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In a vacuum, two particles have charges of q1 and q2, where q1 = +3.5 µC. They are separated by a distance of 0.26 m, and particle 1 experiences an attractive force of 3.4 N. What is q2 (magnitude and sign)?

2007-03-12 16:14:42 · 1 answers · asked by Cameron 2 in Science & Mathematics Physics

1 answers

F=k q1 q2 /R^2
where k=8.988 E+9 N m^2 C^-2

q2=F R^2/(k q1)
q2=3.4 (.26)^2 / ( 8.988 E+9 x 3.5 E-6)
q2=7.3 E-6 or
q2=7.3 uC \
q2 is negative due to attraction to positive q1.

I hope it was useful.

2007-03-13 04:31:46 · answer #1 · answered by Edward 7 · 1 0

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