Two 4.00-kg spheres carry identical charges and are placed a distance d from eac

Question

Two 4.00-kg spheres carry identical charges and are placed a distance d from each other.

Part A

How large should the charge on each sphere be so that the repulsive electric force between them balances the attractive gravitational force between them?

Express your answer to two significant digits and include the appropriate units.

Part B

What is the smallest mass that can balance in this way, given that the smallest possible charge is  e=1.602×1019C?

Express your answer to two significant digits and include the appropriate units.

Explanation / Answer

Part A:

let us consider, F (g) be gravitational force that each sphere exerts on the other...

And, let F (e) be the electric force due to both spheres on the each other..

Now,
F (g) = F (e)

=> (G * m^2)/r^2 = (k * q^2)/r^2, where r is the distance between two spheres
=>6.67 * 10^(-11) * 4.0^2 = 9 *10^9 * q^2

=> q = 3.44 x 10^-9 C

Answer in two significant digits = 3.4 x 10^-9 C

Part B:

In the given situation:

6.67 * 10^(-11) * m^2 = 9 *10^9 * (1.602x10^-19)^2

=> m^2 = 3.463 x 10^-18

Hence, m = 1.86 x 10^-9 Kg

Answer in two significant digits = 1.9 x 10^-9 Kg

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