1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding the electric potential of point charge

  1. May 21, 2008 #1
    [SOLVED] Finding the electric potential of point charge

    1. The problem statement, all variables and given/known data
    I want to find the electric potential at a distance r from a point charge. Here's what I'm doing:

    We have

    [tex]
    V = - \int {E \cdot {\rm{d}}{\bf{l}}}
    [/tex]

    I find that

    [tex]
    V = - \int_\infty ^r {k_e \frac{q}{{r'^2 }}{\bf{r}} \cdot {\rm{d}}{\bf{l}}} = - \int_\infty ^r {k_e \frac{q}{{r'^2 }}\cos \left( \phi \right){\rm{dl}}} = \int_\infty ^r {k_e \frac{q}{{r'^2 }}{\rm{dl}}} = - k_e \frac{q}{{r }}
    [/tex]
    I used that I integrate from infinity to r, so dl is pointing opposite to E. I get a negative potential, which is of course wrong for a point charge. Where's my error?
     
  2. jcsd
  3. May 21, 2008 #2

    Doc Al

    User Avatar

    Staff: Mentor

    E and dl (which is dr) both point outward, so [itex]\cos\phi = 1[/itex].
     
  4. May 21, 2008 #3
    Why does dl point outward when I am integrating from infinity to r?
     
  5. May 21, 2008 #4

    Doc Al

    User Avatar

    Staff: Mentor

    Because r increases outward, regardless of the direction of integration. Note that:

    [tex]\int_a^b f(x) dx = - \int_b^a f(x) dx[/tex]

    (I'm sure you're well aware of that!)
     
  6. May 21, 2008 #5
    I don't get it, and it's elementary calculus.

    So dl always points in the direction of increasing r?
     
  7. May 21, 2008 #6
    Actually, if I use

    [tex]
    \int_a^b f(x) dx = - \int_b^a f(x) dx
    [/tex]

    it works out perfectly, because I integrate from r to infinity, so the angle between E and dl is zero. But again, I cannot see why dl is also in the direction of E when we go from infinity to r.
     
  8. May 21, 2008 #7
    Electric potential is the work it takes per unit charge to bring that charge from zero potential (usually infinity) to the point in question.

    In a line integral, dl is never negative. The limits (the beginning and end of the path) define whether the work done is negative.

    [tex]V=\frac{W_{_{by applied}}}{q_{_{test}}}[/tex]

    [tex]V=\frac{\int_a^b \vec{F}_{_{app}}\cdot\vec{dl}}{q_{_{test}}}[/tex]

    Since the applied force is opposite to coulumb repulsion, we can change that to:

    [tex]V=\frac{\int_a^b -\vec{F}_{_{e}}\cdot\vec{dl}}{q_{_{test}}}[/tex]

    or
    [tex]V=-q_{_{test}}\frac{\int_a^b \vec{E}\cdot\vec{dl}}{q_{_{test}}}[/tex]
    [tex]V=-\int_a^b \vec{E}\cdot\vec{dl}[/tex]

    In this case, we start at infinity and end at r, so the limits become

    [tex]V=-\int_\infty^r \vec{E}\cdot\vec{dl}[/tex]

    And then, solving it:

    [tex]V=-\int_\infty^r \left(E\hat{r}\right)\cdot \left(dr\hat{r}\right)[/tex]

    [tex]V=-\int_\infty^r E dr\left(\hat{r}\cdot\hat{r}\right)[/tex]

    [tex]V=-\int_\infty^r E dr[/tex]

    [tex]V=-\int_\infty^r k\frac{q}{r^2} dr[/tex]

    [tex]V=-\left[-k\frac{q}{r}\right]_\infty^r[/tex]

    [tex]V=\left[k\frac{q}{r}\right]-\left[k\frac{q}{\infty}\right][/tex]

    Or simply

    [tex]V=k\frac{q}{r}[/tex]

    -Ataman
     
  9. May 22, 2008 #8
    So I should never care which way E and dl are pointing to eachother? The only thing that can make the minus-sign go away is if I reverse the limits?
     
  10. May 22, 2008 #9

    Doc Al

    User Avatar

    Staff: Mentor

    Let me see if I can clarify things a bit.

