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In a vacuum, two particles have charges of q1 and q2, where q1 = +4.2C. They are separated by a distance of 0.31 m, and particle 1 experiences an attractive force of 4.5 N. What is the value of q2, with
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In a vacuum, two particles have charges of q1 and q2, where q1 = +4.2C. They are separated by a distance of 0.31 m, and particle 1 experiences an attractive force of 4.5 N. What is the value of q2, with its sign?

User WizardZ
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1 Answer

4 votes

Answer:

11.44pC.

Step-by-step explanation:

F = kq1q2/r^2

q2 = Fr^2/kq1

= 4.5 x 0.31^2/4.2x 9 x 10^9

= 1.144 x 10^-11C

= 11.44pC

The sign is minus since they attracted to each other

User Jfsantos
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