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In a vacuum, two particles have charges of q1 and q2, where q1 = +3.11 μC. They are separated by a distance of 0.241 m, and particle 1 experiences an attractive force of 3.44 N.
In a vacuum, two particles have charges of q1 and q2, where q1 = +3.11 μC. They are separated by a distance of 0.241 m, and particle 1 experiences an attractive force of 3.44 N. What is the magnitude and sign of q2?
Answer:Charge of particle 2,Explanation:Given that,Charge 1,The distance between charges, r = 0.241 mForce experienced by particle 1, F = 3.44 NWe need to find the magnitude of electric charge 2. It can be calculated using formula of electrostatic force. It is given by :orSo, the magnitude of electric charge 2 is. Since, the force is attractive then the magnitude of charge 2 must be negative....
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