How Do You Calculate the Charges on Two Repelling Particles?

[fallen186]

Homework Statement

Two point particles separated by 0.4 m carry a total charge of 185 µC.
(a) If the two particles repel each other with a force of 80 N, what are the charges on each of the two particles?

Homework Equations

F = k* (q1*q2)/(r^2)

The Attempt at a Solution

I tried using altering the formula so it would be 80 = (185-x)*(x) /(.4^2). I tried putting this in my calculator in which y1 = (185-x)*(x)/(.4^2), y2= 80 yet it tells me it intersects at x = 184.9. Um what did i do wrong? I would appreciate the help :).

 

[AEM]

Two quick questions: What happened to k? and Shouldn't you do something with that micro coulombs that follows the 185?

 

[thomate1]

As a scientist, it is important to carefully check your calculations and equations to ensure accuracy in your results. In this case, it seems that you may have made a mistake in setting up the equation for the force. The correct equation should be F = k*q1*q2/(r^2), where q1 and q2 are the charges on the two particles and r is the distance between them.

Using this corrected equation, we can rearrange to solve for the charges on the particles:

80 = (9*10^9)*(q1*q2)/(.4^2)
80 = (9*10^9)*(185-x)*x/(.4^2)
80 = (1.6875*10^11)*(185-x)*x
80 = 31218750000 - (1.6875*10^11)*x^2
(x^2) = (31218750000 - 80)/(1.6875*10^11)
x = √(31218699920/1.6875*10^11)
x = 0.00007858 or -0.00007858

Since the charges of the particles cannot be negative, we can conclude that the charges on the two particles are approximately 0.00007858 C and 184.99992142 C.

It is also important to note that these values are approximations and may vary slightly depending on the level of precision used in the calculations. As a scientist, it is important to always double check your work and use appropriate levels of precision in order to obtain accurate results.