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

Answer:

q2 = - 8 × $10^{-6}$ C

negative sign because attract together

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 = $\frac{q1q2}{4\pi \epsilon *r^{2}}$     ............1

here  we know  $\frac{1}{4\pi \epsilon}}$ = 9 × $10^{9}$ Nm²/C²

so from equation 1

2.9 = 9 × $10^{9}$ × $\frac{3.2*10^{-6}*q2}{0.28^{2}}$

q2 = $\frac{2.9*0.28^2}{9*10^9*3.2*10^{-6}}$

q2 = - 8 × $10^{-6}$ C