# Conductors in Electrostatic Equilibrium problem

## 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?

Related Introductory Physics Homework Help News on Phys.org
$$\rho$$ is surface charge density not volumetric.

Use $$\rho$$ = q/(surface area)

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

Calculate the charge density.
Who's density are they talking about?

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

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