Can someone help? I cannot figure out this problem to save my life?

Question

A positive charge of 5.20 mC is fixed in place. From a distance of 4.60 cm a particle of mass 5.60 g and charge +3.20 mC 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.

This is my work but it's still wrong for some reason.

q1 = 5.20 x 10^-6 C, q2 = 3.20 x 10^-6 C, r1 = 0.046 m, U = 72.0 m/s

εo = 9 x 10^9

the potential energy of the two charges when they are at distance r1 = 0.046 m apart is

U1 = (1/4πεo)(q1*q2/r1)

where (1/4πεo) = 9*10^9 Nm2/C2

the potential energy of the two charges when they are at distance r2 apart is

U2 = (1/4πεo)(q1*q2/r2)

then the work done is W = U2 - U1

but from the work energy thereom the work done is equal to the change in kinetic energy

U2 - U1 = ( 1/2)mV2 - (1/2)mu2

final velocity is V = 0 m/s then

U2 - U1 = -(1/2)mu^2

(1/4πεo)(q1*q2/r2) - (1/4πεo)(q1*q2/r1) = -(1/2)mu^2

1/r2 - 1/r1 = - (1/2)mu^2 / ((1/4πεo)(q1*q2))

1/r2 = 1/r1 - (1/2)mu^2 / ((1/4πεo)(q1*q2))

r2 = ___ m (i got 0.013 m)

Answer 1

you write this??

U2 - U1 = ( 1/2)mV2 - (1/2)mu2

===========================

energy conservation> before= after

U1 + 0.5 mu^2 [farther] = U2 + 0.5 mv^2 [closer]

v=0

U1 + 0.5 mu^2 [farther] = U2 [closer]

U2 [closer] = U1 + 0.5 mu^2 [farther]

(1/4πεo)(q1*q2/r2) = (1/4πεo)(q1*q2/r1) + (1/2)mu^2

1/r2 = 1/r1 + [mu^2 /2*q1 q2 (9*10*9)] >>>>>> shouldn't this be true

+++++++++

check units of charges >> mC = milli coulomb

uC = micro coulomb >>> you are using

===================

q1 = 5.2*10^-6 , q2 = 3.*10^- 6 >>> your values

mu^2/2*q1 q2 (9*10*9) = [0.0056*72*72 / 2*5.2*3.2*10^-12*9*10^9]

= [1/0.0103]

----------- my calculation works like

1/r2 = [1/0.046] + [1/0.0103]  = 118.8265

r2 = 0.0084 m

r2 = 0.84 cm

==================

your calculation

1/r2 = [1/0.046] - [1/0.0103]  >>>>>>>>> negative with KE

1/r2 = 21.74 - 97.09 = - 75.347

r2 = - 0.01327 m [what about this -ve sign >> indicating that q2 has collided with q1 and crossed it]

your problem lied in energy conservation>> signs

if use use> milli C >> r2 = 0.0097 centimeter >>> very very close approach