Solving Gauss' Law: E Field & Volume Charge Density

[suspenc3]

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.

 

[Saketh]

For the second problem, just use Gauss's law.

$$\int_S \vec{E} \cdot \, \vec{dA} = \frac{q}{\epsilon_0}$$

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.)

 

[m2003]

Hello, it looks like you have attempted to solve the first problem using the equation E = σ/ε0, however, this equation is only applicable for a charged plate. In this case, we have two charged plates facing each other, which creates a parallel plate capacitor. To solve for the magnitude of charge on each plate, we can use the equation E = Q/(ε0A), where Q is the charge on each plate, ε0 is the permittivity of free space, and A is the area of the plates. Rearranging this equation, we get Q = ε0EA. Plugging in the given values, we get Q = (8.85x10^-12 F/m)(55 N/C)(1.0 m^2) = 4.868x10^-10 C. Therefore, the magnitude of charge on each plate is 4.868x10^-10 C.

For the second problem, we can use Gauss' Law to determine the electric field at a point P within the sphere. Gauss' Law states that the electric flux through a closed surface is equal to the enclosed charge divided by the permittivity of free space. In this case, we can imagine a closed spherical surface centered at the point P, with radius r. The electric flux through this surface is given by Φ = EA, where E is the electric field and A is the area of the surface. Since the sphere has a uniform volume charge density, the enclosed charge is simply ρV, where V is the volume of the sphere. Therefore, Gauss' Law can be written as EA = ρV/ε0. Solving for E, we get E = ρV/(ε0A). Since the volume of a sphere is (4/3)πr^3 and the area of a sphere is 4πr^2, we can rewrite this equation as E = ρ(4/3)πr^3/(ε0(4πr^2)). Simplifying, we get E = ρr/3ε0, which is the desired equation.