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<title>Electricity - Gauss's Law - Physics 299</title>
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<h1> <img src="ULPhys1.gif" align="texttop" height="50"
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<h1>Gauss's Law<br>
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<font color="#ff0000"><i>"</i></font><font color="#ff0000"><i>
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Equations are just the boring part of mathematics. I attempt
to see things in terms of geometry."</i></font><br>
Stephen Hawking<br>
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<img src="netbar.gif" align="middle" height="40" width="100%"> <br>
<ul>
<li><img alt="gauss" src="gauss.jpg" align="left" height="101"
width="83">Gauss's Law is the first of Maxwell's equations we
will consider.&nbsp; At first the whole concept of Gauss's Law
will seem to be very abstract and confusing,&nbsp;<img
alt="confused" src="confused_smiley.gif" height="22"
width="15"> &nbsp; hopefully at least some of the confusion
will pass as you become more familiar with the idea.&nbsp;&nbsp;</li>
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<ul>
<li>At the outset it is important to realize that <b>Gauss's Law
and Coulomb's Law are different statements of the same
physical concept.</b>&nbsp; Which of the two is used in any
particular situation depends on the particular application and
what you are asked to determine.</li>
</ul>
<ul>
<li>Before stating Gauss's Law we must first define the concept of
<b>FLUX</b> - in particular the flux of the electric field.</li>
</ul>
<div align="left">
<ul>
<li><img alt="Electric flux" src="elec_gauss_figure1.jpg"
align="right" height="433" width="415">At every point on a
surface we can calculate an "element" of the electric flux
given by</li>
</ul>
<div align="center"><img alt="defn of E flux"
src="elec_gauss_eqn1.jpg" height="38" width="117"><br>
<blockquote>
<div align="left">so that the total electric flux passing
through a surface, S, is given by,<br>
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<div align="center"><img alt="elec gauss eqn2"
src="elec_gauss_eqn2.png" height="56" width="133"><br>
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<ul>
<li>Gauss's Law then states that,</li>
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<div align="center"><img alt="red tick" src="tickred1.gif"
align="top" height="48" width="48"> &nbsp;&nbsp; <img
alt="elec gauss 3" src="elec_gauss_eqn3.jpg"
height="84" width="233">&nbsp;&nbsp;&nbsp; <img
alt="red tick" src="tickred1.gif" align="top"
height="48" width="48"><br>
<blockquote>
<div align="left">where the circle on the integral
means that the surface is <i><b>closed</b></i> and
q<sub>inside</sub> is the net charge inside this
closed surface.<br>
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<ul>
<li><img alt="exclamation"
src="exclamation-icon.gif" height="30"
width="31"> A closed surface has a definite
inside and outside differentiated by the surface,
e.g. the surface of a sphere.</li>
<li><img alt="exclamation"
src="exclamation-icon.gif" height="30"
width="31"> The <b>dA</b> vector of a closed
surface is always directed from the inside to the
outside of the surface.</li>
<li><img alt="exclamation"
src="exclamation-icon.gif" height="30"
width="31"> The exact location of the charges
inside the closed surface is not important, all
that matters is the net charge.</li>
<li><img alt="exclamation"
src="exclamation-icon.gif" height="30"
width="31"> &#949;<sub>0</sub> is the "Permittivity
of the Vacuum" a constant whose value is 8.85 x 10<sup>-12</sup>
C<sup>2</sup>/(N.m<sup>2</sup>) where the Coulomb
constant, k = 1/(4&#960;&#949;<sub>0</sub>).&nbsp; Note that
if the charges are not located in vacuum &#949;<sub>0</sub>
must be replaced by the permittivity of the medium
in question.&nbsp; </li>
<li><img alt="exclamation"
src="exclamation-icon.gif" height="30"
width="31"> The proof of Gauss's Law is beyond
the scope of this course.&nbsp; Suffice to say the
inverse square dependence on distance of Coulomb's
Law is critical.</li>
</ul>
<ul>
<li>Before using Gauss's Law to evaluate electric
fields a brief qualitative discussion is
worthwhile.&nbsp; Consider the situation of two
point charges below.&nbsp; Application of Gauss's
Law over each of the closed surfaces:</li>
</ul>
<blockquote>
<ul>
<li><img alt="elec gauss figure 2"
src="elec_gauss_figure2.png" align="right"
height="523" width="359"> S<sub>1</sub>:&nbsp;
At every point on this surface both <b>E </b>and
<b>dA</b> are directed "outwards", such that the
scalar product <b>E<EFBFBD>dA</b> = EdAcos&#952; is always
positive.&nbsp; Thus the integral over the
surface S<sub>1</sub> will be positive, as it
must be if Gauss's Law is to be satisfied, since
the net charge enclosed is positive.</li>
<li>S<sub>2</sub> :&nbsp; <b>E </b>is directed
"inwards", <b>dA</b> "outwards", leading to a
negative value for the flux through S<sub>2</sub>,
consistent with the fact that the net charge
enclosed is negative.</li>
<li>S<sub>3</sub>:&nbsp; Some of this surface has
E directed "inwards" the remainder has <b>E</b>
directed "outwards".&nbsp; <b>dA</b> is
"outwards" everywhere on the surface.&nbsp;
Therefore the flux integral has both positive
and negative contributions.&nbsp; Since there is
no net charge enclosed by S<sub>3</sub> by
Gauss's Law the net flux will be zero.</li>
<li>S<sub>4</sub>: Once again there are negative
and positive contributions to the flux integral,
so that we can write Gauss's Law,</li>
</ul>
<div align="center"><img alt="elec_gauss_eqn4"
src="elec_gauss_eqn4.png" height="86"
width="381"><br>
<div align="left"><br>
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<ul>
<li>&nbsp;Electric flux through a closed box <a
href="http://www.youtube.com/watch?v=5ENl4vn82bc">animation.</a></li>
</ul>
<ul>
<li>Coulomb's Law, Electric Field, Electric Flux and Gauss's Law <a
href="http://www.veoh.com/collection/APPhysics/watch/v15544578N9Hg8YBK">video</a>.<br>
</li>
</ul>
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<font color="#ff0000"><i>Did you hear about the French post-doc
who went to work at the Fermi Lab, but never went in because
the sign over the door always said it was closed.</i></font><br>
<br>
<img src="celticbar.gif" height="22" width="576"> <br>
&nbsp;
<p><i>Dr. C. L. Davis</i> <br>
<i>Physics Department</i> <br>
<i>University of Louisville</i> <br>
<i>email</i>: <a href="mailto:c.l.davis@louisville.edu">c.l.davis@louisville.edu</a>
<br>
&nbsp; </p>
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