Calculate the volume charge density of the atmosphere

[sonutulsiani]

Homework Statement

In the atmosphere and at an altitude of 250 m, you measure the electric field to be 150 N/C directed downward, and you measure the electric field to be 170N/C directed downward at an altitude of 400 m. Calculate the volume charge density of the atmosphere in the region between altitudes of 250 m and 400m, assuming it to be uniform.

Homework Equations

The Attempt at a Solution

I just want to know. When I substitute the values in E = (sigma/epsilon 0) and then sigma = Q/Area.. So I get Q = 150 (epsilon 0) A. Now this I will substitute in the formula rho = Q / V.

Now my question is what will be Area? What value of r should I write in 4 pi r^2 ?
Will it be radius of Earth or the altitude of 250 m? If it's radius of earth, where will I write 250m then?

 

[kuruman]

This is a Gauss's Law application. If you can picture the Gaussian surface that you should use, you will know what to do with r and what its value(s) ought to be. So what kind of Gaussian surface should you use to "calculate the volume charge density of the atmosphere in the region between altitudes of 250 m and 400m, assuming it to be uniform"?

 

[sonutulsiani]

Hmm may be cylinder? But how in the world do I get the idea of gaussian surface?!

 

[kuruman]

Hmm may be cylinder? But how in the world do I get the idea of gaussian surface?!

Choose a Gaussian surface that reflects the symmetry of the problem. The Earth is more closely approximated to a sphere than a cylinder. Also, the expression E = σ/ε₀ applies only in cases where there is a surface charge. There is no surface charge in the atmosphere at heights of 250 m or 400 m or any height. There is only volume charge density in the atmosphere.

 

[sonutulsiani]

So what it should be E = rho / epsilon 0?
And I got what you are saying. I got the charge in the sphere, Q inside as = 4 E pi epsilon 0 r^(2)

Is that right? If that is right, then r=radius of the Earth only

 

[sonutulsiani]

Ok I tried it once again.

This time I took rho = Q / V

And then substituted Q inside = 4 E pi epsilon 0 r^(2)
with rho V = 4 E pi epsilon 0 r^(2)

=> rho 4/3 pi r^3 = 4 E pi epsilon 0 r^(2)

Is it correct?? In that way I can find rho, but then I am still confused about the 'r' s

 

[sonutulsiani]

Wait the r will be radius of Earth plus 250m ?

 

[kuruman]

You know that the E field at R_E+400 m is 170 N/C and that the E field at R_E+250 m is 150 N/C. Both fields are radially in.

Can you calculate the electric flux through the area of a sphere of radius R_E+400 m and then through a sphere of radius R_E+250 m?

The sum of the two would be the total flux through a shell of inner radius R_E+250 m and outer radius R_E+400 m. What should this total flux be equal to according to Gauss's law?