Answer:
70.28 m/s²
Explanation:
Step 1: identify the given parameters
mass of each particle (m) = 6.7X10^-5 kg
Charge of the particles = +5X10^-6 C and -5X10^-6 C
Distance between the particles = 0.68 m
Step 2: calculate the force of attraction between the particles
Applying coulomb's law
[tex]F = \frac{k*q_{1}*q_{2}}{r^{2} }[/tex]
where k is a constant = 9X10^9 Nm²/c²
[tex]F = \frac{9X10^9* 5X10^{-6}*5X10^{-6}}{0.68^{2}}[/tex]
F = 0.4866N
Step 3: calculate the acceleration of each particle
F = m*a
a = F/m
a = 0.4866/(6.7X10^-5)
a = 7,262.69 m/s²
Step 4: calculate the velocity of each particle when the separation is one-half the original value
[tex]\frac{1}{2} X 0.68 = 0.34 m[/tex]
V² = U² + 2as
where U is initial velocity = 0
V² = 0 + 2 X 7,262.69 X 0.34
V² = 4,938.629
V = √4,938.629
V = 70.28 m/s²
the rate of each particle when the separation between them is one-half the original value = 70.28 m/s²
