Question

Two positively charged particles are separated on an axis by 0.1 meters, and q1 has a 6 µC charge and q2 has a 2 µC charge. Recall that k = 8.99 × 109 N • meters squared over Coulombs squared.. What is the force applied between q1 and q2 in N? In which direction does particle q2 want to go?

Answer 1

Answer:

10.8 N & away from particle q1, in a straight line to the right

Explanation:

•°•°•°

Answer 2

Answer:

Force between two charges is 10.8 N

both charges are positive nature charge so they both repel each other and hence q2 will move away from charge q1

Explanation:

As we know that the force between two charge particles is given by Coulomb's law of electrostatics

So we will say it as

$F = \frac{kq_1q_2}{r^2}$

here we know that

$q_1 = 6\mu C$

$q_2 = 2\mu C$

$d = 0.1 m$

$F = \frac{8.99 \times 10^9(6 \mu C)(2\mu C)}{0.1^2}$

$F = 10.8 N$

Since both charges are positive nature charge so they both repel each other and hence q2 will move away from charge q1