# Homework Help: Sphere below the Earth

1. Nov 30, 2013

### Numeriprimi

Hello, everyone :-)

We have a wooden sphere at a height of h = 1 m above the surface of the Earth which has a perimeter of RZ = 6 378 km and a weight of MZ = 5.97 · 10^24 kg. The sphere has a perimeter of r = 1 cm and is made of a wood which has the density of ρ = 550 kg·m − 3. Assume that the Earth has an electric charge of Q = 5 C. What is the charge q that the sphere has to have float above the surface of the Earth? How does this result depend on the height h?

We can use Coulomb's and Newton's law.

I can say the force of F = m_s*g = 4/3 * π * r^3 * ρ * g = the force of Coulomb's law
4/3*π*r^3*ρ*g = |Q_1|*|Q_2|/(4*π*e_0*e_1)
|Q_1| = 4/3*π*r^3*ρ*g*4*π*e_0*e_1/|Q_2|

Ok, this is my solution. Is it OK?

2. Nov 30, 2013

### Simon Bridge

I take it English is a second language. Here, let me help...
A 'perimeter" is something a 2D surface may have - something you could put a fence around.
6378km is the radius of the Earth.
1cm is the radius of the wooden sphere.

That would be the mass of the Earth - "weight" is the force of gravity.
Weight is sometimes given in mass units for objects close to the surface of the Earth.

It helps if you write down your reasoning.
I'll have to see what I can deduce from what you wrote.

4/3*π*r^3*ρ*g = |Q_1|*|Q_2|/(4*π*e_0*e_1)

$\frac{4}{3}\pi r^3\rho g$ is the weight of the wooden sphere in the approximation that h<<RZ ... which would be good.

You are reasoning that the coulomb force must be equal and opposite to this?

$\frac{Q_1 Q_2}{4\pi\epsilon_0\epsilon_1}$... is not the coulomb force.

Lets have q on the wooden sphere and Q on the Earth.
How far away is the charge Q from the charge q?
How does the force between two charges depend on the distance between them?
What is $\epsilon_1$ for?