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# A rod is 2.5 m long. Its charge is $-2 \times 10^{-7}\ \mathrm{C}.$ The observation location is 4 cm from the rod, in the midplane. In the expression $E=\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{r \sqrt{r^{2}+(L / 2)^{2}}}$ what is r in meters?

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In the expression for the field, $r$ is the distance of the point of observation from the center of the rod. Therefore $r=4$ cm $=0.04$ m.

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