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<title>Electricity - Quantitative use of Gauss's Law - Physics 299</title>
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<h1> <img src="ULPhys1.gif" height="50" align="texttop"
width="189"></h1>
</center>
<center>
<h1>Quantitative Use of Gauss's Law <br>
</h1>
</center>
<center><img src="celticbar.gif" height="22" width="576"><br>
<br>
<font color="#ff0000"><i>"</i></font><font color="#ff0000"><i>
<meta http-equiv="content-type" content="text/html;
charset=windows-1252">
It has been said that democracy is the worst form of
government except all the others that have been tried."</i></font><br>
Winston Churchill<br>
</center>
<img src="netbar.gif" height="40" align="middle" width="100%"> <br>
<div align="center">&nbsp;<br>
&nbsp;&nbsp; <img alt="elec gauss 3" src="elec_gauss_eqn3.jpg"
height="84" width="233">&nbsp;&nbsp;&nbsp;&nbsp;</div>
<ul>
<li>Gauss's Law is valid for any closed surface (a Gaussian
surface) and any distribution of charges.&nbsp; If the electric
field is known at every point on the surface S the integral can
in principle be evaluated and will be seen to be equal to the
sum of the enclosed charges divided by &#949;<sub>0</sub>.
&nbsp;&nbsp; However, only in certain very symmetric situations,
where we can infer a great deal of information about the
electric field, can it be used to actually calculate <b>E</b>.&nbsp;
In such cases Gauss's Law provides a short cut to determining <b>E</b>.&nbsp;
The key is to be able to "extract" the <b>E</b> from the flux
integral.</li>
</ul>
<ul>
<li>We will consider three possible geometric situations in which
we can obtain <b>E</b> from Gauss's Law:</li>
<ul>
<li>Spherical symmetry - three dimensions</li>
<li>Rectangular symmetry - two dimensions</li>
<li>Cylindrical symmetry - one dimension</li>
</ul>
</ul>
<div align="center"><img alt="divider bar"
src="divider_ornbarblu.gif" height="64" width="393"><br>
</div>
<div align="center"><big><font color="#3333ff"><u><big><b>SPHERICAL
SYMMETRY</b></big></u></font></big></div>
<ul>
</ul>
<ul>
<li><big><b>Single Point Charge</b></big></li>
</ul>
<blockquote><img alt="elec gauss figure 5"
src="elec_gauss_figure5.jpg" height="313" align="right"
width="237">Consider a single point charge +Q and a spherical
surface, S,&nbsp; of radius r and center at the location of
+Q.&nbsp; From the symmetry of this situation we can conclude
that, everywhere on the surface S, <b>E</b> has the same value
and is directed radially outwards (normal to the surface).&nbsp;
This is the same as the direction of <b>dA</b>.&nbsp; Therefore,<br>
<div align="center"><img alt="elec gauss eqn5"
src="elec_gauss_eqn5.png" height="62" width="515"><br>
<div align="left">so that,<br>
<div align="center"><img alt="elec gauss eqn6"
src="elec_gauss_eqn6.png" height="64" width="117"><br>
<div align="left">which is exactly Coulomb's Law !!<br>
<br>
As has already been stated - <font color="#ff0000"><big><b>Gauss's
Law and Coulomb's Law are different statements of
the same physical principle.<br>
<font color="#330033"><br>
<br>
</font></b></big></font></div>
</div>
</div>
</div>
</blockquote>
<div align="center">
<div align="left">
<div align="center">
<div align="left">
<ul>
<li><big><b>Spherical Charge Distribution with Uniform
Charge Density</b></big></li>
</ul>
<blockquote>Charge<b> </b>is distributed uniformly
throughout the volume of the sphere (this means that the
sphere must be a non-conductor since as we have seen the
charge on a conductor must reside on the surface) such
that the total charge Q is given by,<br>
<br>
<div align="center"><img alt="elec gauss eqn7"
src="elec_gauss_eqn7.png" height="41" width="117"><br>
<br>
<div align="left">where &#961; is the (volume) charge
density, in units of Coulombs/m<sup>3</sup>.<br>
<br>
What is the electric field at any point either outside
or inside the sphere ?<br>
Due to the symmetry of this configuration we can
conclude that <b>E</b> is directed radially outwards
everywhere and can (at most) depend only on the
