A point charge of -5 uC is located at x=4m, y=-2m. A second point charge of 12uC is located at x=1m, y=2m
(a) Find the magnitude and direction of the electric field at x=-1m, y=0
(b) calculate the magnitude and direction of the force on an electron at x=-1m, y=0
Given,
q_{1}=-5$\mathrm{\mu C}$
q_{2}=12 $\mathrm{\mu C}$
a)
${\mathrm{E}}_{1}=\frac{1}{4{\mathrm{\pi \epsilon}}_{0}}\frac{{\mathrm{q}}_{1}}{{\mathrm{r}}_{1\mathrm{p}}}\phantom{\rule{0ex}{0ex}}{\mathrm{r}}_{1}=(\mathrm{x}=4,\mathrm{y}=-2)\phantom{\rule{0ex}{0ex}}\mathrm{p}=(\mathrm{x}=1,\mathrm{y}=0)\phantom{\rule{0ex}{0ex}}{\mathrm{r}}_{1\mathrm{p}}=\sqrt{{(4--1)}^{2}+{(-2)}^{2}}=3\phantom{\rule{0ex}{0ex}}{\mathrm{E}}_{1}=9\times {10}^{9}\times \frac{-5\times {10}^{-6}}{3}=-15\times {10}^{3}\phantom{\rule{0ex}{0ex}}{\mathrm{r}}_{2\mathrm{p}}=\sqrt{{(-1-1)}^{2}+{(-2)}^{2}}=2.83\mathrm{m}\phantom{\rule{0ex}{0ex}}{\mathrm{E}}_{2}=9\times {10}^{9}\times \frac{12\times {10}^{-6}}{2.83}=38.162\times {10}^{3}\phantom{\rule{0ex}{0ex}}\mathrm{E}={\mathrm{E}}_{1}+{\mathrm{E}}_{2}=-15\times {10}^{3}+38.162\times {10}^{3}\phantom{\rule{0ex}{0ex}}\mathrm{E}=23.162\times {10}^{3}\mathrm{N}/\mathrm{C}$
Direction of the Electric field is towards left.
b)
r_{1p}=3 m
r_{2p}=2.83 m
q=1.6$\times {10}^{-19}\mathrm{C}$
${\mathrm{F}}_{1\mathrm{p}}=\frac{1}{4{\mathrm{\pi \epsilon}}_{0}}\frac{{\mathrm{q}}_{1}\mathrm{q}}{{{\mathrm{r}}_{1\mathrm{p}}}^{2}}\phantom{\rule{0ex}{0ex}}{\mathrm{F}}_{2\mathrm{p}}=\frac{1}{4{\mathrm{\pi \epsilon}}_{0}}\frac{{\mathrm{q}}_{2}\mathrm{q}}{{{\mathrm{r}}_{2\mathrm{p}}}^{2}}\phantom{\rule{0ex}{0ex}}\mathrm{F}={\mathrm{F}}_{1\mathrm{p}}+{\mathrm{F}}_{2\mathrm{p}}=\frac{1}{4{\mathrm{\pi \epsilon}}_{0}}\left[\frac{{\mathrm{q}}_{1}\mathrm{q}}{{{\mathrm{r}}_{1\mathrm{p}}}^{2}}+\frac{{\mathrm{q}}_{2}\mathrm{q}}{{{\mathrm{r}}_{2\mathrm{p}}}^{2}}\right]=9\times {10}^{9}\left[\frac{-5\times {10}^{-6}\times 1.6\times {10}^{-19}}{{3}^{2}}+\frac{12\times {10}^{-6}\times 1.6\times {10}^{-19}}{2.{83}^{2}}\right]\phantom{\rule{0ex}{0ex}}=1.358\times {10}^{-15}\mathrm{N}$
Using the parallelogram law the resultant force is in the direction of positive x axis.