Answer :
Answer:
The distance that separates the two particles is 7.42 cm.
Explanation:
Given;
the mass of each particle, m = 3 mg = 3 x 10⁻⁶ kg
the magnitude of charge of each particle, q = 6.0 nC
speed of each particle, v = 5.0 m/s
[tex]F = ma = \frac{kq^2}{r^2}[/tex]
a = v/t
where;
a is the acceleration of the two particles
v is the final velocity
t is time
v = u + gt
5 = 0 + 9.8t
5 = 9.8t
t = 5/9.8
t = 0.51 s
a = v/t
a = 5/0.51 = 9.8 m/s²
Total force on the two particles = (2m)a = (2* 3 x 10⁻⁶)9.8
F = 5.88 x 10⁻⁵ N
Substitute in the value of F in the above equation and calculate r
[tex]F = \frac{kq^2}{r^2}[/tex]
where;
k is coulomb's constant = 8.99 x 10⁹ Nm²/c²
r is the distance of separation between the two particles
[tex]F = \frac{kq^2}{r^2} \\\\r^2 = \frac{kq^2}{F}\\\\r = \sqrt{ \frac{kq^2}{F}} = \sqrt{ \frac{8.99*10^9*(6*10^{-9})^2}{5.88*10^{-5}}} = 0.0742 \ m[/tex]
Therefore, the distance that separates the two particles at the instant when each has a speed of 5.0 m/s, is 7.42 cm.