Question

A charge of -3.14 μC is fixed in place. From a horizontal distance of 0.0517 m, a particle of mass 9.74 x 10-3 kg and charge -9.33 μC is fired with an initial speed of 76.7 m/s directly toward the fixed charge. How far does the particle travel before its speed is zero?

Answer 1

Given that
Q = -3.14*10^-6 C, d = 0.0517 m, m = 9.74*10^-3 kg, q' = -9.33*10^-6 C, v = 76.7 m/s
find the distance the particle travel before its speed is zero.
initial energy = kqq'/d + mv^2/2
final energy = kqq'/x
kqq'/d + mv^2/2 = kqq'/x
1/d + mv^2/(2kqq') = 1/x
so x = 1/[1/d + mv^2/(2kqq')]
   x = 1/[1/0.0517+9.74*10^-3*(76.7)^2/(2*9*10^9*3.14*10^-6*9.33*10^-6)]
    x = 0.00781 m
the required distance = d - x = 0.0517-0.00781 = 0.04389 m