Question

Two identical point charges are initially 5.3 cm from each other. If they are released at the same instant from rest, how fast will each be moving when they are very far away from each other? Assume they have identical masses of 1.0 mg.

Answer 1

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)