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Chapter 24
1.
2.
3.
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
q1q2
F12  ke 2 rˆ12
r

The electric field is defined as
F
E
qo

and is represented through field lines.
The force a charge experiences in an
electric filed
Fe  qE
Two examples
Example 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:
 
 E  EA  EA cos   E  A
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?
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
E  Ei Ai cos θi  Ei  Ai

In general, this becomes
E  lim
Ai 0
E 

 E  A

E  dA

i
i
surface

Review on math:
Integral over a surface.

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/C
Example 1: flux through a cube of a
uniform electric field

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


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
q
E  ke

r2
The field lines are directed radially outwards
and are perpendicular to the surface at every
point, so
 
 E   E  dA   E n dA   EdA  E  dA  E  4r 2

Combine these two equations, we have
q
q
2
 E  E  4r  k e 2 4r  4k e q 
0
r
2
The Gaussian Surface and
Gauss’s Law

Closed surfaces of various shapes can surround the
charge

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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
 
q
 E   E  dA 
0


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
 q1  q2  ...
 


 E   E  dA   (E1  E 2  ...)  dA 
0

Gauss’s Law connects electric field with its source charge
Gauss’s Law – Summary

  qin
Gauss’s law states  E  E  dA 

0


qin is the net charge inside the Gaussian surface
E represents the electric field at any point on the
surface

E 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 E for any charge configuration, in practice it is
limited to a few symmetric situations
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?