1. Norah Ali Al-moneef King Saud university
Chapter electric flux
24.1 Electric Flux
24.2 Gauss’s Law
24.3 Application of Gauss’s Law to Various Charge Distributions
A Spherically Symmetric Charge Distribution (An insulating solid sphere )
The Electric Field Due to a Point Charge
The Electric Field Due to a Thin Spherical Shell
The Electric Field Due to A Cylindrically Symmetric Charge Distribution
the electric field due to an infinite plane
24.4 Conductors in Electrostatic Equilibrium
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2. Norah Ali Al-moneef King Saud university
We will use a “charge density” to describe the distribution of charge.
This charge density will be different depending on the geometry 2
Norah Ali Almoneef
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3. Norah Ali Al-moneef King Saud university
Electric Flux
Electric flux is the product of the magnitude of the electric field and the surface area, A, perpendicular to the field
ΦE = EA
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4. Norah Ali Al-moneef King Saud university
24.1 Electric Flux
The electric flux is proportional to the number of electric field lines penetrating some surface
The field lines may make some angle θ with the perpendicular to the surface
Then ΦE = EA cos θ
The flux is a maximum
when the surface is
perpendicular to the field
The flux is a minimum
(zero) when the surface is (parallel) to the field
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5. Norah Ali Al-moneef King Saud university
Electric Flux
If the field varies over the surface, Φ = EA cos θ is valid for only a small element of the area
In the more general case, look at a small area element
In general, this becomes
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6. Norah Ali Al-moneef King Saud university
Electric Flux
The surface integral means the integral must be evaluated over the surface in question
The value of the flux depends both on the field pattern and on the surface
SI units: N.m2/C
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7. Electric Flux, Closed Surface
For a closed surface, by convention, the A vectors are perpendicular to the surface at each point and point outward
(1) θ < 90o, Φ > 0
(2) θ = 90o, Φ = 0
(3) 180o > θ > 90o, Φ < 0
The net flux through the surface is proportional to the number of lines leaving the surface minus the number entering the surface
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8. Norah Ali Al-moneef King Saud university
Example : flux through a cube of
a uniform electric field
The field lines pass through two surfaces perpendicularly and are parallel to the other four surfaces
For side 1, ΦE = -El 2
For side 2, ΦE = El 2
For the other sides,
ΦE = 0
Therefore, Φtotal = 0
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9. Norah Ali Al-moneef King Saud university
24-2 Electric lines of flux and Derivation of Gauss’ Law using Coulombs law
Consider a sphere drawn around a positive point charge. Evaluate the net flux through the closed surface.
Net Flux =
For a Point charge E=kq/r2
Gauss’ Law
The flux of electric field through a closed surface is proportional to the charge enclosed.
Gauss’ Law 9
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10. Norah Ali Al-moneef King Saud university
Gaussian surfaces of various shapes can surround the charge (only S1 is spherical)
The electric flux is proportional to the number of electric field lines penetrating these surfaces, and this number is the same
Thus the net flux through any closed surface surrounding a point charge q is given by q/εo and is independent of the shape of the surface
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11. Norah Ali Al-moneef King Saud university
Gauss’ Law
If the charge is outside the closed surface of an arbitrary shape, then any field line entering the surface leaves at another point
Thus the electric flux through a closed surface that surrounds no charge is zero
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12. Norah Ali Al-moneef King Saud university
Gauss’ Law
Since the electric field due to many charges is the vector sum of the electric fields produced by the individual charges, the flux through any closed surface can be expressed as
Although Gauss’s law can, in theory, be solved to find for any charge configuration, in practice it is limited to symmetric situations
One should choose a Gaussian surface over which the surface integral can be simplified and the electric field determined
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13. Norah Ali Al-moneef King Saud university
Problem 24-9
The following charges are located inside a submarine: 5.00 μC, –9.00 μC, 27.0 μC, and –84.0 μC.
Calculate the net electric flux through the hull of the submarine.
