Gauss's law, determing average volume charge density

[Breedlove]

Homework Statement

In the air over a particular region at an altitude of 500 m above the ground, the electric field is 120 N/C directed downward. At 600 m above the ground, the electric field is 100 N/C downward. What is the average volume charge density in the layer of air between these two elevations? Is it positive or negative?

Homework Equations

$$\Phi$$_net=$$\stackrel{Q}{\epsilon}$$
$$\rho$$=$$\stackrel{Q}{v}$$
$$\Phi$$=$$\int$$E(dot)dA

The Attempt at a Solution

I substituted $$\rho$$V for Q in
$$\Phi$$_net=$$\stackrel{Q}{\epsilon}$$
and then set that equal to EA from $$\Phi$$=$$\int$$E(dot)dA, and solved for $$\rho$$, getting $$\rho$$=E/(h$$\epsilon$$)
h is coming from A/V

Am I going about this the right way? My next step would be to find $$\rho$$ at both elevations and average them. Are the steps that I have taken this far correct?

Thanks for whatever help you can offer!

Oh, and anything that looks like a superscript or subscript is just an error on my part, I didn't mean for the equations to look like they had them. I'm still pretty new with it.
Thanks!

 

[Vuldoraq]

Looks like a good way to go about the calculation. As a safety check make sure your units on both sides of your equation are consistent, that way you know when your definitely on the wrong track! (P.S. I think your units are good in this case).

 

[nlightner]

As a scientist, it is important to approach problems in a systematic and logical manner. In this case, you have correctly identified the relevant equations and have attempted to use them to solve for the average volume charge density. However, it seems that you may have made a mistake in your substitution of Q for \rhoV.

The correct equation for Gauss's law is \Phi=\intE(dot)dA=\frac{Q}{\epsilon_0}, where Q is the total charge enclosed by the surface and \epsilon_0 is the permittivity of free space. In this case, our surface is the layer of air between 500 m and 600 m, and the electric field is constant within this region. Therefore, we can rewrite this equation as \Phi=EA=\frac{Q}{\epsilon_0}. Solving for Q, we get Q=\epsilon_0EA.

Next, we need to find the electric field strength at each elevation. We can use the given values of 120 N/C and 100 N/C to find the electric field at 500 m and 600 m respectively. Substituting these values into the equation Q=\epsilon_0EA, we get Q_1=\epsilon_0(120)(500)=60000\epsilon_0 and Q_2=\epsilon_0(100)(600)=60000\epsilon_0.

Now, we can find the average volume charge density by taking the difference between the two charges and dividing by the volume of the layer of air. The volume of the layer of air can be found by taking the difference between the two elevations and multiplying by the cross-sectional area, which is equal to the area of a circle with a radius of 500 m (since the layer of air is a cylindrical shape). Therefore, the volume is equal to \pi(500)^2(100)=25000000\pi m^3.

Plugging in the values, we get \rho_{avg}=\frac{Q_2-Q_1}{\pi(500)^2(100)}=\frac{60000\epsilon_0-60000\epsilon_0}{25000000\pi}=\frac{0}{25000000\pi}=0.

This means that the average volume charge density in the layer of air between 500 m and 600 m is 0. Since there is no net charge within this region, the charge density is neither positive nor negative