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? in which direction does particle q2 want to go?

Answer 1

(a) Force between the two charges

The electrostatic force between the two charges is given by:

$F=k\frac{q_1 q_2}{r^2}$

where k is the Coulomb's constant, q1 and q2 the two charges, r their separation.

In this problem:

$q_1 =6 \mu C=6 \cdot 10^{-6}C$

$q_2=2 \mu C=2 \cdot 10^{-6}C$

$r=0.1 m$

Substituting into the equation, we find

$F=(8.99 \cdot 10^9 Nm^2C^{-2})\frac{(6 \cdot 10^{-6}C)(2 \cdot 10^{-6}C)}{(0.1 m)^2}=10.8 N$

(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.

Answer 2

The answer is:

10.8 N

Away from q1 in a straight line