Is the number of electric field lines leaving the submarine greater than, equal to, or less than the number entering it?
(b) Since the net electric flux is negative, more lines enter than leave the surface.
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14. Problem solving guide for Gauss’ law
Use the symmetry of the charge distribution to determine the pattern of the field lines.
Choose a Gaussian surface for which E is either parallel to or perpendicular to dA.
If E is parallel to dA, then the magnitude of E should be constant over this part of the surface. The integral then reduces to a sum over area elements.
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15. Gauss’ Law and Coulomb law
Example 24.4 The Electric Field Due to a Point Charge
Starting with Gauss’s law, calculate the electric field due to an isolated point charge q.
The field lines are directed radially outwards and are perpendicular to the surface at every point, so
The angle  between E and dA is zero at any point on the surface, we can re-write Gauss’ Law as
E has the has same value at all points on the surface
E is can be moved out
Integral is the sum of surface area
A spherical Gaussian surface centered on a point charge q
Coulomb’s Law
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16. Norah Ali Al-moneef King Saud university
Example 24.5 A Spherically Symmetric Charge Distribution
An insulating solid sphere of radius a has a uniform volume charge density ρ and carries total charge Q.
Find the magnitude of the E-field at a point outside the sphere
Find the magnitude of the E-field at a point inside the sphere
For r > a
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17. Norah Ali Al-moneef King Saud university
b--Find the magnitude of the E-field at a point inside the sphere r < a
Now we select a spherical Gaussian surface with radius r < a. Again the symmetry of the charge distribution allows us to simply evaluate the left side of Gauss’s law just as before. r a  Q
The charge inside the Gaussian sphere is no longer Q. If we call the Gaussian sphere volume V’ then“V
Volume charge density“: ρ = charge / unit volume is used to characterize the charge distribution   
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18. Field Due to a Spherically Symmetric Charge Distribution
Inside the sphere, E varies linearly with r (E → 0 as r → 0)
The field outside the sphere is equivalent to that of a point charge located at the center of the sphere
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19. Norah Ali Al-moneef King Saud university
Example 24.6 The Electric Field Due to a Thin Spherical Shell
A thin spherical shell of radius a has a total charge Q distributed uniformly over its surface Find the electric field at points
(A) outside and
(B) inside the shell.
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20. Norah Ali Al-moneef King Saud university
Let’s start with the Gaussian surface outside the sphere of charge, r > a
We know from symmetry arguments that the electric field will be radial outside the charged sphere
If we rotate the sphere, the electric field cannot change
Spherical symmetry
Thus we can apply Gauss’ Law and get
… so the electric field is
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21. Norah Ali Al-moneef King Saud university
Let’s let’s take the Gaussian surface inside the sphere of charge, r < a
We know that the enclosed charge is zero so
We find that the electric field is zero everywhere inside spherical shell of charge
Thus we obtain two results
The electric field outside a spherical shell of charge is the same as that of a point charge.
The electric field inside a spherical shell of charge is zero.
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22. Norah Ali Al-moneef King Saud university
The Electric Field Due to A Cylindrically Symmetric Charge Distribution
Find the E-field a distance r from a line of positive charge of infinite length and constant charge per unit length λ.
Symmetry Þ E field must be ^ to line and can only depend on distance from line
Select a cylinder as Gaussian surface. The cylinder has a radius of r and a length of ℓ
is constant in magnitude and parallel to the surface (the direction of a surface is its normal!) at every point on the curved part of the surface (the body of the cylinder.
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23. Norah Ali Al-moneef King Saud university
Line of Charge
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24. Norah Ali Al-moneef King Saud university
Example 24.8 A Plane of Charge
Find the electric field due to an infinite plane of positive charge with uniform surface charge density σ.
