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GR0877 Problem 37

PROBLEM STATEMENT: This problem is still being typed.

SOLUTION: (E) Clearly, the electric field is in the -x direction, therefore, only (C) and (E) survive. The magnitude of the electric field at P is E = 2E_q \cos \theta = 2E_{-q} \cos \theta where E_q = E_{-q} are the magnitudes of the electric field at point P due to the charges q and -q, respectively; \theta is the angle between x-axis and the line passing through the point P and charge +q. \displaystyle \cos \theta = \frac{l/2}{\sqrt{r^2 + l^2/4}} \displaystyle \implies \displaystyle E = \frac{1}{4 \pi \epsilon_0} \frac{2q}{r^2 + l^2/4} \frac{l/2}{\sqrt{r^2 + l^2/4}}. For r \gg l we have \displaystyle E \approx \frac{1}{4 \pi \epsilon_0} \frac{ql}{r^3}.

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  1. betelguese05
    07.10.2012 at 09:54

    I would like to suggest an alternate solution: This is clearly a dipole, and if you just remember that the dipole field falls off (long range) as 1/r^3, then the only solution satisfying this is E.

  2. Eric
    07.10.2012 at 21:13

    I believe you meant to make the first q in the statement E_q = E_{-q} a subscript.

  3. Arturodonjuan
    28.08.2016 at 10:16

    Quick Solution:

    The direction is very easily obtained with logic/geometry, and that knocks out answer choices (A), (B), and (D). Then, remember that this is an electrical dipole, so far away the electric field will go off as 1/r^3. This leaves (E) as the answer.

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