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

Strategy For the purposes of this example, we are treating the electron and proton as two point particles, each with an electric charge, and we are told the distance between them; we are asked to calculate the force on the electron. We thus use Coulomb’s law.