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Higher potential in electric field

  1. May 14, 2012 #1
    1. The problem statement, all variables and given/known data
    The electric field on the x axis due to a point charge fixed at the origin
    is given by E = (b/x^2)[itex]\hat{x}[/itex]
    , where b = 6.00 kV·m and x ≠ 0 . (a) Find the magnitude
    and sign of the point charge. (b) Find the potential difference between the points
    on the x-axis at x = 1.00 m and x = 2.00 m. Which of these points is at the higher
    potential.


    2. Relevant equations
    dV = -E*dl
    U=q*V

    3. The attempt at a solution
    V = E/x
    V1 = 6000V/m / 1.0m = 6000V
    V2 = 6000V/m / 2.0m = 3000V

    This is where I have my problem, what does higher potential mean? I thought it means that if I take a random charge and put it in at x=1.0 or x=2.0, the one with most potential energy is the higher potential.
    U=q*V, V1>V2, so shouldnt that be the higher potential?
    www. uccs.edu/~rtirado/Ch23%20ISM.pdf This says that V2 has the higher potential (Problem 36, about 1/3 down)

    What if they were negative?
    www. personal.utulsa.edu/~alexei-grigoriev/index_files/Homework3_solutions.pdf
    Here -0.5kV has a higher potential than -8kV, why? (Also problem 36)
     
    Last edited: May 14, 2012
  2. jcsd
  3. May 14, 2012 #2

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    Hello Pentafem,

    Welcome to Physics Forums!

    Part (a) requires that you find the magnitude and sign of the charge, q. It seems like you skipped that part.

    For part (b), your calculations,

    V1 = 6000V/m / 1.0m = 6000V
    V2 = 6000V/m / 2.0m = 3000V​

    are not correct for a point charge. The way in which you are applying that calculation is only correct for a uniform electric field. But the electric field in this problem is not uniform. Try using a different formula for the potential of a point charge, or you can derive the formula yourself by integrating dV = -E*dl.

    Essentially, yes you are correct. :approve: (I wouldn't use the term 'random' charge though. How about 'arbitrary', positive test charge.)
    Your links did not work for me, at least not directly. But google did find a some some sort of similar document where the electric field is given as [itex] \vec E = bx^3 \hat \imath [/itex]. That's a completely different problem. The result of that integration is very different than this problem, where [itex] \vec E = b/x^2 \ \hat \imath[/itex].
     
  4. May 14, 2012 #3


    Why arent they correct?
    Integral of -b/x^2 from x1 to x2 is b/x2-b/x1, if we set x1 as reference point at infinity with U = 0, then we get V(x) = b/x. Which makes my calculations correct. The answer also confirm that (Im not allowed to make links directly, sorry, you have to remove the space between the www. and the rest.

    In the answer they also say:
    "Because V1=V2 + 3.00 kV , the point at x = 2.00 m is at the higher potential."

    Which makes me go crazy. You see it right here, V1 is obviously larger than V2, it says right in the answer sheet, why, in the same sentence does it say that x=2.00 has higher potential? Is the answer wrong or am I missing something crucial?


    As far as the other exercise goes, you are indeed correct, that is a different integral. But I get
    V(x) = -1/4 bx^4, // b=2kV/m^4
    so if you plug inn 1.0m, you get -1/2kV, and for x=2.0m you get -8kV

    The answer sheet says:
    "Because V2 =V1 − 7.50kV , the point at x = 1.00 m is at the higher potential."
    This supports my calculations, but does not answer my prime question; why, once again, is the lower voltage potential (0.5kV < 7.5kV) is the one with higher potential?
     
  5. May 14, 2012 #4

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    My mistake. You are correct. :approve:

    For some reason when I glanced at your equations I thought I saw a multiplication sign rather than a dividing sign. You were correct. Sorry about that.
    You are correct, that for this problem, where [itex] \vec E = b/r^2 \ \hat x [/itex], and b is positive, the point at x = 1.0 m has the higher potential.

    I'm just saying, don't confuse it with the solution in your link. That problem does not involve a point charge, and instead [itex] \vec E = br^3 \ \hat i [/itex], which gives a different result.

    Maybe it's a mistake in the book? Maybe the problem statement forgot about a negative sign in b? I'm not sure.
    Don't forget about the minus sign. Just like -1 is greater than -2; -0.5 kV > -8 kV.

    [Edit: when determining potential (and the direction of the electric field), the test charge is assumed to be positive.]
     
    Last edited: May 14, 2012
  6. May 15, 2012 #5
    Ok, thanks, I think I got this now.

