Question #148476
Two copper spheres are currently 1.2 meters apart. one sphere has a charge of +2.2 × 10-4 c and the other has a charge of –8.9 × 10-4 c. what is the force between the charged spheres? is the force attractive or repulsive?
1
Expert's answer
2020-12-04T11:14:01-0500

The force between two charged bodies is given by Coulomb's law:


F=kq1q2r2F = k\dfrac{|q_1||q_2|}{r^2}

where q1=2.2×104Cq_1 = 2.2\times 10^{-4}C and q2=8.9×104Cq_2 = -8.9\times 10^{-4}C are the charges of the first and second body respectively, r=1.2mr = 1.2m is the distance between the bodies, and k9×109 Nm2/C2k\approx 9\times 10^9\space N\cdot m^2/C^2 is Coulomb's constant.

Thus, obtain:


F=9×1092.2×1048.9×1041.221224 NF =9\times 10^9\dfrac{| 2.2\times 10^{-4}|| -8.9\times 10^{-4}|}{1.2^2} \approx 1224\space N

Answer. 1224 N.


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