Electrostatics - Charge and Gravitational Force?

Question

How much charge would have to be put on objects that are 3.84 x 10^8 m apart so that it creates the gravitational force of 1.99 x 10^20 N?

Answer 1

Ok, Force of gravity, which will be denoted as F is:

F = 1.99 X 10^20 N

You're given the radius of the charge, denoted as r:

r = 3.84 X 10^8 m

The force of the electric field, also denoted as F, is the following expression:

F = (1/4*pi*Eo) (Q*q/r^2) = q*E

(*for the sake of simplicity, and since the problem did not specify that you will have two different charges, I will say that:

Q*q = Q^2 )

Likewise for the F of gravity:

F = G (Mm/r^2) = m*g

Since both equations equal the force, you set them equal to each other like so:

(1/4*pi*Eo) (Q^2/r^2) = G (Mm/r^2)

Ok, so the left expression is force of electricty and right expression is force of gravity. We've been give the numerical value for the force of gravity, and the radius for electric force. I'm going to substitute that in, and here's what it will look like:

(1/4*pi*Eo)(Q^2/(3.84 x10^8 m)^2) = 1.99 x 10^20 N

And now, I solve for Q*q:

Q^2= (1.99 x 10^20 N) (3.84 x 10^8 m)^2 / (8.988×10^9 N m^2 C^-2 )

where (1/4*pi*Eo) = 8.988×109 N m^2 C^-2

After calculation:

Q^2 = 3.26 x 10^27 C^2

ok, remember C = Coulombs.

Now, i'm going to take the square root of both sides:

[Q^2] ^ (1/2) = [3.26 x 10^27 C^2] ^(1/2)

Finally, the charge, that you must have for the conditions you set, is :

Q = 5.714 x !0^13 C

Answer 2

Electrical clarge is totally different from gravity.