Question #159542

If the electric forces of repulsion between two 2.80-C charges have magnitude 17.0 N, how far apart are they?


1
Expert's answer
2021-01-29T11:58:02-0500
F=kq2r217=91092.82r2r=6.44104 mF=k\frac{q^2}{r^2}\\17=9\cdot10^9\frac{2.8^2}{r^2}\\r=6.44\cdot10^4\ m


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