Question

A 3.5 x 10-6 C charge is located 0.28 m from a 2.8 x 10-6 C charge. What is the magnitude of the force being exerted on the smaller charge, to the nearest tenth of a Newton?

Answer 1

The electrostatic force between two charges is given by Coulomb's law:

`F=k_e (q_1 q_2)/(r^2)`
where
ke is the Coulomb's constant
q1 is the first charge
q2 is the second charge
r is the separation between the two charges

By substituting the data of the problem into the equation, we can find the magnitude of the force between the two charges:

`F=(8.99 \cdot 10^(9) Nm^2C^(-2) ) ((3.5 \cdot 10^(-6) N/C)(2.8 \cdot 10^(-6)N/C))/((0.28m)^2)=1.2 N`