Physics: Two particles, each having a mass of 3.0 mg and having equal but opposite charges of magnitude of 6.0 nC, are released simultaneously from rest when they are a very large distance apart. What distance separates the two at the instant when each has a speed of 5.0 m/s?

X/7 = 18/21Lets Reduce:21/7=3 so we know for every 3 in 18/21 we have one in x/7This means 6/7=18/21Check:18/6=321/7=3...

Two particles, each having a mass of 3.0 mg and

makaileep7449

21.01.2022

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

$F = ma = \frac{kq^2}{r^2}$

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

$F = \frac{kq^2}{r^2}$

where;

k is coulomb's constant = 8.99 x 10⁹ Nm²/c²

r is the distance of separation between the two particles

$F = \frac{kq^2}{r^2} \\\\r^2 = \frac{kq^2}{F}\\\\r = \sqrt{ \frac{kq^2}{F}} = \sqrt{ \frac{8.99*10^9*(6*10^{-9})^2}{5.88*10^{-5}}} = 0.0742 \ m$

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.

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