Work Done on a particle

[Dillion]

Homework Statement

A -2.0*10^-3 C charge is 0.2 m away from a -6.0*10^-3 C charge. How much work is must be done on the first charge to move it to a distance of 0.9m?

Homework Equations

F = qs*qt*k/r^2

W = F * d* cos theta

The Attempt at a Solution

(-2.0*10^-3)(-6.0*10^-3)(9*10^9)/0.2^2
=2700000 N
(-2.0*10^-3)(-6.0*10^-3)(9*10^9)/0.7^2
=220408.1633 N

W = 2700000N * .2 = 540000
W = 220408.1633 * .7 = 154285.7143

The sum of those is 694285.7143


This is obviously not right because the answer is supposed to be -4.2*10^5

 

[Hiranya Pasan]

I think It is easy to use potential method, first calculate the potential difference 0.2m to 0.9m due to the second charge,
the use the eqn, W=qV