Velocity of colliding electron and surface

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Homework Statement



An insulating sphere has a radius of 5.00 mm and has a total charge of 4.25 x 10-12 Coulombs, distributed uniformly over its surface. And electron starts from rest at a distance of 9.00mm from the surface and accelerates towards it. How fast in the electron moving when it crashes into the sphere's surface?


Homework Equations



PEe=(kq1q2)/r

W=deltaPEe

V(electric potential around point charge)=(kq)/r

The Attempt at a Solution



I'm afraid I just don't know where to begin. I've been playing around with these equations but I can't get a velocity to fall out.
 
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What is the initial PE and KE of the electron?

What is the final PE and KE of the electron?

Apply the conservation of energy equation and find the final velocity of the electron.
 
Okay, so using PEi+KEi=PEf+KEf...

k, and the charges are constant (I got 6.13x10^21 for the numerators of the PEf and PEi).
Dividing this by the changes in distance I got 1.233x10^-18 at 5mm for PEf and 6.81x10^-19 at 9mm for PEi. Since the electron is starting from rest, I get..

PEi-PEf=.5mv^2

subtracting PEf from PE i I got -1.23 and divided by the mass of an electron (9.12x10^-31) however this will leave me taking the square root of a negative number to get the velocity so I must be doing something wrong.

Also the book says our answer should be 1.31x10^6
 
When the electron is at 9mm, the PE of the system PEi = k*q*e/(9*10^-3)

When the electron is at 5mm, the PE of the system PEf = k*q*e/(5*10^-3)

Now PEf - PEi = 1/2*m*v^2.

Substitute the values and find v.
 
Ah, okay, I was subtracting final from initial. Thank you!