dr. jie zouphy 13611 chapter 24 gauss’s law (cont.)
TRANSCRIPT
Dr. Jie Zou PHY 1361 1
Chapter 24
Gauss’s Law (cont.)
Dr. Jie Zou PHY 1361 2
Outline
Gauss’s law (24.2) Application of Gauss’s law to
various charge distributions (24.3)
Dr. Jie Zou PHY 1361 3
Gauss’s law
Gauss’s law describes a general relationship between the net electric flux through a closed surface and the charge enclosed by the surface. Closed surface: often called a
gaussian surface.
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Let’s begin with one example.
A spherical gaussian surface of radius r surrounding a point charge q. The magnitude of the electric
field everywhere on the surface of the sphere is E = keq/r2.
The electric field is to the surface at every point on the surface.
Net electric flux through such gaussian surface is
00
22 4
1444
q
qqkrr
qkdAEEdAd e
eE AE
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A non-spherical closed surface surrounding a point charge
As we discussed in the previous section, the electric flux is proportional to the number of electric field lines passing through a surface.
The number of lines through S1 is equal to the number of lines through the nonspherical surfaces S2 and S3.
The net flux through any closed surface surrounding a point charge q is given by q/0 and is independent of the shape of that surface.
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A point charge located outside a closed surface
Any electric field line that enters the surface leaves the surface at another point.
The net electric flux through a closed surface that surrounds no charge is zero.
Revisit Example 24.2
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Example: Problem #14, P. 762
Calculate the total electric flux through the paraboloidal surface due to a constant electric field of magnitude E0 in the direction shown in the figure.
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Now let’s consider a more general case.
The net electric flux through S is E = q1/0.
The net electric flux through S’ is E = (q2 + q3)/ 0.
The net electric flux through S” is E = 0.
Charge q4 does not contribute to the flux through any surface because it is outside all surfaces.
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Gauss’s law Gauss’s law: the net flux through any
closed surface is
qin = the net charge inside the gaussian surface. E = the (total) electric field at any point on the
surface, which includes contributions from charges both inside and outside the surface.
Pitfall prevention: Zero flux is not zero field.
0in
E
qdAE
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Application of Gauss’s law to various charge distributions
Example 24.5 A spherically symmetric charge distribution: An insulating solid sphere of radius a has a uniform volume charge density and carries a total positive charge Q.
(A) Calculate the magnitude of the electric field at a point outside the sphere.
(B) Find the magnitude of the electric field at a point inside the sphere.
Answer: (A) E = keQ/r2, r > a;
(B) , r < a. r
a
QkE e 3
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Application of Gauss’s law to various charge distributions
Example 24.7 A cylindrical symmetric charge distribution: Find the electric field a distance r from a line of positive charge of infinite length and constant charge per unit length .
Answer:r
kr
E e
22 0
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Homework
Ch. 24, P. 762, Problems: #12, 14, 29.