Applying Coulomb's Law

[Kennedy111]

Homework Statement

What is the distance between two charges of -5.00 C each if the force of electrostatic repulsion acting on them is 4.00 x 10^3 N?

q1 = -5.00 C
q2 = -5.00 C
Fe = 4.00 x 10^3
k = 8.99 x 10^9 Nm^2/C^2

Homework Equations

Fe = kq1q2/r^2

The Attempt at a Solution

I rearranged the equation above and got:

r = (sqrt) kq1q2/Fe

r = (sqrt) (8.99 x 10^9 Nm^2/C^2)(5.00 C)(5.00 C) / (4.00 x 10^3 N)
= 7495.832175 m

To three significant digits my answer would be 7.50 x 10^ 3 m

 

[collinsmark]

Hi Kennedy111,

Homework Statement

What is the distance between two charges of -5.00 C each if the force of electrostatic repulsion acting on them is 4.00 x 10^3 N?

q1 = -5.00 C
q2 = -5.00 C
Fe = 4.00 x 10^3
k = 8.99 x 10^9 Nm^2/C^2

Homework Equations

Fe = kq1q2/r^2

The Attempt at a Solution

I rearranged the equation above and got:

r = (sqrt) kq1q2/Fe

r = (sqrt) (8.99 x 10^9 Nm^2/C^2)(5.00 C)(5.00 C) / (4.00 x 10^3 N)
= 7495.832175 m

To three significant digits my answer would be 7.50 x 10^ 3 m

Your math and approach look fine to me.

But just for the sake of practicality, you might wish to double check that you're using the correct units from the beginning. Five Coulombs (whether negative or positive) is a lot of charge. Most charges in electrostatics problems I'm used to use units of μC.

I'm just sayin', in the system you're modeling, as you've expressed it, there is still a four kilo-Newton force between the charges, even after separating them by seven and a half kilometers (several miles). That's a lot of charge!