Question

A particle of charge q is fixed at point P, and a second particle of mass m and the same charge q is initially held a distance r1 from P. The second particle is then released. Determine its speed when it is a distance r2 from P. Let q = 3.1 µC, m = 47 mg, r1 = 0.83 mm, and r2 = 2.5 mm.
_________m/s

Answer 1

Answer:

v=1617.77m/s

Explanation:

A particle of charge q is fixed at point P, and a second particle of mass m and the same charge q is initially held a distance r1 from P. The second particle is then released. Determine its speed when it is a distance r2 from P. Let q = 3.1 µC, m = 47 mg, r1 = 0.83 mm, and r2 = 2.5 mm.

_________m/s

from the law of applied electric force which states that the force of attraction of electric charge particles is directly proportional to the two charge particles and inversely proportional to the square of their distances apart

dk= -dU

kinetic energy equals potential energy

$\frac{1}{2} mv^{2} -\frac{1}{2} mu^{2} =-(\frac{kq^2}{r2} -\frac{kq^2}{r1} )$

$\frac{1}{2} 47*10^-6v^{2} -\frac{1}{2} 47*10^-6*0^{2} =-(\frac{9*10^9*(3.1*10^-6)^2}{(2.5*10^-3)} -\frac{9*10^-9*(3.1*10-6)^2}{0.83*10^-3} )$

23.5*10^-6v^2=96.1-34.596)

v^2=61.504/23.5*10^-6)

v^2=2617191.48

v=1617.77m/s

v=1.617km/s