Topics: • Symmetry • The Concept of Flux • Calculating Electric Flux • Gauss’s Law • Using Gauss’s Law • Conductors in Electrostatic Equilibrium
PHYS 1022: Chap. 28, Pg 2
1 New Topic PHYS 1022: Chap. 28, Pg 3
For uniform field, if E and A are perpendicular, define
If E and A are NOT perpendicular, define scalar product
Unit of electric flux: N . m2 / C Electric flux is proportional to the number of field lines passing through the area. And to the total change enclosed PHYS 1022: Chap. 28, Pg 4
2 + +
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+ +
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3 r + ½r + 2r
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r + ½r + 2r
PHYS 1022: Chap. 28, Pg 8
4 A metal block carries a net positive charge and put in an electric field. 1) 0 What is the electric field inside? 2) positive 3) Negative 4) Cannot be determined
If there is an electric field inside the conductor, it will move the charges in the conductor to where each charge is in equilibrium. That is, net force = 0 therefore net field = 0!
PHYS 1022: Chap. 28, Pg 9
A metal block carries a net charge and 1) All inside the conductor put in an electric field. Where does 2) All on the surface the net charge reside ? 3) Some inside the conductor, some on the surface
The net field = 0 inside, and so the net flux through any surface drawn inside the conductor = 0. Therefore the net charge inside = 0!
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5 PHYS 1022: Chap. 28, Pg 11
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6 PHYS 1022: Chap. 28, Pg 13
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7 Group Work
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Group Work
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8 New Topic PHYS 1022: Chap. 28, Pg 17
Gauss' Law The total electric flux through a closed surface is equal to the total charge enclosed by that surface, divided by ε0.
-12 2 2 ε0=8.85x10 C /N.m is called permittivity of free space The k in Coulomb’s law 2 F=kq1q2/r is related to ε0 by
PHYS 1022: Chap. 28, Pg 18
9 New Topic PHYS 1022: Chap. 28, Pg 19
Gauss’s law can be used to find out the electric fields around symmetric charge distributions
PHYS 1022: Chap. 28, Pg 20
10 1. Gauss’s law applies only to a closed surface, called a Gaussian surface. 2. A Gaussian surface is not a physical surface. It need not coincide with the boundary of any physical object (although it could if we wished). It is an imaginary, mathematical surface in the space surrounding one or more charges. 3. We can’t find the electric field from Gauss’s law alone. We need to apply Gauss’s law in situations where, from symmetry and superposition, we already can guess the shape of the field.
PHYS 1022: Chap. 28, Pg 21
Some charge distributions have translational, rotational, or reflective symmetry. If this is the case, we can determine something about the field it produces: The symmetry of an electric field must match the symmetry of the charge distribution.
For example, the electric field of a cylindrically symmetric charge distribution a) cannot have a component parallel to the cylinder axis. b) cannot have a component tangent to the circular cross section.
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11 PHYS 1022: Chap. 28, Pg 23
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12 PHYS 1022: Chap. 28, Pg 25
Whenever you can, make the electric field parallel to the area vector (normal vector).
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13 PHYS 1022: Chap. 28, Pg 27
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14 PHYS 1022: Chap. 28, Pg 29
We place a positive charge q on a solid conducting sphere with radius R. Find the electric field at any point inside or outside the sphere. + This problem has spherical symmetry. For a spherical surface + + of any radius, the surface integral becomes R + + + + Choose the sphere to be inside the + conductor (r Choose the sphere to be outside the conductor (r>R). Since the entire 2 charge is enclosed: 4π r E=q/ ε0. So PHYS 1022: Chap. 28, Pg 30 15 Positive electric charge Q is distributed uniformly throughout the volume of an insulating sphere with radius R. Find the E field both inside and outside of the sphere. This problem has spherical symmetry. The E field must be +Q radially outward. Choose Gaussian surface as a sphere. R Inside: Outside like a point charge: PHYS 1022: Chap. 28, Pg 31 If the inner shell has a net charge +q, and the outer shell has no net charge, how is the charge distributed on the inside and outside wall of each sphere? find the E field everywhere. +q r a b c r>d, E=kq/r2 PHYS 1022: Chap. 28, Pg 32 16 PHYS 1022: Chap. 28, Pg 33 17