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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. 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. 07.10.2012 at 21:13

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

• 08.10.2012 at 11:52

Yep. Thanks and fixed.

3. 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.