Work needed to move alpha particle

[fruitbubbles]

Homework Statement

Four electrons are located at the corners of a square 10.0 nm on a side, with an alpha particle at its midpoint.
How much work is needed to move the alpha particle to the midpoint of one of the sides of the square?

Homework Equations

W = PE_o-PE_f
PE (if several point charges) = q_o*k*Σ(q/r)
q_o will be the alpha particle (2*(1.6*10^-19 C))
, and the other q will be the 4 electrons

The Attempt at a Solution

[/B]
I use the Pytagorean Theorem to find the original distance from each electron to the alpha particle, and I get (.00707 m). Since all four electrons are the same charge and same distance, I multiple 4*(e^-/.00707) and then by k and q_o, and get a PE_o of -2.6*10^-25J. To get PE_f, I calculate the distance, but this time since it will be in the midpoint of one of the sides of the square, the alpha particle will be .005 m from two of the electrons, and again using the Pythagorean Theorem I calculate that it is 0.0112 m from the other two. I plug all that into the equation for PE to get PE_f, which I get as -2.66*10^-25 J (awfully close to the original PE, which seems kinda weird). Then I do W = PE_o-PE_f and get 6*10^-27J of work done, but the answer is -6.08*10^-21 J. I'm not really sure where I went wrong.

 

[Doc Al]

I use the Pytagorean Theorem to find the original distance from each electron to the alpha particle, and I get (.00707 m).

The dimensions are given in nm. 1 nm = 10^-9 m.

 

[fruitbubbles]

The dimensions are given in nm. 1 nm = 10^-9 m.

Oops, my mistake. I see that now I'm getting the right power, but I'm not sure why the answer should be negative when my answer is positive. Is it because since the positive alpha particles is getting closer to two electrons (which attract the positive charges), the PE decreases?

 

[haruspex]

, but I'm not sure why the answer should be negative when my answer is positive. Is it because since the positive alpha particles is getting closer to two electrons (which attract the positive charges), the PE decreases?

Yes.

 

[fruitbubbles]

Yes.

I don't understand why..just setting up the equation W = PE₀-PE_f wouldn't just take care of the sign, since we are calculating the work. Like, why does there have to be..further thinking about it after calculating W?

 

[haruspex]

I don't understand why..just setting up the equation W = PE₀-PE_f wouldn't just take care of the sign, since we are calculating the work. Like, why does there have to be..further thinking about it after calculating W?

The trouble with just applying equations like that is that you have to remember exactly what the terms stand for. What precisely is W in that equation?