    Just to check that I got the concept. If we take x1=-1.0m and x2=2.0m, we would get
    First exercise: ( V(x) = 6000/x )
    v1=-6000V
    v2=-3000V, where V2 has the higher potential

    Second exercise: ( -1/4 2000*x^4 )
    V1=0.5kV
    V2=8kV, where V2 has the higher potential, right?
     
  7. May 15, 2012 #6

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    Be careful here. Let's go back to that original integral.

    [tex] \Delta V = -\int \vec E \cdot \vec{dl} [/tex]

    Above, [itex] \vec E [/itex] and [itex] \vec {dl} [/itex] are both vectors with magnitude and direction. But the dot product, [itex] \vec E \cdot \vec{dl}[/itex], is a scalar!

    What does this mean? Well let's apply it to our particular problem here.

    [tex] V = -\int_r ^{\infty} \frac{b}{r^2} \hat r \cdot \vec{dr} [/tex]

    But note that for our particular problem with a point charge at the origin, [itex] \vec{dr} [/itex] always points in the same direction as [itex] \hat r [/itex]. Rewriting the equation,

    [tex] V = -\int_r ^{\infty} \frac{b}{r^2} \hat r \cdot dr \ \hat r = -\int_r ^{\infty} \frac{b}{r^2} dr \ \hat r \cdot \hat r[/tex]

    Noting that [itex] \hat r \cdot \hat r [/itex] = 1,

    [tex] V = -\int_r ^{\infty} \frac{b}{r^2} dr [/tex]

    And we're left only with scalars. Now [itex] r [/itex] is the distance from the charge at the origin to the positive test charge: not the displacement. In other words, [itex] r [/itex] is now positive scalar.

    Evaluating the integral,

    [tex] V = -\int_r ^{\infty} \frac{b}{r^2} dr = \frac{b}{r}[/tex]

    So when you plug in a value for x to find the potential, you need to plug in the scalar distance, not the vector displacement.

    (In other words, the potential at x = -1, is V(-1) = 6000/1. A positive result [since b is positive (in this case) and r is a positive scalar].)

    The fact that the r should be treated as a positive distance, such that you can ignore the sign of x, is a result of this specific problem. If [itex] E [/itex] was defined some other way, you can't automatically assume that x is equal to some scalar distance. It's just that you can for this specific problem.

    (And by the way, I just want to reiterate, when the coursework stated "Because V1=V2 + 3.00 kV , the point at x = 2.00 m is at the higher potential," I'm pretty sure that is a mistake. It's obvious that V1 is 3.00 kV larger than V2, so the point at x = 1.00 m has the higher potential.)

    For the other exercise we have

    [tex] \vec E = b x^3 \hat \imath[/tex]

    Notice that x here is not defined as a distance such as r was in the last problem. So later on, when when we plug in numbers for negative values of x, we need to keep the negative. But we'll come to that in a second.

    So we have the integral,

    [tex] \Delta V = -\int bx^3 \hat \imath \cdot dx \hat \imath = -\int bx^3 dx [/tex]

    But we haven't picked the limits of integration yet. It doesn't make sense to integrate from x to ∞, similar to what we did for the last problem. The electric field blows up as x approaches ∞, so those are not good limits in this case.

    What you have done is to find the potential at point x, relative to the origin. Which is a fine thing to do, I suppose. (Or we could calculate the path from x1 to x2 as your coursework did. And your coursework's choice of limits is probably a better way, and more direct. But either way works.)

    [tex] \Delta V = = -\int_0^x bx^3 dx = -\frac{1}{4}bx^4[/tex]

    Now remember, in this particular problem we need to keep the sign for x when evaluating the potential. That's because x ≠ |r| like it was in the last problem. But it just so happens in this problem that x4 is always positive. That means the potential, relative to the origin, is always negative.

    So getting back your evaluation,
    You forgot the negative sign.

    V1 = -0.5 kV
    V2 = -8 kV

    And V1 has the higher potential.

    Once again it looks like a mistake in the coursework. Take a look at the line right above the final answer in your coursework pdf. It states,

    "Simplify to obtain: [itex] V_2 - V_1 = -7.5 \ \mathrm{kV} [/itex]"​
    which is correct. But then on the very next line it states

    "Because [itex] V_2 = V_1 + 7.5 \ \mathrm{kV} [/itex] ..."​
    which is simply wrong. Any basic algebra from the previous equation shows that V1 has the higher potential.

    I appreciate your frustration with this. It looks to me as though your coursework pdf has a few mistakes in it.
     
  8. May 15, 2012 #7
    Thank you for explaining this so well! I do understand this much better now, although my initial confusion was because of errors in the coursework.
     
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