How Far Does a Charged Particle Travel Before Stopping?

[lee_chick]

Sorry for the poor title - I'm not entirely sure how to describe the type of problem!

Homework Statement

A charge of -4.00 x 10^-6 C is fixed in place. From a horizontal distance of 55.0cm a particle of mass 2.50 x 10^-3 kg and charge -3.00 x 10^-6 C is fired with an initial speed of 15.0m/s directly towards the fixed charge. How far does the particle travel before it stops and begins to return back?

Homework Equations

V = k(Q/r)

PE = (QV)/2

KE = (mv²)/2

The Attempt at a Solution

I've tried this a few times and haven't found anything that works so here is my latest attempt:

A__________________________B_______________________C
A = fixed charge
B = unknown distance
C = Starting point for moving charge

Q_a = -4.00 x 10^-6 C
Q_c = -3.00 x 10^-6 C
v_0c = 15m/s
v_1c = 0m/s
m_c = 2.50 x 10^-3 kg
r_c = 0.55m
r_b = ?

r_b = k(Q_a/V_c)
V_c = 2PE/Q_c
PE = -KE
KE = (mv²)/2

KE = (2.50 x 10^-3 kg x -15m/s)/2 = -0.009375
PE = 0.009375

V_c = 2(0.009375)/-3.00 x 10^-6 C = -6.25 x 10³

r_b = 9 x 10⁹ (-4.00 x 10^-6 C/-6.25 x 10³) = 5.76 m

 

[rl.bhat]

KE = 1/2*m*V^2
PE = k*Qa*Qc/rb
In this problem ra is not needed.

 

[Doc Al]

Homework Equations

V = k(Q/r)

PE = (QV)/2

Don't confuse the PE between two point charges with the energy stored in a capacitor. Look up the correct formula for PE.

KE = (mv²)/2

Good.

Hint: Compare the PE at point C with the PE at point B.

 

[lee_chick]

Thanks - I knew I was doing something wrong and just couldn't seem to sort out the problem correctly!