Question

How strong is the repulsive force exerted on two point charges that each carry 1.0 E-6 C of negative charge and are 0.30 meters apart? Coulomb’s Constant is 9.0 E9 N*m^2/C^2 Remember to identity all of your data, write the equation, and show your work.

Answer 1

Answer

Force between two charged bodies q₁ and q₂ is given by the formula below.

F = Kq₁q₂/r²

Where k = constant

            r = distance between them

F = (9.0 E9 × 1.0 E-6 × 1.0 E-6)/0.30²

  = 9.0E-3/0.30²

   = 9.0E-3/0.09

   = 0.1 N

Answer 2

Coulomb's law states that the force between two charged particles is directly  proportional to the multiplication of the two charges and inverse with the square of the distance between them.

Mathematically,

$F=k\frac{q_{1}q_{2}}{r^2}$

where: F is the force in (N), $q_{1},q_{2}$ are the charge of the two particles in (C), r is the distance between them and k is Coulomb's constant

Thus

$F= 9.0*10^9* \frac{1.0*10^-6*1.0*10^-6}{0.3^2} =0.1N$