Question #207050

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?


1
Expert's answer
2021-06-15T10:00:46-0400

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


F=kq1q2r2.F=k\frac{q_1q_2}{r^2}.

Express the distance:


r=kq1q2F=(9109)(5106)(30106)0.9=1.22 m.r=\sqrt{\frac{kq_1q_2}{F}}=\sqrt{\frac{(9·10^9)(5·10^{-6})(30·10^{-6})}{0.9}}=1.22\text{ m}.


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