Answer to Question #305429 in Physics for zhaiiy

Question #305429

Two identically charged one-peso coins are 1.5 m apart on a table. What is the charge of the coins if each of them experiences a repulsive force of 2.0 N?


1
Expert's answer
2022-03-03T12:05:50-0500

The Coulomb's law says

F=kq1q2r2=kq2r2F=k\frac{q_1q_2}{r^2}=k\frac{q^2}{r^2}

The absolute value of charge

q=Fr2/kq=2.01.529109=2.2105Cq=\sqrt{Fr^2/k}\\ q=\sqrt{\frac{2.0*1.5^2}{9*10^9}}=2.2*10^{-5}\:\rm C


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