Answer to Question #207050 in Physics for hav

Question

a force of .90n to the left acts on a 5 microcoulombs charge.it is caused by a separate 30 microcoulombs charge...where is the 30 microcoulombs charge located relative to the 5 microcoulombs charge?

Answer 1

We can apply Coulomb's law to figure out the answer:

$$F=k\frac{q_1q_2}{r^2}.$$

Express the distance:

$$r=\sqrt{\frac{kq_1q_2}{F}}=\sqrt{\frac{(9·10^9)(5·10^{-6})(30·10^{-6})}{0.9}}=1.22\text{ m}.$$