Charges, Electric Fields, and Work?

[ObviousManiac]

Homework Statement

a +70µC charge is placed 50 cm from a -50µC charge. How much work would be required to move a -.5µC test charge from a point 10 cm from the +70µC charge to a point 20 cm from the -50µC charge?

Homework Equations

F=k(q$_{1}$q$_{2}$/r$^{2}$)
PE=Fd

...that's all I could get, I could be off by a longshot

The Attempt at a Solution

I tried solving for PE by finding the Force acting upon the test charge in its initial and final positions, and then using the difference of those two positions as the force in the PE=Fd formula, where d = .2 m

All I can say is that I was way off my teacher's given answer, which is 2.66 J.

 

[Spinnor]

Could you do the easier problem, with only 1 fixed charge 1 moving charge, if so you just add the work done separately for the two fixed charges of your problem? What would the work be? You have a formula, PE=Fd, which works for some problems? You need the integral form of your formula, you have an integral you must do as the force changes with distance. Can you integrate 1/r^2?

 

[ObviousManiac]

Could you do the easier problem, with only 1 fixed charge 1 moving charge, if so you just add the work done separately for the two fixed charges of your problem? What would the work be? You have a formula, PE=Fd, which works for some problems? You need the integral form of your formula, you have an integral you must do as the force changes with distance. Can you integrate 1/r^2?

Not needed, I figured it out.

Using PE = K(q1q2/r)

I just found ∆PE for the two positions:

PE (first position) = K(((-.5µC * 70µC)/.1) + ((-.5µC * -50µC)/.4))
= -2.58 J

PE (second position) = K(((-.5µC * 70µC)/.3) + ((-.5µC * -50µC)/.2))
= .075 J

∆PE = PE2 - PE 1
= (.075) - (-2.58)
= .075 + 2.58
= 2.66 J