Question

Two point charges each carrying a charge of +3.5 E−6 C are located 3.5 meters away from each other.

How strong is the electrostatic force between the two points (k = 9.0 E9 Nm2/C2)?
Is this force a repulsive force or an attractive force?


Remember to identify all data (givens and unknowns), list equations used, show all your work, and include units and the proper number of significant digits to receive full credit.

Answer 1

Answer: the answer is 0.009N

Explanation: as we know, force =KqQ/R^2

F= 9*10^9*3.5*10^-6*3.5*10^-6/(3.5)^2

F=9*10^-3N

Answer 2

Given:
Charge on each point, q = +3.5E-6 C
Distance between the points, r = 3.5 m
Coulomb's constant, k = 9.0E9 Nm²/C²

The equation for the electrostatic force between two point charges is given by:

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

where F is the force between the charges, q1 and q2 are the magnitudes of the charges, r is the distance between the charges, and k is Coulomb's constant.

Substituting the given values, we get:

F = (9.0E9 Nm²/C²) * [(3.5E-6 C) * (3.5E-6 C)] / (3.5 m)²

F = 2.3E-2 N

Therefore, the electrostatic force between the two point charges is 2.3E-2 N.

Since both charges are positive, the electrostatic force between them is a repulsive force.