A particle of charge +3.00\times 10^{-6} C is 12.0 cm distant from a second particle of charge -1.50\times 10^{-6} C. Calculate the magnitude of the electrostatic force between the particles.

#### Solution:

Show me the final answer↓

We will use Coulomb’s law to figure out the electrostatic force between the particles. Coulomb’s law states:

F=k\dfrac{(q_1)(q_2)}{r^2}

(Where F is the electrostatic force, k is Coulomb’s constant, q_1 is the charge of the first particle, q_2 is the charge of the second particle, and r is the distance between the two charges)

Substitute the values we know:

F=k\dfrac{(q_1)(q_2)}{r^2}

F=(8.99\times 10^9)\dfrac{(3\times 10^{-6})(1.5\times 10^{-6})}{0.12^2}

F=(8.99\times 10^9)\dfrac{(3\times 10^{-6})(1.5\times 10^{-6})}{0.12^2}

(Note that we only use the magnitudes of the charges since there is an attraction between them. Also note that 12 cm = 0.12 m)

F=2.81 N

#### Final Answer:

F=2.81 N