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Question

Suppose there is a uniform electric field in the (+x)direction. If a negative charge moves in the (-x)direction; does the electric potential energy increase or decrease? Explain.

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Solution

$Theelectricpotentialenergywillinceasebecauseoffollowingreason,\phantom{\rule{0ex}{0ex}}Elelctricpotentialenergy=-\int eE.dr=-\frac{1}{4\pi {\epsilon }_{0}}×e×{\int }_{{r}_{A}}^{{r}_{B}}\frac{q}{{r}^{2}}.dr\phantom{\rule{0ex}{0ex}}Whereeiselectroniccharge,qthechargeofelectricfieldEand{r}_{B}<{r}_{A}.\phantom{\rule{0ex}{0ex}}Onintegrationpotentialenergybecomes,\phantom{\rule{0ex}{0ex}}U=-\frac{eq}{4\pi {\epsilon }_{0}}×{\left[\frac{-1}{r}\right]}_{{r}_{A}}^{{r}_{B}}=\frac{eq}{4\pi {\epsilon }_{0}}×\left(\frac{1}{{r}_{B}}-\frac{1}{{r}_{A}}\right).\phantom{\rule{0ex}{0ex}}Since{r}_{A}>{r}_{B},thereforeelectricpotentialenergyispositive.$

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