Question

A positive charge of 5.40 µC is fixed in place. From a distance of 3.00 cm a particle of mass 5.80 g and charge +3.90 µC is fired with an initial speed of 74.0 m/s directly toward the fixed charge. How close to the fixed charge does the particle get before it comes to rest and starts traveling away?
What are the steps and formulas to obtain this answer?

Answer 1

In this problem, we have to use the priciple of energy conservation, since no external forces acting on the sytem.

 

Initially,

Potential Energy of two-charge system = kq₁q₂/r = (9x10⁹) * (5.4x10^-6) * (3.9x10^-6) / (3x10^-2) = 6.32 J

Kinetic Energy of particle of mass 5.8 g = (1/2)mv² = (1/2) * (5.8x10^-3) * (74)² = 15.88 J

 

Finally, let the particle stop at a distance d from the charge.

Potential Energy of two-charge system = kq₁q₂/r = (9x10⁹) * (5.4x10^-6) * (3.9x10^-6) / (d)

Kinetic Energy of particle of mass 5.8 g = 0

 

 

Eqauting the total intial energy and final energy, we get,

6.32 + 15.88 = [(9x10⁹) * (5.4x10^-6) * (3.9x10^-6) / (d)] + 0

d = 0.19/22.2 m = 8.56 x 10^-3 m (or 8.56 x 10^-1 cm)

 

Thus, particle stops at a distance of 8.56 x 10^-1 cm from the charge.