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Gauss' Law Again

  • Thread starter suspenc3
  • Start date
  • #1
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Homework Statement


1.)Two large metal plates of area [tex]1.0m^2[/tex] face each other. They are 5 cm apart and have equal but opposite charges on their inner surfaces. If the magnitude E of the electric field between the plates is 55N/C, what is the magnitude of the charge on each plate?Neglect edge effects

2.)A non conducting sphere has a uniform volume charge density [tex]\rho[/tex]. Let r be the vector from the center of the sphere to a general point P within the sphere. Show that the electric field at P is given by [tex]E=\rho r/3\epsilon_0[/tex]

Homework Equations


[tex]E=\sigma/\epsilon_0[/tex]


The Attempt at a Solution


1.)[tex]E=\sigma/\epsilon_0[/tex]
[tex]E\epsilon_0=\sigma=(55N/c)(8.85x10^{-12}F/m)=4.868x10^{-10}C/m^2[/tex]
[tex]E=\sigma/2\epsilon_0=\frac{4.868x10^{-10}C/m^2}{2(8.85x10^{-12}F/m} = 27.5N/C[/tex]

Im guessing that I did this wrong, it seemed too easy.

2.)I dont know how to start this one.
 
Last edited:

Answers and Replies

  • #2
261
2
For the second problem, just use Gauss's law.

[tex]
\int_S \vec{E} \cdot \, \vec{dA} = \frac{q}{\epsilon_0}
[/tex]

The important thing is to be careful in selecting your surfaces. Here's a suggestion -- use a Gaussian surface that is a spherical shell concentric with the sphere itself.

So the left side simplifies essentially to EA, where A is just [itex]4\pi r^2[/itex], where A is the surface area of our Gaussian surface (r < R).

Now, how do you find q? That is, how do you find the charge enclosed by our Gaussian surface? (I'll leave this part to you.)
 

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