Challenge Problem For Physics Prodigies Only!?

Question

A positive charge of 5.20 microC is fixed in place. From a distance of 4.60 cm a particle of mass 5.60 g and charge +3.20 microC is fired with an initial speed of 72.0 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?

**HINT: The particle comes to rest when it has no kinetic energy. The change in kinetic energy will be equal to the increase in electrical potential energy. Note that the particle has some potential energy at the starting point.

Answer 1

Hi, Lilbac2003--Your "hint" takes all the fun out of it   :)

The charges are the same sign, duh...otherwise the answer is 0.

Solution:

At the point of closest approach,

KE = PE(close) - PE(far)

1/2 mv^2 = kQ1Q2(1/xclose - 1/xfar)

with a little algebra...

xclose = 1/[(mv^2/(2kQ1Q2)) + (1/xfar)]

with k=9E9 Nm^2/C^2; m=5.6E-3 kg; Q1=3.2E-6C; Q2=5.2E-6C; v=72m/s and xfar=.046 meters, we get...

xclose = .012 meters or 1.2 cm  (if I plugged and chugged correctly, but the set up and method are A OK).

-Fred