The attractive electrostatic force between the point charges 8.11×10−6 C and Q has a magnitude of 0.495 N when the separation between the charges is 5.40 m.
find the sign and magnitude of the charge Q
From Coulomb's inverse square law the force of attraction or repulsion is given by
Here in our problem 0.495 = 9x109x8.22x10-6 xQ/(5.4)2
Therefore the charge on the second body is
Q = 0.495x(5.4)2/ (9x109x8.11x10-6)
Q = 198x10-3C
Since the force is attractive the charges should be opposite in charge. Hence this charge should be negative.
Therefore Q=-198x10-3C
The attractive electrostatic force between the point charges 8.11×10−6 C and Q has a magnitude of...
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