Solving Gauss' Law: E Field & Volume Charge Density

AI Thread Summary
The discussion focuses on solving two problems related to Gauss' Law. For the first problem involving two charged plates, the electric field E is calculated using the formula E = σ/ε₀, leading to a charge density of approximately 4.868 x 10^-10 C/m². The second problem involves a non-conducting sphere with uniform volume charge density, where the electric field at a point P inside the sphere is derived using Gauss's Law. A suggestion is made to use a spherical Gaussian surface to simplify the calculations, emphasizing the importance of correctly determining the enclosed charge. Overall, the thread highlights the application of Gauss' Law in different contexts to find electric fields and charge densities.
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Homework Statement


1.)Two large metal plates of area 1.0m^2 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 \rho. 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 E=\rho r/3\epsilon_0

Homework Equations


E=\sigma/\epsilon_0


The Attempt at a Solution


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

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

2.)I don't know how to start this one.
 
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For the second problem, just use Gauss's law.

<br /> \int_S \vec{E} \cdot \, \vec{dA} = \frac{q}{\epsilon_0}<br />

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 4\pi r^2, 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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