Electric Potential near a point charge.

In summary: F dot dr. You are calculating the work done by the electric field in the opposite direction of the direction the charge is moving. This will create a negative value in the integral.
  • #1
AlexChandler
283
0
Hello,
I am having trouble in proving the equation of the electric potential near a point charge Q. There are two different ways I can think to go about deriving the equation V=q/(4*Pi*E0*r). The first way is to calculate the change in potential energy of a test charge q0 in coming from infinite distance to a point P near the point charge Q, considering U=0 at infinity as a reference point. Then setting the change in potential energy equal to the negative of the work done by the electric field. The second way seems to me to be exactly the same mathematically. I am considering the change in potential energy in moving a test charge charge q0 from the point P to infinity. Again the change in potential energy should be equal to the opposite of the work done by the electric field along the path. Setting U=0 at infinity as a reference point, I find that the potential energy at P near the point charge Q is equal to the work done by the electric field along the path to infinity. Using the second method, I arrive at the right equation. But using the first method, I find a negative sign in the equation and I cannot get rid of it. I hope that somebody may be able to tell me what I have done wrong, or if there is something that I am not understanding correctly. I have attached a picture of my work on paper. I hope I have done a good job explaining the problem and that the picture is of decent enough quality.
Thanks so much, hope to hear a response soon.
-Alex
 

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  • #2
The problem begins on the first line. The two equations cannot be the same. One time the scalar product is negative, and one time it is positive. The minus sign will flip the integral boundaries, but not the scalar product. I think the scalar product will do exactly what you are trying to force with the minus sign. We are running against the field so the result is automatically negative. No need for an extra minus.
 
  • #3
Yes you are right! thank you. I am still confused as to the cause of this however.
Considering a positive point charge Q, the work done by the E field produced by Q along a radial line moving a positive point charge q0 from point a to point b, where a is closer to Q than b, should be the integral from a to b of F dot dr. Evaluating that integral I get a correct answer for the work, coming out positive. However if I calculate the work done by the E field produce by Q along the same line, but now from b to a, I would think the integral should be this: the integral from b to a of F dot dr... where now F is in the opposite direction as dr. But now, as you have said, switching the bounds creates a negative sign, and also the opposite direction of F and dr creates another negative sign... so the integral again comes out to be positive, but it is clear that is should come out negative! If you could point out the fault in my second integral I would be very very grateful, as this has been causing quite a lot of stress for me for the past few days, and I have an exam on monday so I would love to get this cleared up before then.
I have attached a picture as well if it may be of some help to see what I am trying to describe.
Thanks!
Alex
 

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1. What is electric potential near a point charge?

The electric potential near a point charge is the amount of electric potential energy per unit charge at a specific point in space. It is a measure of the electric potential energy that a charged particle would have if placed at that point.

2. How is electric potential near a point charge calculated?

The electric potential near a point charge can be calculated using the equation V = kQ/r, where V is the electric potential, k is the Coulomb's constant, Q is the charge of the point charge, and r is the distance from the point charge to the point of interest.

3. What factors affect the electric potential near a point charge?

The electric potential near a point charge is affected by the magnitude of the charge, the distance from the point charge, and the medium in which the point charge and the point of interest are located. In a vacuum, the electric potential decreases as the distance from the point charge increases. In a medium with a dielectric constant, the electric potential may also be affected by the polarization of the medium.

4. How does the electric potential near a point charge relate to the electric field?

The electric potential near a point charge is directly related to the electric field at that point. The relationship is given by the equation E = -∇V, where E is the electric field, V is the electric potential, and ∇ is the gradient operator. This means that the electric field points in the direction of decreasing electric potential.

5. Can the electric potential near a point charge be negative?

Yes, the electric potential near a point charge can be negative. This indicates that the electric potential energy of a charged particle at that point would be lower than at a point infinitely far away from the point charge. However, the electric potential at a point is not affected by its sign, as it is a scalar quantity and only the difference in electric potential between two points is physically meaningful.

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