Assume that we have a thin, infinite non-conducting sheet of positive charge
The charge density in this case
is the charge per unit area,
From symmetry, we can see
that the electric field will be
perpendicular to the surface of the sheet
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25. Norah Ali Al-moneef King Saud university
To calculate the electric field using Gauss’ Law, we assume a Gaussian surface in the form of a right cylinder with cross sectional area A and height 2r, chosen to cut through the plane perpendicularly.
Because the electric field is perpendicular to the plane everywhere, the electric field will be parallel to the walls of the cylinder and perpendicular to the ends of the cylinder.
Using Gauss’ Law we get
the electric field from an infinite
non-conducting sheet with charge density 
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26. 24-4Conductors in Electrostatic Equilibrium
By electrostatic equilibrium we mean a situation where
there is no net motion of charge within the conductor
The electric field is zero everywhere inside the conductor
Any net charge resides on the conductor’s surface
The electric field just outside a charged conductor is perpendicular to the conductor’s surface
On an irregularly shaped conductor, the surface charge density is greatest at locations where the radius of curvature of the surface is smallest.
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27. 24-4Conductors in Electrostatic Equilibrium
The electric field is zero everywhere inside the conductor
Why is this so?
Place a conducting slab in an external field, E.
If there was a field in the conductor the charges would accelerate under the action of the field.
Ein
The charges in the conductor move creating an internal electric field that cancels the applied field on the inside of the conductor E E
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28. Norah Ali Al-moneef King Saud university
When electric charges are at rest, the electric field within a conductor is zero.
On an irregularly shaped conductor, the surface charge density is greatest at locations where the radius of curvature of the surface is smallest.
The electric field is stronger where the surface is more sharply curved.
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29. Charge Resides on the Surface
Choose a Gaussian surface inside but close to the actual surface
The electric field inside is zero
There is no net flux through the gaussian surface
Because the gaussian surface can be as close to the actual surface as desired, there can be no charge inside the surface
Since no net charge can be inside the surface, any net charge must reside on the surface
Gauss’s law does not indicate the distribution of these charges, only that it must be on the surface of the conductor
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30. E-Field’s Magnitude and Direction
Choose a cylinder as the Gaussian surface
The field must be perpendicular to the surface
If there were a parallel component to E, charges would experience a force and accelerate along the surface and it would not be in equilibrium
The net flux through the surface is through only the flat face outside the conductor
The field here is perpendicular to the surface
Applying Gauss’s law
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31. Under electrostatic conditions the electric field inside a solid conducting sphere is zero. Outside the sphere the electric field drops off as 1 / r2, as though all the excess charge on the sphere were concentrated at its center.
Electric field = zero (electrostatic) inside a solid conducting sphere
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32. Norah Ali Al-moneef King Saud university
homework
problem1 - Charge in a Cube Q
Q=3.76 nC is at the center of a cube. What is the electric flux through one of the sides?
Problem 2 :
Two large thin metal plates with surface charge densities of opposite signs but equal magnitude of ×10–20 Cm–2 are placed parallel and close to each other. What is the field
To the left of the plates?
(ii) To the right of the plates?
(iii) Between the plates?
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33. Norah Ali Al-moneef King Saud university
problem3 :
A spherical shell of radius 10 cm has a charge 2×10–6 C distributed uniformly over its surface. Find the electric field
(a) Inside the shell
(b) Just outside the shell
(c) At a point 15 cm away from the centre
problem4 :
A 5 nC charge is placed inside of a cubic surface. The cube is measured to be 0.5 m per side. What is the flux through the cube?
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34. Norah Ali Al-moneef King Saud university
problem5 :
Ex) What is the electric flux through this disk of radius
r = 0.10 m if the uniform electric field has a magnitude E = 2.0x103 N/C?
problem6 :
Consider a thin spherical shell of radius 14.0 cm with a total charge of 32.0 μC distributed uniformly on its surface. Find the electric field
10.0 cm and
20.0 cm from the center of the charge distribution.
