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Homework Help: Finding the electric field in Griffiths

  1. Aug 3, 2005 #1
    This is referring to problem 2.25 in Griffiths. Find the potential at a distance 'z' above the midpoint between two equal charges, q, a distance 'd' apart. Compute the electric field in each case.
    In the first part the charges are equal. The electric potential is:
    [tex]V = \frac{1}{4\pi \epsilon_0} \frac{2q}{{\sqrt{z^2 + (d^2/4)}}}[/tex]
    In this case, electric field given by [itex]\vec E = -\nabla V[/itex] is:
    [tex]\vec E = \frac{1}{4\pi \epsilon_0} \frac{2qz}{[{z^2 + (d^2/4)}]^{(3/2)}}} \hat z[/tex]
    Matches with the result known in problem 2.2(a).

    In the 2nd case we consider oppositely charges +q & -q. In this case the electric potential V = 0 which naively(:wink:) suggests [itex]\vec E = -\nabla V = 0[/itex] which is in contradiction to a previous known result in problem 2.2(b) which suggests:
    [tex]\vec E = \frac{1}{4\pi \epsilon_0} \frac{qd}{[{z^2 + (d^2/4)}]^{(3/2)}} \hat z[/tex]

    How is this discrepancy accounted for?
  2. jcsd
  3. Aug 3, 2005 #2
    I don't have the book with me right now. It would help if you posted the problem from 2.2 for comparison.
  4. Aug 3, 2005 #3


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    Why does V=0 even naively suggest that [itex]\nabla V[/itex], it's gradient is zero?
  5. Aug 5, 2005 #4
    Alright, here are both the questions:

    Prob2.25: Find the electric potential at a distance 'z' above the midpoint between two charges +q and +q with 'd' being the distance between them. Also find the potential if the charges are -q and +q. Compute [itex]\vec E = -\nabla V[/itex] and compare your answer with prob.2.2. Explain carefully any discrepancy.

    Prob2.2: Find the electric field at a distance 'z' above the midpoint between two charges +q and +q with 'd' being the distance between them. Repeat the problem with the charges as -q and +q.

    This should be helpful to solve my problem.
  6. Aug 5, 2005 #5
    For 2.25, calculate the potential using generalized coordinates first. That is,

    [tex]V=kq \left( \frac{1}{\sqrt{(x+\frac{d}{2})^2+y^2+z^2}} + \frac{1}{\sqrt{(x-\frac{d}{2})^2+y^2+z^2}} \right) [/tex]

    Switch to minus for the second part. Take the above expression's derivative and then plug in (0,0,z). The reason for this is that you lose the derivatives with respect to x and y if you plug in (0,0,z) before taking the derivative.

    As a note, it may also help if you think about electric field lines. For two opposite charges next to each other, the field lines go from the positive to the negative, like in this picture:

    http://electron9.phys.utk.edu/phys136d/modules/m4/images/dipole.gif [Broken]

    Now, if you were to orient the two charges on the x-axis and your point were somewhere away from the midpoint on the z-axis, as you can see, the resultant field is not in the z direction. In fact, it's in the x-direction. Knowing this intuitively, you should expect the x-component to not go away. The only way to find it is to first use generalized coordinates.

    I hope that helps some.
    Last edited by a moderator: May 2, 2017
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