Electrostatic & Gravitational Forces?

Question

Two dust particles each of mass 13 micro grams are floating on a gentle sream of air. What equal (positive) charge would each dust particle have to carry in order that their electrostatic repulsion should exactly balance their mutual gravitational attraction?

This has been driving me nuts... If there was a distance specified I'd have it in an instant. Can anyone help point me in the right direction?

Answer 1

Don't worry about the fact that you don't know what the distance is. Don't even think about plugging in numbers this early. Come up with an equation that expresses the relationship between the two forces, and at some point perhaps the distance will cancel.

Answer 2

The electrostatic repulsion between 2 like charges q and gravitational attraction between 2 like masses m are:

(kq²)/r²
(Gm²)/r²

so that the difference is

(kq² - Gm²)/r²

this means if kq² - Gm² > 0, then it's always repulsive regardless of r. Likewise, if kq² - Gm² < 0, then it's always attractive regardless of r. For typical dust particles, the latter is the case, which is why interstellar gas clouds usually end up clumping together. The condition you are looking for is kq² - Gm² = 0.

Answer 3

Distance does not matter since both forces depend on r^2 - writing the equations out you can see the distances cancel:


q^2 / (k r^2) = G m^2 / r^2 ... k = 1/(4 pi epsilson)

q^2 = G k m^2

q = m sqrt ( G K )

Answer 4

The force is inverse square in distance for both and is not relevant.

Gm2 = kq^2
q = m * sqrt(G/k)

I am sure you can compute the answer.

Answer 5

considering the two forces selection as a million/r² yet are in opposite instructions, the two forces could be equivalent in the event that they have been nearer jointly or farther aside. so as that they might in simple terms pass under the impact of in spite of different forces have been performing on the two bodies.

Answer 6

the gravity of 13 micrograms would be too weak to even break the resistance cause by friction. so any electrostatic charge will be enough regardless of the distance. infact none should be enough aswell.