Conductors in Electrostatic Equilibrium problem

[dimpledur]

Homework Statement

In fair weather, over flat ground, there is a downward electric field of about 150 N/C. A) assume that the Earth is a conducting sphere with charge on its surface. If the electric field just outside is 150 N/C pointing radially inwards, calculate the total charge on the Earth's surface. B) at an altitude of of 250 m above the Earth's surface, the field is only 120 N/C. Calculate the charge density.

Homework Equations

E=\frac{k\left|q\right|} {r^2}

\rho=\frac{q} {V}

The Attempt at a Solution

I only need help with part b)

\left|q\right|=\frac{Er^2} {k}

\left|q\right|=\frac{(120 N/C)[6.371(10)^6 m +250 m]^2} {8.99(10)^9 Nm/C^2}

\left|q\right|=5.42(10)^5 C

We then have,

\rho=\frac{5.42(10)^5 C} {(4/3)(\pi)[(6.371(10)^6 m +250 m)^3-(6.371(10)^6 m)^3]}

\rho=4.25(10)^{-12} C/m^3

However, the answer should be \rho=1(10)^{-12} C/m^3 according to the back of the book. Suggestions?

 

[cupid.callin]

\rho is surface charge density not volumetric.

Use \rho = q/(surface area)

 

[dimpledur]

In my textbook \rho is for volumetric charge density.. Additionally, the answer is given per unit volume, so surface area is not suffice.

 

[cupid.callin]

Calculate the charge density.

Who's density are they talking about?

 

[dimpledur]

I'm assuming that they want to know the charge density of the air

 

[cupid.callin]

well if there's charge in air also and you need only density in air then net E at 120m is due to both charge on Earth and air in between.

So find field due to Earth at 250m above Earth and subtract it form E given and then find ρ