# [Electromagnetism] How much work is required in moving Q3 to infinity?

1. Jan 30, 2012

### Sean1218

1. The problem statement, all variables and given/known data

How much work is required in moving Q3 to infinity while Q1 and Q2 remain in their positions?

Q3-------a-------|
|-----------------| b
Q1-------------Q2

a = 16.0 cm
b = 6.0 cm
Q1 = 5.70 μC
Q2 = -5.70 μC
Q3 = 1.8 μC

2. Relevant equations

W=ΔPE
PE=kQ1Q2/r

3. The attempt at a solution

Just added the potential energies associated with Q3 which is the work needed to move Q3 to infinity (i.e. bring said potential energies to 0).
kQ1Q3/r13 + kQ2Q3/r23
=(8.9875e9)(5.7e(-6))(1.80e(-6))/0.06 + (8.9875e9)(-5.7e(-6))(1.8e(-6))/sqrt(0.06^2 + 0.16^2)
=0.997 J

Also tried calculating ΔPE = final - initial (I think I messed up the formula, but somehow I got the negative of the first one)
(8.9875e9)(5.7e(-6))(-5.7e(-6))/.16 + 0.827790168702400828475759973771078927721678492943077184393)
= -0.997 J

Trying ΔPE again:
kQ1Q2/r12 - (kQ1Q3/r13 + kQ2Q3/r23 + kQ1Q2/r12)
=-kQ1Q3/r13 - kQ2Q3/r23
=-(8.9875e9)(5.7e(-6))(1.8e(-6))/0.06 - (8.9875e9)(-5.7e(-6))(1.8e(-6))/sqrt(0.06^2 + 0.16^2)
= -0.997 J

What am I supposed to do? Neither of these work.

2. Jan 30, 2012

### Redbelly98

Staff Emeritus
Yes. And you realize ΔPE is PEfinal - PEinitial, right?

No, the work needed is ΔPE, which is PEfinal - PEinitial.
You have calculated PEinitial here. You want PEfinal-PEinitial.
Your second answer (the negative one) looks correct to me. However if I use a rounded-off version of k, 8.99e9 N*m2/C2, then the answer rounds off to -0.998 J.

p.s. please don't report results, even intermediate calculations, to 50 significant figures. It just clutters up your work unnecessarily for people who are trying to follow your calculation.

3. Jan 30, 2012

### Spinnor

Are you submitting this to a computer? I'm not sure how many significant digits your answer should have, but have you tried -1.0?