Question

Particle q1 has a positive 6 µC charge. Particle q2 has a positive 2 µC charge. They are located 0.1 meters apart. Recall that k = 8.99 × 109 N•. What is the force applied between q1 and q2?

Answer 1

Answer:

Therefore,

The force applied between q1 and q2 is 10.788 Newton.

Explanation:

Given:

q1 = 6 µC

q2 = 2 µC

r = 0.1 m

$k = 8.99\times 10^{9}\ Nm^{2}/C^{2}$

To Find:

Force applied between q1 and q2,

F = ?

Solution:

Coulomb's Law :

Coulomb's law states that: The magnitude of the electrostatic force of attraction or repulsion between two point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them.

The Force is given by

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

Where,

$k= Coulomb's\ constant = 8.99\times 10^{9}\ Nm^{2}/C^{2}$

$q_{1}\ and\ q_{2} = Charges$

r = seperation  distance

Substituting the values we get

$F =8.99\times 10^{9}\times \dfrac{6\times 10^{-6}\times 2\times 10^{-6}}{0.1^{2}}$

$F=10.788\ N$

Therefore,

The force applied between q1 and q2 is 10.788 Newton.

Answer 2

Answer:

10.8 N

Explanation: