Solution to 2008 Problem 37

The exact electric field at point P is

\begin{align*}\mathbf{E} = \frac{q}{4 \pi \epsilon_0} \cdot \frac{r \hat{\mathbf{y}} - (l/2) \hat{\mathbf{x}} }{\left(r^2 + l...
where the first term is from the positive charge and the second term is from the the negative charge. The y-components of the electric fields due to the two charges cancel, so
\begin{align*}\mathbf{E} = -\frac{lq}{4 \pi \epsilon_0\left(r^2 + l^2/4 \right)^{3/2}} \hat{\mathbf{x}}\end{align*}
When r is much greater than l, \left(r^2 + l^2/4 \right)^{3/2} \approx r^3, so
\begin{align*}\mathbf{E} = \boxed{-\frac{lq}{4 \pi \epsilon_0r^3} \hat{\mathbf{x}}}\end{align*}
Thus, answer (E) is correct.

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