Calculating work done by E field/voltage

[xcvxcvvc]

Homework Statement

Four point particles with charges .6 microC, 2.2 microC, -3.6microC, and 4.8 microC are placed at the corners of a square of side 10cm. what is the external work needed to bring a charge of -5 microC from infinity to the center of the square? (Assume the speed of the -5 microC charge is kept constant)

Homework Equations

V=kq/r(from infinity)
w = Q * v
Vtotal = V1 + v2...Vn

The Attempt at a Solution

9x10^9 * (-5 x 10^-12) / squareroot(.1^2 + .1^2) * (.6 + 2.2 - 3.6 + 4.8) = -1.28 J
answer in book: -2.55J
my answer is nearly a perfect factor of 2 away. What's going on here?

 

[thaer_dude]

I'm not too sure what you did in your calculations but..
First step is to do a diagram with the 4 charges each at a corner of the square. Then you have to find the potential energy at the center of the square (which is acquired by adding the V=kq/r for each charge). Now that you have the potential energy at the center of the square you can just apply w=qv to get the correct answer.

 

[xcvxcvvc]

I'm not too sure what you did in your calculations but..
First step is to do a diagram with the 4 charges each at a corner of the square. Then you have to find the potential energy at the center of the square (which is acquired by adding the V=kq/r for each charge). Now that you have the potential energy at the center of the square you can just apply w=qv to get the correct answer.

that's what i did, and there is no need for a diagram due to the simplicity of a cube


let me explain my calculations: w = k Qq/r

i factored the k out, i factored the r out, and the Q(the charge brought to the middle) is also factored out. I also factored out a x10^-6 (the -12 comes from that), so all that is left is the sum of the charges on the corner.

 

[thaer_dude]

your method is fine, you must have done a calculation error somewhere. just redo it on a piece of paper.

 

[xcvxcvvc]

your method is fine, you must have done a calculation error somewhere. just redo it on a piece of paper.

I have! Could you do it yourself to confirm my answer? Erroneous book answers are not unheard of.

 

[thaer_dude]

yeah, I had just done it and I got -2.55J as the answer right before I posted my reply. here:

W=(kQq)/r
get it into W=(k)*(Q)/(r)*(sum of qs) form
W=(9*10^9)*(-5microC)/(0.0707m) *(.6microC + 2.2microC - 3.6microC + 4.8microC)
W=-636396*(4microC)
W=-2.55J

 

[thaer_dude]

I'm kind of sleepy so i couldn't pinpoint the problem earlier but I'm pretty sure you just used the wrong value of r, that's why I recommended to draw a diagram. its the middle of a square with 10 cm sides so if you do pythagoras you have to use .05m and not .1m like you did

 

[xcvxcvvc]

I'm kind of sleepy so i couldn't pinpoint the problem earlier but I'm pretty sure you just used the wrong value of r, that's why I recommended to draw a diagram. its the middle of a square with 10 cm sides so if you do pythagoras you have to use .05m and not .1m like you did

omg... of course. a factor of two was no coincidence! I forgot to divide my length by two!