第1章静止电荷的电场(之二)

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Gauss’ Law :? —— In cases of high symmetry, this will be MUCH EASIER than the brute force method. Furthermore, Gauss’ law serves as a guide for understanding more complicated problems.
a
a

Fundamental Law of Electrostatics

Coulomb’s Law Force between two point charges Gauss’ Law Relationship between electric flux and charges The two laws are fundamental laws of electrostatics. Coulomb’s law Gauss’ law Gauss’ law Coulomb’s law
Chapter 1, ACT 1

Examine the electric field lines produced by the charges in this figure.

Which statement is true? q1 q2
(a) q1 and q2 have the same sign (b) q1 and q2 have the opposite signs and (c) q1 and q2 have the opposite signs and

Consider three imaginary spheres centered on:
a) +q (blue) b) -q (red) c) midpoint (yellow)

All lines leave a),
flux through a) is positive All lines enter b),
Electric Flux…

In more general situations, the electric field may vary over the surface. Therefore, the electric flux Fe through the surface is given by:

The electric flux Fe through the surface is defined as the product of electric field strength E and surface area S.
Electric Flux…

If the surface is not perpendicular to the field. The normal to the surface of area S is at an angle to the uniform electric field. The electric flux Fe through the surface is given by:
Electric Flux…

Let’s look at the formula (1.16) on page 23.
Electric Flux…

First consider an electric field that is uniform in both magnitude and direction. The rectangular surface is perpendicular to the field. The normal to the surface of area S is parallel to the uniform electric field.
dS 4 r
closed surface
q r
P
E
Derivation of Gauss’ Law…

Electric Flux

Electric flux : Define: Number of field lines passing through an area of interest. Units: N·m2/C. The number of lines per unit area is proportional to the magnitude of . What’s the relationship between electric flux and ?

The Story Thus Far

We want to be able to calculate the electric fields from various charge arrangements. Two ways:

Brute Force: Add up / integrate contribution from each charge —— Often this is pretty difficult. Ex: infinite line of charge
left
right
The Sign Problem...

For a closed surface, we choose the direction of dS pointing out.
A differential surface element, with its vector
dS
Electric Dipole: Flux
Electric Flux…

Electric flux Fe through the closed surface S is given by:
Ex: Flux through a cube or a sphere
The Sign Problem

For an open surface we can choose the direction of dS two different ways. to the left or to the right flux would be different these two ways different by a minus sign
c
a

flux through b) is negative

Equal amounts of leaving
b
and entering lines for c),
flux through c) is zero
Chapter 1, Act 2

Imagine a cube of side a positioned in a region of constant electric field as shown Which of the following statements about the net electric flux Fe through the surface of this cube is true?
Chapter 1(continued)
c
a
r a a
b
a
r
Today’s Agenda
Electric Flux（电通量） Gauss’ Law（高斯定理） Application of Gauss’ Law Uniformly Charged Sphere Infinite Line of Charge Infinite Sheet of Charge Two Infinite Sheets of Charge

There is an easier way. Gauss’ Law states the net flux Φe is proportional to the NET enclosed charge qin . The NET charge is zero in the cube. But, what is Gauss’ Law ???


( Coulomb’s law Gauss’ law ) From Coulomb’s law, we know that the magnitude of q the electric field of a positive point charge is E 2 4 r 0 and the direction of E is radial outward. The net electric flux through the closed surface of radius r is:
(a) Fe = 0
(b) Fe 2a2
(c) Fe 6a2
Chapter 1, Act 2 Solution

Therefore, the total flux through the cu Fe = 0
(b) Fe 2a2
(c) Fe 6a2
Chapter 1, Act 2 Solution
the electric field.
E dS on the bottom face is negative. ( dS is down; E is up)

E dS on the top face is positive. ( dS is up; E is up)

(c) q1 and q2 have the opposite signs and Gauss’ law will make it easier!