How much work is required to bring a charge to the origin from a distant point?

[fisixC]

Homework Statement

A charge −1.3 × 10e−5
C is fixed on the x-axis
at 8 m, and a charge 1 × 10e−5
C is fixed on
the y-axis at 5 m, as shown on the diagram.
A −4 × 10e−6 C charge is brought from a very
distant point by an external force and placed
at the origin
Calculate the work that had to be done by an
external force to bring Q to the origin from
the distant point.
Answer in units of J.

Homework Equations

Potential energy = kq1q2/r

The Attempt at a Solution

So I set up a conversation of energy problem:
KEi + Ui + W= Uf + KEf

So I know that it starts a very long ways away so this means the Ui(initial potential energy) is basically zero and since it is being brought into the system it isn't moving to begin with so KEi(kinetic energy initial) is zero. Also it is brought to rest in the end meaning that KEf is zero. So we are left with:

W = Uf

Work being equal to the sum of the final potential energy:

U1f = k(4e-6)(1e-5)/5 = 0.0719004J
U2f = k(4e-6)(1.3e-5)/8 = 0.058419075 J

So summed I get: W = 0.130319475J which is incorrect

Any ideas?

 

[SammyS]

It looks like you ignored the signs of the charges.

 

[fisixC]

So you are saying that U2f is negative so that answer should be:

U1f - U2f instead of addition?

 

[fisixC]

That is not the correct answer either for when I subtracted them I got the value of 0.013481325J

 

[fisixC]

Solved: positive is negative and negative is positive so:

U2f - U1f = -0.013481325J

 

[SammyS]

So you are saying that U2f is negative so that answer should be:

U1f - U2f instead of addition?

Actually -- to get technical -- U_2f is negative. You still do addition !

So, U_1f + U_2f = 0.0719004 J + -0.058419075 J

 

[fisixC]

Solved: positive is negative and negative is positive so:

U2f - U1f = -0.013481325J

This is correct.