بار الکتریکی $q=+۲\mu C$ را از نقطهٔ A درون میدان الکتریکی یکنواخت $E={{۱۰}^{۶}}\frac{N}{C}$، در مسیر خطچین نشان داده شده به نقطهٔ M رساندهایم. تغییر انرژی پتانسیل بار در این جابهجایی (${{U}_{M}}-{{U}_{A}}$) چند ژول است؟
$\left\{ \begin{matrix} {{U}_{B}}-{{U}_{A}}=-\left| q \right|E\left| d \right|\operatorname{Cos}0=-\left| 2\times {{10}^{-6}} \right|\times {{10}^{6}}\left| \frac{50}{100} \right|(+1)=-1J\,\,\,\,\,\,\,\,\,\,\,\,\, \\ {{U}_{C}}-{{U}_{B}}=-\left| q \right|E\left| d \right|\operatorname{Cos}{{90}^{{}^\circ }}=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\ {{U}_{D}}-{{U}_{C}}=-\left| q \right|E\left| d \right|\operatorname{Cos}{{180}^{{}^\circ }}=-\left| 2\times {{10}^{-6}} \right|\times {{10}^{6}}\left| \frac{30}{100} \right|(-1)=+0/6J \\ {{U}_{M}}-{{U}_{D}}=-\left| q \right|E\left| d \right|\operatorname{Cos}{{90}^{{}^\circ }}=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\\end{matrix} \right.\Rightarrow {{U}_{M}}-{{U}_{A}}=-1+0+0/6+0=-0/4J$