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Question

A rod is 2.5 m long. Its charge is 2×107 C.-2 \times 10^{-7}\ \mathrm{C}. The observation location is 4 cm from the rod, in the midplane. In the expression E=14πε0Qrr2+(L/2)2E=\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{r \sqrt{r^{2}+(L / 2)^{2}}} what is r in meters?

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Answered 2 years ago
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In the expression for the field, rr is the distance of the point of observation from the center of the rod. Therefore r=4r=4 cm =0.04=0.04 m.

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