Answer to Question #253414 in Electricity and Magnetism for shamma

Question #253414

A positive and a negative charge, each of magnitude 2.6 x 10–5 C, are separated by a distance of 25 cm. Find the force on each of the particles

1
Expert's answer
2021-10-19T15:09:48-0400

Given:

q=2.6105Cq=2.6*10^{-5}\:\rm C

r=0.25mr=0.25\:\rm m

k=9109Nm2/C2k=9*10^9\:\rm N\cdot m^2/C^2


The Coulomb law says

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

F=9109(2.6105)20.252=97NF=9*10^9*\frac{(2.6*10^{-5})^2}{0.25^2}=97\:\rm N

The force is attraction.


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