so that the cylindrical symmetry is attained.
Surface charge Consider a positively charged
plate as in Figure 6. The electric field points away from
both sides of the plate in the direction perpendicular of
the plate. If the plate is large enough, the field can be
taken to be uniform. Hence, the reasonable Gaussian
surface is a “pillbox” through the plate. The flux through
the curved surface of the pillbox is zero, leaving the flux
through the two flat surfaces each of area 𝐴. From Gauss’
law,
𝐄 ∙ 𝑑𝐀 = 𝐸
𝑆
𝑑𝐴 = 𝐸(2𝐴) =
𝑆
Figure 6: Gaussian “pillbox”
𝑄𝑖𝑛
.
𝜀0
Therefore, the electric field on the flat surface of the
pillbox is given by
𝐸 =
𝑄𝑖𝑛
𝜎
=
,
2𝜀0 𝐴
2𝜀0
where 𝜎 is charge per unit area of the plate.
Note that the electric field strength from a infinitely large plated does not depend on the distance from
the plate; the electric field is uniform near the plate.
Parallel charged plates Consider two parallel
charged plates of equal magnitude and opposite signs in
Figure 7. Each plate produced the field according to the
formula 𝐸 = 𝜎 2 𝜀0 . By considering the directions of the
fields from the plates, the fields cancel outside the plates
and add up between the plates. Therefore, the field outside the plates is zero and the field between the plates is
given by
𝐸 =
𝜎
.
𝜀0
Example 6 The electric charge is spread on the surface
of a conductor. Use Gauss’ law to explain why the electric field inside conductors is always zero.
Figure 7: Electric field between two
parallel plates
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Example 7 Two parallel plates with equal surface charge
density 𝜎 but opposite signs are brought close to each
other. Determine the pressure on each plate.
*Example 8 An insulating solid sphere of radius 𝑅 has
electric charge 𝑄 uniformly spread inside. Let 𝑟 be distance measured from the center of the sphere.
a) Determine the electric field 𝐸(𝑟) when 𝑟 < 𝑅 and
𝑟 ≥ 𝑅.
b) Sketch the graph 𝐸(𝑟).
Example 9 A spherical insulator of radius 𝑎 is placed at
the center of a hollow spherical conductor. The inner
and outer radii of the conductor are 𝑏 and c respectively.
A charge 𝑄 is given to the insulator. Find the formulae
for electric field 𝐸(𝑟) for 𝑎 < 𝑟 < 𝑏, 𝑏 < 𝑟 < 𝑐 and 𝑟 > 𝑐.