    First, let's understand the meaning of the minus sign in the line integral definition of potential:

    [tex]V = - \int {E \cdot {\rm{d}}{\bf{l}}} [/tex]

    If I move along the curve of the line integral in such a way that I oppose the electric field (and thus [itex]\vec{E}\cdot\vec{d\ell}[/itex] is negative), that represents a positive increase in potential: the external minus sign cancels the minus sign of the dot product to give a positive potential.

    The tricky part is translating the line integral (and dot product) into an ordinary definite integral. In so doing you parameterize the curve in terms of the variable r. Note that dr is a positive change in the direction of increasing r. Now imagine integrating from infinity to some point r = a. In this case E points outward, in the direction of increasing r. As I move inward along r, I'm actually moving in the negative r direction thus my differential element of distance is -dr. So if I integrate in that direction I get a positive value for potential:

    [tex]V = \int_{\infty}^{a}E(r) (-dr) = - \int_{\infty}^{a}E(r) dr[/tex]

    [tex]V = - \int_{\infty}^{a}\frac{kq}{r^2} dr = \frac{kq}{a}[/tex]

    Of course, if I reverse direction I get a negative potential difference.

    (I hope this helps a bit rather than adds to the confusion.)
     
  11. May 22, 2008 #10
    It actually helped.

    In my book it says

    [tex]
    V_a - V_b = \int_{a}^{b}Er \cdot dr
    [/tex]

    Now I want to test it all. Using your reasoning, I want to find the potential between two plates a and b separated by a distance d. Lets just say that E goes from b -> a so the zero-potential point is at a.

    [tex]
    V_b - V_a = \int_{d}^{0}E(r) dr = -Ed
    [/tex]

    Now where is my error?
     
  12. May 22, 2008 #11

    Doc Al

    User Avatar

    Staff: Mentor

    Since E points from b to a, opposite to the direction of your integration, the field should actually be -E. (Assuming the field is a constant of magnitude E, the component in the direction of r is -E.)

    So:
    [tex]
    V_a - V_b = \int_{a}^{b}E(r) \cdot dr = \int_{a}^{b}(-E) dr = -Ed
    [/tex]

    Thus, taking V_a as 0:

    [tex]
    V_a - V_b = - V_b = -Ed
    [/tex]

    [tex]
    V_b = Ed
    [/tex]
     
  13. May 23, 2008 #12
    Ok, now I am only confused about one thing.

    Why do you not include a minus in

    [tex]
    V = \int_{\infty}^{a}E(r) (-dr) = - \int_{\infty}^{a}E(r) dr
    [/tex]?

    I mean, the minus in [tex]
    V = - \int {E \cdot {\rm{d}}{\bf{l}}}
    [/tex] is not included in your above expression?

    EDIT: I actually asked my professor about this question today, but he was kinda busy, but from what I could tell, I am not to worry about the direction of dl when using the formula

    [tex]
    V = - \int {E \cdot {\rm{d}}{\bf{l}}}
    [/tex]
    but only worry about the direction of E with respect to the coordinate-system. Am I correct?
     
    Last edited: May 23, 2008
  14. May 23, 2008 #13

    Doc Al

    User Avatar

    Staff: Mentor

    I don't include the outside minus sign because the direction of integration is opposite to the direction of the field.

    The only thing that I worry about is whether my end point is at a higher or lower potential than the starting point. You can tell that by the direction of the field.
     
    Last edited: May 23, 2008
  15. May 23, 2008 #14
    Ok, now I am thoroughly confused.

    You first remove the minus sign because we integrate in the opposite direction of the electric field, and then you dot then and get a minus. Isn't that doing the same thing twice? In the example with the parallel plates you just dotted and got the minus.

    I really appreciate you guys helping me, and being so patient. But sometimes I wish consistency in physics-litterature could be explained a little better.
     
  16. May 23, 2008 #15

    Doc Al

    User Avatar

    Staff: Mentor

    let's get serious ;-)

    To be honest with you, I never attempt to do the line integral to ordinary integral step "formally", I just know the appropriate sign that the answer must have. But I'm going to give it one more shot, as I think I might be confusing you (and myself) and would like to straighten it out.