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35. Norah Ali Al-moneef King Saud university
problem7 :
A sphere of radius 8.0 cm carries a uniform volume charge density r = 500 nC/m3. What is the electric field magnitude at r = 8.1 cm?
Problem 8:
A long, straight wire is surrounded by a hollow metal cylinder whose axis coincides with that of the wire. The wire has a charge per unit length of λ, and the cylinder has a net charge per unit length of 2λ. From this information, use Gauss’s law to find (a) the charge per unit length on the inner and outer surfaces of the cylinder and (b) the electric field outside the cylinder, a distance r from the axis.
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36. Norah Ali Al-moneef King Saud university
Problem 9:
A uniformly charged, straight filament 7.00 m in length has a total positive charge of 2.00 μC. An uncharged cardboard cylinder 2.00 cm in length and 10.0 cm in radius surrounds the filament at its center, with the filament as the axis of the cylinder. Using reasonable approximations, find (a) the electric field at the surface of the cylinder and
(b) the total electric flux through the cylinder.
Problem 10:
A square plate of copper with 50.0-cm sides has no net charge and is placed in a region of uniform electric field of 80.0 kN/C directed perpendicularly to the plate. Find
the charge density of each face of the plate and
the total charge on each face.
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37. Norah Ali Al-moneef King Saud university
Problem 11:
A square plate of copper with 50.0-cm sides has no net charge and is placed in a region of uniform electric field of 80.0 kN/C directed perpendicularly to the plate. Find
the charge density of each face of the plate and
the total charge on each face.
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38. Norah Ali Al-moneef King Saud university
Summary:
Gauss’ Law
· Gauss’ Law depends on the enclosed charge only
If there is a positive net flux there is a net positive charge enclosed
If there is a negative net flux there is a net negative charge enclosed
If there is a zero net flux there is no net charge enclosed
Gauss’ Law works in cases of symmetry
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39. GAUSS LAW – SPECIAL SYMMETRIES
SPHERICAL
(point or sphere)
CYLINDRICAL
(line or cylinder)
PLANAR
(plane or sheet)
CHARGE
DENSITY
Depends only on radial distance
from central point
Depends only on
perpendicular distance from line
Depends only on perpendicular distance from plane
GAUSSIAN
SURFACE
Sphere centered at point of symmetry
Cylinder centered at axis of symmetry
Pillbox or cylinder
with axis
perpendicular to plane
ELECTRIC
FIELD E
E constant at surface
E ║A - cos  = 1
E constant at curved surface and E ║ A
E ┴ A at end surface
cos  = 0
E constant at end surfaces and E ║ A
E ┴ A at curved surface
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40. Norah Ali Al-moneef King Saud university
Geometry
Charge Density
Gaussian surface
Electric field
Linear
l = q/L
Cylindrical, with axis along line of charge
Sheet or Plane
s = q/A
Cylindrical, with axis along E.
Spherical
r = q/V
Spherical, with center on center of sphere
Line of Charge
Conducting
Nonconducting
r  R
r < R
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41. Norah Ali Al-moneef king saud university
1- The electric field has a magnitude of 3.0 N/m at a distance of 60 cm from a point charge. What is the charge?
(a) 1.4 nC
(b) 120 pC
(c) 36 mC
(d) 12 C
(e) 3.0 nC
2- An electron traveling horizontally from North to South enters a region where a uniform electric field is directed downward. What is the direction of the electric force exerted on the electron once it has entered the field?
(a) downward
(b) upward
(c) to the east
(d) to the west
(e) to the north
4/17/2017
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42. Norah Ali Al-moneef king saud university
A conducting sphere has a net charge of −4.8 × 10−17 C. What is the approximate number of excess electrons on the sphere?
(a) (b) (c) (d) (e) 500
Two point charges, 8x10-9 C and -2x10-9 C are separated by 4 m. The electric field magnitude (in units of V/m) midway between them is:
9x B) 13, C) 135,000
D) 36x E) 22.5
4/17/2017
Norah Ali Al-moneef king saud university