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Lecture 4 - Electric fields around conductors, Electric Flux and Gauss's law

Field due to a uniform disk of charge

To find the field due to a uniform disk of charge which has an areal charge density $\sigma$ we can use our result for a ring of charge and consider the field as an integral over a set of these rings. If we take a ring of charge $dQ$

$dE=\frac{1}{4\pi\epsilon_{0}}\frac{z}{(z^{2}+r^{2})^{3/2}}\,dQ$

We need to relate the charge element to the width of the ring $dr$

$dQ=\sigma 2\pi r\,dr$

$dE=\frac{1}{4\pi\epsilon_{0}}\frac{z\sigma 2\pi r}{(z^{2}+r^{2})^{3/2}}\,dr=\frac{z\sigma r}{2\epsilon_{0}(z^{2}+r^{2})^{3/2}}\,dr$

$E=\frac{z\sigma}{2\epsilon_{0}}\int_{0}^{R}\frac{r}{(z^{2}+r^{2})^{3/2}}\,dr=\frac{z\sigma}{2\epsilon_{0}}[-\frac{1}{(z^{2}+r^{2})^{1/2}}]_{0}^{R}=\frac{\sigma}{2\epsilon_{0}}[1-\frac{z}{(z^{2}+R^{2})^{1/2}}]$

The limit as $R\to\infty$ or when $z<<R$ that $E=\frac{\sigma}{2\epsilon_{0}}$ is valid for any infinite plane of charge.

Field due to two sheets of charge

$E=\frac{\sigma}{2\epsilon_{0}}$ is valid for any infinite plane of charge, so what is the field inside and outside of two parallel sheets of charge with opposite sign?

Field inside a conductor

If we consider a static situation inside a conductor the field should be zero. This is because if there is an electric field the free electrical charges will move in response to the field because they experience a force. If there is a net charge on a metal arranges itself so that the field inside is equal to zero. The result is that the charges are arranged on the surface of the conductor.

If we take for example a long thin rectangular conductor with a net positive charge we can see that this will have zero field inside if the charge is on the surface from our previous result that the field due to a infinite plane of charge is $\frac{\sigma}{2\epsilon_{0}}$. This does not depend on the distance from the charge. The field points away from the positive charges and so the field from the charges on each surfaces exactly cancels out everywhere inside the conductor.

Electric fields lines are always perpendicular to a conducting surface, any component of the field parallel to the surface would lead to a movement of charge along the surface.

Charge inside a conducting container

If we consider a charge inside a conducting sphere, the inside surface of the sphere must develop a charge to stop the field from developing inside the conductor. To maintain neutrality the outer sphere must develop an equal and opposite charge. The field outside the sphere is then the same as if the sphere was not there are at all.

Faraday cage

If we place a metal object in an electric field the interior of the conductor has no field. It is also true that if we then make a hole in the middle of the conductor there is also no field inside (as long as there is no net charge in the hole.) This principle is used in the Faraday cage which is a metal enclosure used to shield people or equipment from static electric fields.

figure_21_39.jpg

Electric Flux

We are now going to define a new quantity, electric flux, which will allow us to introduce Gauss's Law. We will come to see that Gauss's law is often a much more effective way of calculating the electric field than Coulomb's law.

If we consider a uniform electric field $\vec{E}$ passing through a flat area $A$ we can define the electric flux as $\Phi_{E}=EA\cos\theta$

Area vector

We would like to define the electric flux in terms of a vector dot product. In order to do this we represent the area by a vector with magnitude equal to the area and direction normal to the surface.

Now we can go from

$\Phi_{E}=EA\cos\theta$

to

$\Phi_{E}=\vec{E}.\vec{A}$

Electric flux for non-uniform fields and areas

If we have an arbitrary surface and field we can divide it up in to infinitesimally small areas $d\vec{A}$ within which the field is uniform. To find the total flux through a surface we integrate over these area elements

$\Phi_{E}=\int\vec{E}.d\vec{A}$

In order to introduce Gauss's law we need to define a closed surface. A closed surface completely encloses a volume. We denote an integral over a complete closed surface using a special symbol, ie, the electric flux passing through a closed surface is

$\Phi_{E}=\oint\vec{E}.d\vec{A}$

For a closed surface we need to pay attention to the direction of the vector $\vec{A}$ which we define to always point outwards from the surface.

Gauss's Law

Gauss's law is a very useful law that relates the integral we just defined to the amount of charge $Q$ enclosed inside the surface.

$\Phi_{E}=\oint\vec{E}.d\vec{A}=\frac{Q}{\varepsilon_{0}}$

Gauss's law tells us that the net difference of electrical flux going in to and out of a closed surface is determined by the amount of charge the surface encloses. We should be careful to note it does not necessarily imply that there is zero field inside a Gaussian surface that does not enclose any charge.

The reasons that this law is so useful is that it does not matter how the charge is distributed inside the surface. Also we can choose the surface to be whatever we want, if we choose carefully, guided by symmetry, the integrals required can often be very easy.

Coulomb's law from Gauss's law

If we take an isolated point charge of magnitude $+Q$ we can expect the field to point radially outwards. Therefore it makes sense to choose a spherical Gaussian surface centered on the charge so that the field will be the same in magnitude and perpendicular to the surface everywhere it meets the surface.

$\Phi_{E}=\oint\vec{E}.d\vec{A}=\oint EdA=E\oint dA=EA=E4\pi r^{2}$

Gauss's law then tells us that

$E4\pi r^{2}=\frac{Q}{\varepsilon_{0}}\to E=\frac{Q}{4\pi\varepsilon_{0}r^{2}}$

and we have recovered Coulomb's law.

Charged metal sphere

We can now look at a situation we discussed earlier, a hollow metal sphere with a net positive charge.

Because of the spherical symmetry for any spherical Gaussian surface of radius r

$\Phi_{E}=\oint\vec{E}.d\vec{A}=\oint EdA=E\oint dA=EA=E4\pi r^{2}$

and Gauss's law gives us

$\Phi_{E}=\oint\vec{E}.d\vec{A}=\frac{Q}{\varepsilon_{0}}$

Any spherical Gaussian surface we draw inside the sphere of charge does not include any charge and so the net electric flux through the sphere must be zero. By contrast if we draw a spherical gaussian surface outside the charge we can show that the field outside the sphere is $E=\frac{Q}{4\pi\varepsilon_{0}r^{2}}$

phy142kk/lectures/4.txt · Last modified: 2016/02/01 17:33 by kkumar
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