(radial) distance from the center of the sphere.&nbsp;
There are two distinct regions to consider:<br>
<br>
<b><img alt="elec gauss figure 5"
src="elec_gauss_figure6.png" height="215"
align="right" width="155"><u>Outside the
sphere,&nbsp; r &gt; R</u></b><br>
<br>
Applying Gauss's Law over a Gaussian surface (sphere)
of radius r, then,<br>
<div align="center"><img alt="elec gauss eqn 5"
src="elec_gauss_eqn5.png" height="62" width="515"><br>
<br>
<div align="left">so that,<br>
<div align="center"><img alt="elec gauss eqn 6"
src="elec_gauss_eqn6.png" height="64"
width="117"><br>
<br>
<div align="left">In other words, for points
outside the sphere, the sphere behaves as a
point charge located the sphere's center.<br>
<img alt="hot" src="hot.gif" height="43"
align="middle" width="79">&nbsp; We saw
exactly the same type of behavior when
considering the gravitational effect of a
spherical mass.<br>
<br>
</div>
</div>
</div>
</div>
<b><img alt="elec gauss figure 7"
src="elec_gauss_figure7.png" height="230"
align="right" width="167"><u>Inside the sphere, r
&lt; R</u></b><br>
<br>
Applying Gauss's Law over a Gaussian surface (sphere)
of radius r, then,<br>
<br>
<div align="center"><img alt="elec gauss eqn8"
src="elec_gauss_eqn8.jpg" height="63" width="563"><br>
<div align="left">Or in terms of Q and R,<br>
<br>
<div align="center"><img alt="elec gauss eqn9"
src="elec_gauss_eqn9.jpg" height="64"
width="123"><br>
<br>
<div align="left">Note that for r &lt; R only
the charge inside a sphere of radius r
contributes to <b>E</b>.&nbsp; The charge
between r and R has no effect.<br>
<br>
<img alt="exclamation"
src="exclamation-icon.gif" height="30"
width="31"> It is important to realize that
without using Gauss's Law, these results could
be obtained via Coulomb's Law, but would
involve considerably more work - setting
up&nbsp; a non-trivial multiple integral to
consider every point charge in the sphere....<br>
<br>
<div align="center"><img alt="divider bar"
src="divider_ornbarblu.gif" height="64"
width="393"><br>
<big><font color="#3333ff"><u><big><b>CYLINDRICAL
SYMMETRY<br>
<br>
</b></big></u></font></big></div>
</div>
</div>
</div>
</div>
</div>
</div>
</blockquote>
<div align="center">
<div align="left">
<div align="center">
<div align="left">
<div align="center">
<div align="left">
<div align="center">
<div align="left">
<ul>
<li><big><b>Infinite </b><b>Line Charge</b></big>
<big><b>with Linear Charge Density &#955;</b></big></li>
</ul>
<blockquote><img alt="elec gauss figure8"
src="elec_gauss_figure8.png"
height="212" align="right" width="331">Determine
the <b>E</b> field a distance r from the
line charge.&nbsp; (Note that the units of
&#955; are Coulombs/meter)<br>
<br>
Symmetry tells us that <b>E</b> can only
have a component perpendicular to the line
charge, that is perpendicular to the
cylindrical surface shown.<br>
<br>
Applying Gauss's Law over the cylindrical
Gaussian surface, radius r and length l,
as shown, there will in principle be three
contributions - one from the curved
surface and one from each of the two
ends.&nbsp; However, on the ends <b>E</b>
and <b>dA</b> are perpendicular, so that
<b>E<EFBFBD>dA</b> = 0, therefore there is no
contribution to the flux through S.&nbsp;
On the curved surface <b>E</b> and <b>dA</b>
are parallel, thus,<br>
<br>
<div align="center"><img alt="elec gauss
eqn10" src="elec_gauss_eqn10,jpg.jpg"
height="51" width="577"><br>
<div align="left">so that,<br>
<div align="center"><img alt="elec
gauss eqn11"
src="elec_gauss_eqn11.jpg"
height="75" width="117"><br>
<div align="left"><br>
We can extend this analysis to the
case of a uniformly charged
infinite cylinder in a similar
manner to the extension of the
point charge to the spherical
charge distribution above.<br>
<br>
<div align="center"><img
alt="divider bar"
src="divider_ornbarblu.gif"
height="64" width="393"><br>
<big><font color="#3333ff"><u><big><b>RECTANGULAR
SYMMETRY</b></big></u></font></big><br>
</div>
</div>
</div>
</div>
</div>
</blockquote>
<div align="center">
