What Determines the Repulsion Between Two Positively Charged Spheres?

[Ry122]

Each of two small non-conducting spheres is charged positively, the combined charge being 40C . When the two spheres are 50 cm apart, each sphere is repelled from the other by a force of magnitude 2.5 N. Determine the magnitude of the smaller of the two charges

my attempt:
use eq. F=kq1q2/d
2.5=((9x10^9)(40-q2)(q2))/.5
this gives me the wrong answer
what am i doing wrong?

 

[Pyrrhus]

You forgot r is square.

 

[nlightner]

It seems like you have set up the equation correctly, but you may have made a mistake in your calculations or in the units used. Here is a step-by-step solution using the given information:

1. First, we need to convert the distance between the two spheres from centimeters to meters, since the equation for force uses meters as the unit of distance. 50 cm = 0.5 m.

2. Next, we can plug in the given values into the equation for force:
2.5 N = (9x10^9 N*m^2/C^2)(40 C)(q2 C)/ (0.5 m)^2

3. Simplifying, we get:
2.5 N = 720 C^2(q2 C)/ 0.25 m^2

4. Rearranging the equation to solve for q2:
q2 = (2.5 N * 0.25 m^2)/ (720 C^2) = 8.33x10^-7 C

Therefore, the magnitude of the smaller of the two charges is approximately 8.33x10^-7 C. Double check your calculations and units to ensure accuracy.