In a vacuum, two particles have charges of q1 and q2, where q1 = +3.2 µc. they are separated by a distance of 0.28 m, and particle 1 experiences an attractive force of 2.9 n. what is q2 (magnitude and sign)?

q2 = - 8 × C

negative sign because attract together

Explanation:

given data

q1 = +3.2 µC = 3.2 × C

distance r = 0.28 m

force F = 2.9 N

to find out

q2 (magnitude and sign)

solution

we know that here if 2 charge is unlike charge

than there will be electrostatic force of attraction , between them

now we apply coulomb law that is

F =     ............1

here  we know   = 9 × Nm²/C²

so from equation 1

2.9 = 9 × ×

q2 =

q2 = - 8 × C

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