(a) Force between the two charges
The electrostatic force between the two charges is given by:
[tex] F=k\frac{q_1 q_2}{r^2} [/tex]
where k is the Coulomb's constant, q1 and q2 the two charges, r their separation.
In this problem:
[tex] q_1 =6 \mu C=6 \cdot 10^{-6}C [/tex]
[tex] q_2=2 \mu C=2 \cdot 10^{-6}C [/tex]
[tex] r=0.1 m [/tex]
Substituting into the equation, we find
[tex] F=(8.99 \cdot 10^9 Nm^2C^{-2})\frac{(6 \cdot 10^{-6}C)(2 \cdot 10^{-6}C)}{(0.1 m)^2}=10.8 N [/tex]
(b) direction of particle q2
Particle q2 wants to move in the direction of the force acting on it. The direction of the force depends on the relative sign of the two charges: like charges attract each other, opposite charges repel each other. In this case, the two charges are both positive, so they repel each other and q2 tends to move away from particle q1.
(1) The force between the two charges is 10.788 N.
(2) The direction of the particle 2 will be in opposite direction to particle 1 because both charges are positive and similar charges repel.
The given parameters:
The force between the two charges is calculated as follows;
[tex]F = \frac{kq_1 q_2}{r^2} \\\\F = \frac{8.99 \times 10^9 \times 6 \times 10^{-6} \times 2 \times 10^{-6} }{0.1^2} \\\\F = 10.788 \ N[/tex]
The direction of the particle 2 will be in opposite direction to particle 1 because both charges are positive and similar charges repel.
Learn more about Coulomb's force here: https://brainly.com/question/66110
