Question

3. 3d +O +60 An electron and two large positive charges are positioned in a horizontal line, as shown above. 5 pts.) You can solve (a) & (b) in either order. Label your work clearly a. Find the magnitude and direction of the net electric force acting on the electron. Show your work. b. Find the magnitude and direction of the net electric field at the electron's position. (Note: Do NOT consider any field contribution from the electron itself, just from the two large positive charges.) Show your work. Express your final answers.. solely in terms of algebraic variables Q and d, and any necessary physical constants. Do NOT plug in numerical values for the constants!) in SIMPLEST algebraic form.

Answer 1

$According\,\,to\,\,Coulomb's \,\,law\,\,electric\,\,force\,\,\overrightarrow{F}\,\,acting\,\,between\,\,two\,\,charge\,\,\\particles\,\, q_1 \,\,and\,\,q_2\,\,separated \,\,by\,\,a\,\,distance\,\,d.$

$\overrightarrow{F}=\frac{1}{4\pi\varepsilon_0\varepsilon_r }\frac{q_1q_2}{d^2}\hat{r}$
$Where\,\, \frac{1}{4\pi\varepsilon_0\varepsilon_r }\,\,is\,\,Coulomb's\,\,law constant\\ \varepsilon_0\,\, is\,\,vacuum\,\, permittivity\\ \varepsilon_r\,\,is\,\,relative\,\, permittivity\\ \hat{r}\,\,is \,\,direction\,\,vector$
$One\,\,can\,\,substitute\,\,k=\frac{1}{4\pi\varepsilon_0\varepsilon_r }\,\,as\,\,electrostatic\,\,constant$

$\therefore \overrightarrow{F}=k\frac{q_1q_2}{d^2}\hat{r}$

$Now\,\,electric\,\,force\,\,\overrightarrow{F}=q.\overrightarrow{E} \\Where\,\, \overrightarrow{E}\,\,is\,\,electric\,\, field\,\,acting\,\,upon \,\, charge\,\,q\\$

$\therefore \overrightarrow{E}=k\frac{q_2}{d^2}\hat{r}$

${\color{DarkBlue} Now\,\,to\,\,the\,\,problem}$

$\dpi{200} \large a.$

$Electric\,\,force\,\,over\,\,e^-\,\,due\,\,to\,\,the\,\,charge\,\,particle\,\,+Q\,\,towards\,\,+Q\\F_1=k\frac{e^-Q}{d^2}\\\\ Electric\,\,force\,\,over\,\,e^-\,\,due\,\,to\,\,the\,\,charge\,\,particle\,\,+6Q\,\,towards\,\,+6Q\\ F_2=k\frac{e^-6Q}{(3d)^2}\\\\ =k\frac{2e^-Q}{3d^2}$

$Net\,\,electric\,\,force\,\,over\,\,e^-\,\,due\,\,to\,\,both\,\,charge\,\,particles\\towards\,\,the\,\,charge\,\,particle\,\,+Q\\$

  $F_{net}=F_1-F_2$

  $=k\frac{e^-Q}{d^2}-k\frac{2e^-Q}{3d^2}$

  $=k\frac{e^-Q}{3d^2}$

$\dpi{200} \large b.$

$Electric\,\,field\,\,over\,\,e^-\,\,due\,\,to\,\,the\,\,charge\,\,particle\,\,+Q\,\,towards\,\,+Q\\E_1=k\frac{Q}{d^2}\\\\ Electric\,\,field\,\,over\,\,e^-\,\,due\,\,to\,\,the\,\,charge\,\,particle\,\,+6Q\,\,towards\,\,+6Q\\E_2=k\frac{6Q}{(3d)^2}\\\\ =k\frac{2Q}{3d^2}$

$Net\,\,electric\,\,field\,\,over\,\,e^-\,\,due\,\,to\,\,both\,\,charge\,\,particles\\towards\,\,the\,\,charge\,\,particle\,\,+Q\\$

  $E_{net}=E_1-E_2$

  $=k\frac{Q}{d^2}-k\frac{2Q}{3d^2}$

  $=k\frac{Q}{3d^2}$

$\dpi{120} \large {\color{Blue} alternet \,\,a.}$

$We\,\,know\,\,\overrightarrow{F}=q.\overrightarrow{E}$
$\therefore F=e^-E_{net}$

  $=k\frac{e^-Q}{3d^2}\,\,\,\,\,\,towards\,\, the \,\,charge\,\,particle\,\,+Q\,\,[same\,\,direction\,\,as\,\,field]$