Minimum work required to move a point charge

In summary: Or another way of saying it, the work done by the force equals the negative of the initial PE.In summary, the problem is to find the minimum amount of work required by an external force to move the charge q1 to infinity in a system with three point-like charges placed at the corners of a rectangle. The initial total EPE of the system is calculated using the equation EPE system = (kq2q3/a) + (kq2q1/b) + (kq1q3/√a^2 + b^2), and the final EPE is calculated by negating q1 as if it were off to infinity, giving the equation EPE system final = (kq2q3
  • #1
kno
3
0
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.
 
Physics news on Phys.org
  • #2
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 ?
 
  • #3
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?
 
  • #4
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?
 
  • #5
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?
 
  • #6
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.
 

Related to Minimum work required to move a point charge

1. What is the minimum work required to move a point charge?

The minimum work required to move a point charge is equal to the product of the charge and the potential difference between the initial and final positions of the charge. This can also be expressed as the integral of the electric field along the path of the charge.

2. How is the minimum work required to move a point charge related to the electric potential?

The minimum work required to move a point charge is directly related to the electric potential. The electric potential is a measure of the potential energy per unit charge at a given point in space. Therefore, the work required to move a charge is equal to the change in electric potential multiplied by the charge.

3. Can the minimum work required to move a point charge be negative?

Yes, the minimum work required to move a point charge can be negative. This occurs when the electric field and the displacement of the charge are in opposite directions, resulting in a decrease in potential energy and a negative work value.

4. How does the distance between the initial and final positions of the charge affect the minimum work required to move it?

The minimum work required to move a point charge is directly proportional to the distance between the initial and final positions of the charge. In other words, the farther the charge has to be moved, the more work is required to do so.

5. What factors affect the minimum work required to move a point charge?

The minimum work required to move a point charge is affected by several factors, including the magnitude of the charge, the potential difference between the initial and final positions, and the distance between these positions. Additionally, the presence of other charges and the nature of the medium in which the charge is being moved can also affect the minimum work required.

Similar threads

  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
2K
Replies
1
Views
5K
  • Introductory Physics Homework Help
Replies
21
Views
848
  • Introductory Physics Homework Help
Replies
4
Views
6K
  • Introductory Physics Homework Help
2
Replies
56
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
3K
  • Introductory Physics Homework Help
Replies
5
Views
938
  • Introductory Physics Homework Help
Replies
7
Views
4K
Back
Top