Answer:
when they are far away from each other they will move at a velocity of v∞/q = 1.84*10⁶ m/(s*C)
Step-by-step explanation:
if we assume that the charges repel according to a Coulomb's law ( we neglect any relativistic effects) , then the force of repulsion is
F rep = k*q*q/r²
where q= charge
k= Coulomb's constant
r= distance
from Newton's second law :
F rep = m*a= m*dv/dt = m*dv/dr * dr/dt = m*v*dv/dr
k*q*q/r² = m*v*dv/dr
k*q*q/(m*r²) = v*dv/dr
(k*q*q/m) ∫ r⁻² dr= ∫v*dv
-2*k*q*q/m*r = v² - v₁²
for r=r₀ , v=0 ,then
-2*k*q*q/m*r₀ = 0 - v₁²
v₁² = 2*k*q*q/m*r₀
v²= -k*q*q/(m*r) + v₁²= 2*k*q*q/m* (1/r₀-1/r)
then
v²= 2*k*q²/m* (1/r₀-1/r)
when the masses are far away from each other r→∞ , 1/r → 0, then
v∞² = 2*k*q²/(m*r₀)
v∞ = √[2*k*q²/(m*r₀)]
since
k= 8.987*10⁹ N·m²/C² , m= 1 mg = 0.001 kg , r₀= 5.3 cm = 0.053 m
v∞/q = √[[2*k/(m*r₀)] = √[2*8.987*10⁹ N·m²/C²/(0.001 kg*0.053 m)] = 1.84*10⁶ m/(s*C)
v∞/q = 1.84*10⁶ m/(s*C)