<div align="left">
<div align="center">
<div align="left">
<div align="center">
<div align="left">
<ul>
<li><big><b>Infinite plane of
charge</b></big></li>
</ul>
<blockquote><img alt="elec gauss
figure 9"
src="elec_gauss_figure9.jpg"
height="200" align="right"
width="246">Determine the <b>E</b>
field at any distance above or
below an infinite plane with
charge density &#963; (Coulombs/m<sup>2</sup>).<br>
<br>
Symmetry dictates the <b>E</b>
must be perpendicular to the
surface everywhere.<br>
<br>
Applying Gauss's Law over the
cylindrical surface shown,
then the curved surface of the
cylinder&nbsp; contributes
nothing to the flux since <b>E</b>
and <b>dA</b> are
perpendicular.&nbsp; But on
the ends <b>E</b> and <b>dA</b>
are parallel.&nbsp; Therefore,<br>
<br>
<div align="center"><img
alt="elec gauss eqn12"
src="elec_gauss_eqn12.jpg"
height="56" width="536"><br>
<br>
<div align="left">so that,<br>
<div align="center"><img
alt="elec gauss eqn13"
src="elec_gauss_eqn13.jpg" height="96" width="117"><br>
<br>
</div>
</div>
</div>
</blockquote>
<blockquote>That is the electric
field is constant - it does
not depend on how far the
field point is from the plane
!!&nbsp; <br>
<br>
<img alt="exclamation"
src="exclamation-icon.gif"
height="30" width="31"> Note
that this is only true for an
infinite plane of
charge.&nbsp; If the distance
of the field point from the
plane is small compared to the
"size" of the plane, the above
expression is a good
approximation.<br>
<br>
<div align="center"><img
alt="divider"
src="divider_ornbarblu.gif"
height="64" width="393"><br>
</div>
</blockquote>
<div align="center">
<div align="left">
<ul>
<li>In all the above
situations the key to
using Gauss's Law is <b>SYMMETRY</b>.&nbsp;
There must be enough
symmetry in the problem
to know the direction of
<b>E</b> everywhere in
the vicinity of the
charge
distribution.&nbsp;
Knowing the direction of
<b>E</b> the trick is
then to choose a
Gaussian surface over
which to apply Gauss's
Law such that <b>E</b>
can be "taken out" of
the flux integral.&nbsp;
So when using Gauss's
Law to determine <b>E</b>
there are three key
steps:</li>
</ul>
<ul>
<ol>
<li>
<h3>State what you are
assuming about <b>E</b>
based on the
symmetry of the
problem.</h3>
</li>
<li>
<h3>State clearly the
Gaussian surface(s)
you will use - often
most easily done by
sketching the
surface(s) on a
diagram.</h3>
</li>
<li>
<h3>Evaluate the
surface (flux)
integral to
determine E.&nbsp;
The symmetry of E
and choice of
Gaussian surface
should allow "E" to
be "taken out" of
the integral and
thus be determined.<br>
</h3>
</li>
</ol>
</ul>
</div>
</div>
<blockquote> </blockquote>
<blockquote> </blockquote>
</div>
</div>
</div>
</div>
</div>
</div>
<blockquote>
<div align="center">
<div align="left">
<div align="center"> </div>
</div>
</div>
</blockquote>
<ul>
</ul>
</div>
</div>
</div>
</div>
</div>
</div>
</div>
</div>
<blockquote>
<div align="center">
<div align="left"> </div>
</div>
</blockquote>
</div>
</div>
</div>
</div>
<ul>
<font color="#ff0000"><big> </big></font>
</ul>
<font color="#ff0000"><big><b> </b></big></font><img
src="netbar.gif" height="40" width="100%"><br>
<br>
<center><span style="font-size: 12pt; font-family: &quot;Times New
Roman&quot;;"><span style="color: rgb(255, 0, 0); font-style:
italic;"></span></span><font color="#ff0000"><i>An engineer
friend of mine told me of a group of scientists that were
nominated for a Nobel prize. Using dental tools, they were
able to sort out the smallest particles that mankind has yet
discovered. The group became known as " the Graders of the
Flossed Quark."</i><i><span style="font-size: 12pt;
font-family: &quot;Times New Roman&quot;;"></span></i><i><span
style="font-size: 12pt; font-family: &quot;Times New
Roman&quot;;">
<meta http-equiv="content-type" content="text/html;
charset=windows-1252">
</span></i><i> </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>
<p><img src="header-index.gif" height="51" width="92"> </p>
</center>
<p><br>
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