Question

um, and -54.6. of charge +9e and a negative lon of charge -Be (see the figure below). (Take a 5.30 um, b-4.32 -Se i9e Electson x-direction. Measure the angle counter-clockwise from the +x-axis.) (o) Find the magnitude and direction of the resultant force on the electron. (Let right be the + magnitude direction tx-direction. Measure the angle counter-clockwise from the (b) Find the magnitude and direction of the electron's instantaneous acceleration (Let right be the +x-axis.) magntude direction

Answer 1

Charge $+9e$ attracts the electron $-e$ towards it. Force of attraction will be along the line joining charges and is towards the charge $+9e$.

Charge $-8e$ repels the electron $-e$ away from it.Force of repulsion will be along the line joining charges making an angle $\theta$ below the X-axis.

Electric force on electron due to charge $+9e$ is $\vec{F}_1=\frac{(9e)(e)}{4\pi\epsilon_0 a^2}(-\hat{i})$

Electric force on electron due to charge $-8e$ is $\vec{F}_2=\frac{(8e)(e)}{4\pi\epsilon_0 (3.00*10^{-6})^2}(\cos\theta\hat{i}-\sin\theta \hat{j})$

Net force on electron is $\vec{F}=\frac{(9e)(e)}{4\pi\epsilon_0 a^2}(-\hat{i})+\frac{(8e)(e)}{4\pi\epsilon_0 (3.00*10^{-6})^2}(\cos\theta\hat{i}-\sin\theta \hat{j})$

$\vec{F}=8.99*10^9*\left ( \frac{9*(1.6*10^{-19})^2}{(5.30*10^{-6})^2}(-\hat{i})+\frac{8*(1.6*10^{-19})^2}{ (3.00*10^{-6})^2}(\cos54.6\degree\hat{i}-\sin 54.6\degree \hat{j}) \right )$

$\vec{F}=8.99*10^9*\left (- 8.20*10^{-27}\hat{i} +13.18*10^{-27}\hat{i}-18.55*10^{-27}\hat{j}\right )$

$\vec{F}=\left (4.48*10^{-17}\hat{i}-16.6*10^{-17}\hat{j}\right )$

Magnitude is ${F}=1.72*10^{-16}\,N$

Direction is $\theta=360\degree-\tan^{-1}\left ( \frac{16.6*10^{-17}}{4.48*10^{-17}} \right )=285\degree$ counter-clockwise from positive X-axis.

b)

Electron's acceleration $\vec{a}=\frac{\vec{F}}{m}=\frac{4.48*10^{-17}\hat{i}-16.6*10^{-16}\hat{j}}{9.11*10^{-31}}=(4.92*10^{13}\hat{i}-18.2*10^{13})$

Magnitude is $a=18.8*10^{13}\,m/s^2$

Direction is $\theta=360\degree-\tan^{-1}\left ( \frac{18.2*10^{13}}{4.92*10^{13}} \right )=285\degree$ counter-clockwise from positive X-axis.