Repulsion between two small spheres

[objective33]

Homework Statement

Calculate the force of electric repulsion between two small spheres placed 1.0 m apart if each has a deficit of 1.0 x 10^8 electrons.

Homework Equations

N = q/e
qe = mg
q = mg / e
q = mgr / Vb
e = 1.602 x 10^-19 C.
k = 9 x 10^9 - not sure if relevant or not.
Fe = qe
Answer ( as stated in the textbook): 2.3 x 10^-12N

The Attempt at a Solution

Fe = qe?
r = 1. 0 m
e constant used
g = 9.8 N/kg?
Fe = (1.0 x 10^8 C)( 1.6 x 10^-19 C)x (1.0m/9.8N/kg) = 1.63 x 10^-12 N.

 

[objective33]

Through my reasoning, I found the answer.


Charge of one electron = 1.602 x 10^-19 C
Charge of a sphere = 1.602 x 10^-19 x 1.0 x 10^8 C

F = (1/4piÈ) Q1Q2/d^2
(1/4piÈ) is a constant equal to 9 x 10^9
F = 9x10^9 x (1.602x10^-11) x (1.602x10^-11) / 1
F = 2.3 x 10^-12 N

 

[nlightner]

I would like to provide the following response to this content:

The force of electric repulsion between two small spheres can be calculated using the equation Fe = q1q2/r^2, where q1 and q2 are the charges on the spheres and r is the distance between them. In this case, the charge on each sphere can be calculated using the equation q = mg/e, where m is the mass of the sphere and e is the elementary charge. Assuming a mass of 1 gram for each sphere and using the given deficit of 1.0 x 10^8 electrons, we can calculate the charge on each sphere to be 1.0 x 10^-10 C. Plugging this into the equation for force of electric repulsion, along with a distance of 1.0 m, we get a value of 2.3 x 10^-12 N, which matches the answer given in the textbook. It is worth noting that the constant k, which represents the Coulomb constant, is not needed in this calculation since it is already accounted for in the units of charge and distance used. Additionally, the value of g, which represents the acceleration due to gravity, is not relevant in this calculation as it is not related to the force of electric repulsion between the spheres.