Question

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)?

Answer 1

q2 = - 8 ×
`10^(-6)` C

negative sign because attract together

Step-by-step explanation:

given data

q1 = +3.2 µC = 3.2 ×
`10^(-6)` 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 =
`(q1q2)/(4\pi \epsilon *r^(2))` ............1

here we know
`(1)/(4\pi \epsilon)}` = 9 ×
`10^(9)` Nm²/C²

so from equation 1

2.9 = 9 ×
`10^(9)` ×
`(3.2*10^(-6)*q2)/(0.28^(2))`

q2 =
`(2.9*0.28^2)/(9*10^9*3.2*10^(-6))`

q2 = - 8 ×
`10^(-6)` C