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ttiger2k7
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[SOLVED] Work Needed to Move a Charge
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)
W = Fd = qEd
E = kQ/d
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
---
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.
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
---
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.
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