1. Quiz 1 (lecture 4)
Ea<Eb

2. Gauss’ Law
*relates the electric fields at points on a closed Gaussian surface and the net charge enclosed by that surface.
Suppose you know E at every point on the surface and all have the same magnitude & point radially outward. Can guess that a positive charge must be inside. If you know Gauss’ law you can calculate “how much E” is intercepted by the surface. “how much” involves flux of the E field through the surface. ?
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3. Concept of Flux
Air stream with uniform velocity, v, flows through a loop of area A 
can define a flux: v
Two approaches to calculating E-filed due to charge distributions. Although expressed in different ways, the laws are equivalent for describing the relation between charge and electric field.
Apply Coulomb’s Law by writing the differenital form of the law for a differential elment and then integrate
Use Gauss’ Law which uses flux (E dot A) is equal to the enclosed charge. Gauss’ Law is used for symmetric shapes.
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4. Flux of an Electric Field
For a closed surface, points outward
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5. Flux of an Electric Field
Gaussian surface of arbitrary shape immersed in a non-uniform electric field
Two approaches to calculating E-filed due to charge distributions. Although expressed in different ways, the laws are equivalent for describing the relation between charge and electric field.
Apply Coulomb’s Law by writing the differenital form of the law for a differential elment and then integrate
Use Gauss’ Law which uses flux (E dot A) is equal to the enclosed charge. Gauss’ Law is used for symmetric shapes.
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6. Example: Flux through a Cube
Consider a uniform electric field oriented parallel to the x - direction. Find the net flux through the surface of a cube of edge L oriented as shown. y E x z
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7. Example: Flux R 2R q
Consider two Gaussian spheres (of radius R and 2R ) drawn around a single charge as shown. Which of the following statements about the net electric flux through the true surfaces (F2R and FR) is true?
B. FR > F2R
C. FR = F2R
A. FR < F2R
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8. Gauss’ Law
*Relates net flux, , of an electric field through a closed surface to the net charge that is enclosed by the surface.
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9. Gauss’ Law and Coulomb’s Law
Are equivalent and we can derive one from the other.
Gaussian Surface +
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10. Gauss’ Law: Spherical Symmetry
Shell Theorems
A shell of uniform charge attracts or repels a charged particle that is outside the shell as if all the shell’s charge were concentrated at the center of the shell
A shell of uniform charge exerts no electrostatic force on a charged particle that is located inside the shell.
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11. Gauss’ Law: Spherical Symmetry
Prove:
A shell of uniform charge attracts or repels a charged particle that is outside the shell as if all the shell’s charge were concentrated at the center of the shell
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12. Gauss’ Law: Spherical Symmetry
Prove
A shell of uniform charge exerts no electrostatic force on a charged particle that is located inside the shell.
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13. Gauss’ Law: Spherical Symmetry
Demo: 5A-13 No Internal Field
Electric Field inside and outside a shell of uniform charge distribution
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14. Gauss’ Law: Spherical Symmetry
Spherically Symmetric Charge Distribution
Outside
Inside
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15. Gauss’ Law: Spherical Symmetry
Spherically Symmetric Charge Distribution r R E
Outside
Inside
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16. Infinite Line of charge density 
From Symmetry: E-field only depends on distance r from line
Therefore, select the Gaussian surface to be a cylinder of radius r and length h aligned with the x-axis. y Er Er
+ + +
+ + + x h
Apply Gauss’ Law and assume uniform charge density :
On the ends,
On the barrel,
and
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17. Quiz 2 (lecture 4)
Two long, charged concentric cylinders have radii of r1=3.0 cm and r2=6.0 cm. The charge per unit length is 5.0 x 10-6 C/m on the inner cylinder and -7.0 x 10-6 C/m on the outer cylinder. If you were using Gauss’ Law to find the electric field at r = 8.0 cm, where r is the radial distance from the common axis, would the enclosed charge be r1 r2 r
Positive
Zero
Negative
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18. Example: Gauss’ Law: Cylindrical Symmetry
Two long, charged concentric cylinders have a radii of 3.0 cm and 6.0 cm. The charge per unit length is 5.0 x 10-6 C/m on the inner cylinder and -7.0 x 10-6 C/m on the outer cylinder. Find the electric field at (a) r = 4.0 cm and (b) r = 8.0 cm, where r is the radial distance from the common axis.
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19. Example: Gauss’ Law: Spherical Symmetry
A spherically symmetric charge distribution has a charge density given by =a/r where a is a constant. Find the electric field as a function of r.
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20. Quiz 3 (lecture4)
Consider three point charges fixed at the vertices of an equilateral triangle of side d as shown. Two of the charges are positive (+1µC and +2µC) and one is negative (-1µC). Consider a sphere of radius R=d/2 centered on the equilateral triangle as shown. What is the sign of the total electric flux through this sphere produced by the fields of the three charges. +1 +2 -1 d R
(A) The flux through the sphere is negative
(B) The flux through the sphere is zero
(C) The flux through the sphere is positive
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21. Quiz 4 (lecture 4)
Two identical point charges are each placed inside a large cube. One is at the center while the other is close to the surface. Which statement about the net electric flux, Φnet, through the surface of the cube is true?
Φnet is larger when the charge is at the center.
Φnetis larger when the charge is near the surface.
Φnet is the same (and not zero).
Not enough information to tell.
Φnet is zero in both cases. +Q
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22. Quiz 5 (lecture4) R a
E=?
You are told to use Gauss' Law to calculate the electric field at a distance R away from a charged cube of dimension a. . Which of the following Gaussian surfaces is best suited for this purpose?
a sphere of radius R+1/2a
a cube of dimension R+1/2a
a cylinder with cross sectional radius of R+1/2a and arbitrary length
This field cannot be calculated using Gauss' law
None of the above
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23. Quiz 6 (lecture 4)
Three cylinders each have a charge of Q distributed uniformly over the volume of the cylinder. Concentric with each cylinder is a cylindrical Gaussian surface, all three with the same radius. Rank the Gaussian surfaces according to the electric field at any point on the surface. 1 2 3
Gaussian Surface
Cylinder
1 > 2 > 3
1 = 2 = 3
1 < 2 < 3
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24. Vector Mathematics
Scalar product
Vector product
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25. Gauss’ Law: Symmetry ALWAYS TRUE!
In cases with symmetry can pull E outside and get
In General, integral to calculate flux is difficult…. and not useful!
To use Gauss’ Law to calculate E, need to choose surface carefully!
1) Want E to be constant and equal to value at location of interest OR
2) Want E dot A = 0 so doesn’t add to integral
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26. Gauss’ Law Symmetry ALWAYS TRUE!
In cases with symmetry can pull E outside and get
Spherical
Cylindrical
Planar
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