    I think I know what he means and that's a good way to approach it. Let's take a simple example and do it every which way to be sure we have it straight. Let's assume [itex]\vec{E}[/itex] has constant magnitude E (a positive number). From here on, we will let the coordinate system do the work.

    Options:
    (A) Electric field' points right
    (B) Electric field' points left
    (1) Coordinate system (variable r) goes left to right
    (2) Coordinate system (variable r) goes right to left

    Let's say point a is to the left of point b.

    Case A-1:
    Let's find Vb - Va.

    [tex]V_b - V_a = -\int_{a}^{b}\vec{E}\cdot d\ell = -\int_{a}^{b}E dr = - E (b-a) = -Ed[/tex]

    Since the field is in the direction of the coordinate system (r), the dot product is positive. This answer makes sense.

    If we integrated from b to a, we'd get -E(a-b) = +Ed. This answer makes sense, since that's calculating Va - Vb.

    Case B-1:

    [tex]V_b - V_a = -\int_{a}^{b}\vec{E}\cdot d\ell = -\int_{a}^{b}-E dr = E (b-a) = Ed[/tex]

    This makes sense, since now the field opposes the direction of r.

    Case A-2:

    [tex]V_b - V_a = -\int_{a}^{b}\vec{E}\cdot d\ell = -\int_{a}^{b}-E dr = E (b-a) = -Ed[/tex]

    The only difference here is that we changed the coordinates system to point the other way. The potential difference doesn't care about our coordinate choices, so the answer is the same as A-1. Note that the dot product is negative, since the field opposes the direction of r. But the integral of the distance is also negative, since now a > b.

    Case B-2:

    [tex]V_b - V_a = -\int_{a}^{b}\vec{E}\cdot d\ell = -\int_{a}^{b}E dr = - E (b-a) = Ed[/tex]

    Again, this should make sense: The field and coordinate system are aligned, but the distance integral is negative.

    Hopefully this helps a bit (and I didn't make typos). Using this systematic approach (defining things with respect to the coordinate system) should work like a machine for all cases.

    If you still have questions, ask away. Let's get this one done.
     
  17. May 23, 2008 #16
    I like the approach in post #15 - very systematic, which is what I really need it this case. I agree and understand 100% what you have written. I will now apply it to the point-charge example, where we set the direction that points away from the positive point charge to be positive:

    [tex]
    V_r - V_\infty = - \int_{\infty}^{r}\vec{E}\cdot d\vec{r} = \int_{\infty}^{r}Edr
    [/tex]

    Since E points in the positive direction and dr points in the negative, we get the above signs, and now I have my pickle. What part of my reasoning is wrong?
     
    Last edited: May 23, 2008
  18. May 23, 2008 #17

    Doc Al

    User Avatar

    Staff: Mentor

    No, dr does not point in the negative direction. With this systematic approach, all that matters is whether E points in the same direction as the coordinate system (variable r). And it does, so:

    [tex] V_r - V_\infty = - \int_{\infty}^{r}\vec{E}\cdot d\vec{r} = - \int_{\infty}^{r}Edr = - \int_{\infty}^{r}\frac{kq}{r^2}dr = \frac{kq}{r} [/tex]
     
  19. May 24, 2008 #18
    Ok, now I think I have it, but there is one final thing.

    When finding the potential difference, should I do it with the limits:

    [tex]
    V_{highest} - V_{lowest}
    [/tex]?
     
    Last edited: May 24, 2008
  20. May 24, 2008 #19

    Doc Al

    User Avatar

    Staff: Mentor

    It doesn't matter. Vb-Va means the potential of point b with respect to point a. If Vb > Va, you'll get a positive number; if Vb < Va, it will be negative. Obviously, Vb-Va = - (Va-Vb).

    Often, when finding the potential difference all you care about is the magnitude.
     
  21. May 24, 2008 #20

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Pardon me for jumping in.

    In a path integral, the vector dl points in the direction one is integrating. I.e. from the lower integration limit towards the upper limit.

    dr will always point away from the origin, regardless of what the integration limits are.

    So if the lower integration limit is [tex]\infty[/tex], and the integral's path is toward the origin, then dl = -dr.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?