Question

Use Coulombs Law to calculate the force between a 0.05 C charge and a 0.06 C charge when they are 2 meters apart from each other

Answer 1

Answer:

6,740,663.84 N

Explanation:

Coulombs Law states that $F = k\frac{q_{1} q_{2}}{r^{2}}$, where k = coulombs law, q1 and q2 are charge one and charge 2 respectively, and r is distance

Substitute for q1 and q2, multiply .05 and .06 together

$f= k \frac{.003}{r2}$

Substitute for r, simplify $2^{2}$

$f= k \frac{.003}{4}$

Divide .003 by 4

$f=k*.00075$

Substitute in k

$f=(8.9875517873681764 * 109 N.m2.C-2)*.00075$

$force = 6740663.8405261$