What is the Work Needed to Move a Charge?

[ttiger2k7]

[SOLVED] Work Needed to Move a Charge

Homework Statement

Three identical point charges Q1, Q2, and Q3 all having charge 3 microCoulombs are located at the vertices of an equilateral triangle with sides s=3 m. How much work would I have to do to move Q1 to a point on the line connecting Q2 and !3 (which are fixed) if this point is a distance .4 s from Q2 and a distance .6 s from Q3?

http://img129.imageshack.us/img129/1091/trianglejp1.th.gif
(Picture obviously not drawn to scale)

Homework Equations

W = Fd = qEd
E = kQ/d

The Attempt at a Solution

I am using q = 3*10^-6 Coulombs in all of my attempts. The thing that is throwing me off is what to use as the distance and electric field.

I did two different approaches:

Attempt 1
First I tried to find the distance between Q1 and its destination point. I just made a right triangle:
http://img98.imageshack.us/img98/1240/88380904mi7.gif
And the hypotenuse was sqrt(2.34).

Then,

E = kq/d
E = (9*10^9)(3*10^-6)/sqrt(2.34)
E = 17650.45216 N/C

So I plugged in E to the formula for Work

W = qEd
W = (3*10^-6)(17650.45216)(sqrt(2.34))
W = .081 J

Attempt 2
I tried to calculate the electric field at the point Q1 between Q2 and Q3. For distance i used 1.2 (.4*3m) and 1.8 (.6*3m).

First,

E_tot = E1 + E2
E=kq/d^2 (electric field at a point between two charges)
E1=(9*10^9)(3*10^-6)/1.2^2
E1=18750

E2=(9*10^9)(3*10^-6)/1.8^2
E2=-8333.34 (rounded up, because it was .3 repeating, negative because the direction of the field that acting on Q1 is to the left)

E_tot = E1+E2
E_tot = 18750 - 8333.34 = 10416.67

Then,

W = qEd
d will remain the same as in attempt 1.
W = (3*10^-6)(10416.67)(sqrt(2.34))
W = .0478 J

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There you have it. Neither of them were the right answer. Please help. I know I am close but like I said, I am confused as what to use for E and d. Thank you.

 

[Shooting Star]

Work done = difference in PE between final position and initial position.

You know the Potential at the initial and final posn due to Q2 and Q3.

 

[ttiger2k7]

UE at initial point:

UE = k(3*10^-6)^2/3 = .027
Since that is the same for both charges, the answer is .054.

UE at bottom:

UE = k(3*10^-6)^2/1.8 = .045
UE = k((3*10^-6)^2/1.2 = .0675
Tot = .1125

UE_ini - UE_final = .1125-.054 = .0585 J Which was the correct answer.

Thanks a bunch! I really appreciate the quick reply. That helped a lot. :)