Minimum work required to move a point charge

AI Thread Summary
The discussion revolves around calculating the minimum work required to move a point charge, q1, to infinity in a system with three charges arranged in a rectangle. The initial and final electric potential energy (EPE) of the system is calculated, leading to the equation W = -(EPE final - EPE initial) for work done. The key point is that the minimum work occurs when the final kinetic energy (KE) is zero, meaning there is no residual energy after moving the charge. The confusion arises from the interpretation of "minimum work," which simply indicates that the work is minimized when KE is not present. Ultimately, the conclusion is that the final potential energy should equal the initial potential energy when calculating the work done.
kno
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Homework Statement
Three point-like charges are placed at the corners of a rectangle as shown in the figure, a = 22.0 cm and b = 54.0 cm. Find the minimum amount of work required by an external force to move the charge q1 to infinity. Let q1= +3.00 µC, q2= −3.30 µC, q3= −3.60 µC.
Relevant Equations
W = -(EPE final - EPE initial)
W = -q * (V final - V initial)
This is the figure for the problem:
19-74.jpg
1.) Solved for initial total EPE of the system

EPE system = (kq2q3/a) + (kq2q1/b) + (kq1q3/√a^2 + b^2)

2.) Solved for final EPE of the system negating q1 as if it were off to infinity

EPE system final = (kq2q3/a)

3.) Plugged values into equation

W = -(EPE final - EPE intial)

I wasn't sure how to figure out the minimum work required, I think I am missing a step or two.
 
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kno said:
Homework Statement: Three point-like charges are placed at the corners of a rectangle as shown in the figure, a = 22.0 cm and b = 54.0 cm. Find the minimum amount of work required by an external force to move the charge q1 to infinity. Let q1= +3.00 µC, q2= −3.30 µC, q3= −3.60 µC.
Homework Equations: W = -(EPE final - EPE initial)
W = -q * (V final - V initial)

This is the figure for the problem:
View attachment 2498341.) Solved for initial total EPE of the system

EPE system = (kq2q3/a) + (kq2q1/b) + (kq1q3/√a^2 + b^2)

2.) Solved for final EPE of the system negating q1 as if it were off to infinity

EPE system final = (kq2q3/a)

3.) Plugged values into equation

W = -(EPE final - EPE intial)

I wasn't sure how to figure out the minimum work required, I think I am missing a step or two.
The work of an external force can change both the kinetic energy and potential energy of a particle. When is the work minimum when moving q1 to infinity ?
 
I know net work = change in KE. So by that equation work would be closest to zero when there is the smallest change in KE. Since the point charge starts at rest (I'm assuming) then the KE is zero. However I don't see how the kinetic energy can be zero in its final state to produce the smallest amount of work required. I was also unsure how to calculate KE in the first place since I am not given mass or velocity once the point is moved.

In short, is the answer to your question that work would be smallest when KE is largest? AKA KE final will equal EPE initial?
 
kno said:
net work = change in KE
That is the net work done on a body. That is different from the net work done by a force.
Your original work on that is correct. You just seem to be confused about the significance of the "minimum" condition.
There is no requirement given for the charged particle to have any final KE, so what will be its final KE if the work done by the force is minimised?
 
So when the work done is minimized the KE = 0? I thought I solved the problem that way initially by negating the KE in the W = -(EPE final - EPE initial) equation. The only other equation I know for work is W=FdcosΘ however I am not given the direction that the particle moves or its displacement. The last idea I had was the conservation of energy equation where KE initial + PE initial = KE final + PE final. If I use this and assume KE is zero in both cases then I'm left with PE initial = PE final. So, should my final PE be equal to my initial and use this value in the work equation?
 
kno said:
So when the work done is minimized the KE = 0? I thought I solved the problem that way initially by negating the KE in the W = -(EPE final - EPE initial) equation.
Indeed you did, and as I wrote in post #4, that was correct. But you seemed puzzled by the requirement for the work done to be minimised. All it means is, no residual KE.
 
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