chapter 24
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Chapter 24. Review on Chapter 23 From Coulomb's Law to Gauss’s Law Applications of Gauss’s Law. Review on Chapter 23: Coulomb’s Law and the electric field definition. Coulomb’s Law: the force between two point charges The electric field is defined as - PowerPoint PPT PresentationTRANSCRIPT
Chapter 24
1. Review on Chapter 23
2. From Coulomb's Law to Gauss’s Law
3. Applications of Gauss’s Law
Review on Chapter 23: Coulomb’s Law and the electric field definition
Coulomb’s Law: the force between two point charges
The electric field is defined as
and is represented through field lines. The force a charge experiences in an
electric filed
1 212 122
ˆe
q qk
rF r
oq
FE
e qF E
Two examplesExample 23.9 (page 662) Example 23.10 (page 663)
From Coulomb’s Law to Gauss’s law
Try to calculate the electric field of A point charge An infinitely long straight wire with evenly distributed charge A wire loop A round disk An infinitely large plane A solid sphere with evenly distributed charge
Are there other ways to calculate electric field generated from a charge distribution?
Electric field is generated by source charges, are there ways to connect electric field directly with these source charges?
The answer is YES!
Some preparation:Electric Flux through a perpendicular plane
Electric flux is the product of the magnitude of the electric field and the surface area, A, perpendicular to the electric field:
ΦE = EA Compare to a water flux
in a tube: ΦW = –V1A1= V2A2
This sign means water flows into the tube, by convention.
Electric Flux, plane with an angle θ When the field lines make an
angle θ with the direction (i.e., the normal) of the surface, the flux is calculated as:
And the electric field E has to be a constant all over the area A.
Question: when this is not the case, what do you do the get the flux?
AE
cosEAEAE
Review on math: 1. direction of a surface is defined as the
(outwards) normal to that surface.2. Dot product of two vectors.
Electric Flux, General In the more general case,
look at a small area element
In general, this becomes
cosE i i i i iE A θ E A
0
surface
limi
E i iA
E
E A
d
E A
The surface integral means the integral must be evaluated over the surface in question
In general, the value of the flux will depend both on the field pattern and on the surface
When the surface is closed, the direction of the surface (i.e. the normal of it) points outwards.
The unit of electric flux is N.m2/CReview on math: Integral over a surface.
Example 1: 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
Example 2: flux through a sphere with a charge at its center.
From Coulomb’s Law to Gauss’s Law A positive point charge, q, is located at the
center of a sphere of radius r According to Coulomb’s Law, the magnitude
of the electric field everywhere on the surface of the sphere is
The field lines are directed radially outwards and are perpendicular to the surface at every point, so
Combine these two equations, we have
24AE rEdAEEdAdAEd nE
2r
qkE e
0
22
2 444
qqkr
r
qkrE eeE
The Gaussian Surface and Gauss’s Law Closed surfaces of various shapes can surround the
charge Only S1 is spherical The flux through all other surfaces (S2 and S3) are the
same. These surfaces are all called the Gaussian Surface.
Gauss’s Law (Karl Friedrich Gauss, 1777 – 1855): The net flux through any closed surface surrounding a
charge q is given by q/εo and is independent of the shape of that surface
The net electric flux through a closed surface that surrounds no charge is zero
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
Gauss’s Law connects electric field with its source charge
0
AE
qdE
0
2121 A)EE(AE
d...dE
Gauss’s Law – Summary
Gauss’s law states
qin is the net charge inside the Gaussian surface represents the electric field at any point on the
surface is the total electric field at a point in space and may have
contributions from charges both inside and outside of the surface
Although Gauss’s law can, in theory, be solved to find for any charge configuration, in practice it is limited to a few symmetric situations
E
E
E
0
AE
inE
qd
Preview sections and homework 1/27, due 2/3 Preview sections:
Section 24.3 Section 24.4
Homework: Problem 4, page 687. Problem 9, page 687. (optional = do it if you find it fun, or would like to challenge
yourself) Problem 11, page 687. (optional): On an insulating ring of radius R there evenly
distributed 73 point charges, each with a charge Q =+1 μC. The charges are fixed on the ring and cannot move. There is a bug with charge q = -0.1 μC sits at the center of the ring, and enjoys zero net force on it. When one of the charge Q is removed from the ring, what is the net force of the remaining charges exert on the poor bug?