Question

A charged particle (mass $m$ and charge $q$ ) moves along $X$ axis with velocity $V_{0}$. When it passes through the origin it enters a region having uniform electric field $\vec{ E }=- E \hat{ j }$ which extends upto $x=d$. Equation of path of electron in the region $x > d$ is:

• A. $y =\frac{ qEd }{ mV _{0}^{2}}\left(\frac{ d }{2}- x \right)$
• B. $y =\frac{ qEd }{ mV _{0}^{2}}( x - d )$
• C. $y =\frac{ qEd }{ mV _{0}^{2}} x$
• D. $y =\frac{ qEd ^{2}}{ mV _{0}^{2}} x$

Answer 1

Answer: (A)

Let particle have charge $q$ and mass 'm' Solve for $( q , m )$ mathematically
$F _{ x }=0, a _{ x }=0,( v )_{ x }=$ constant
time taken to reach at ' $P'=\frac{ d }{ v _{0}}= t _{0}$ (let) $\ldots$.. (1)
(Along $- y ), y _{0}=0+\frac{1}{2} \cdot \frac{ qE }{ m } \cdot t _{0}^{2} \ldots .(2)$
$v _{ x }= v _{0}$
$v = u + at $ (along -ve 'y')
speed $v_{y 0}=\frac{q E}{m} \cdot t_{0}$
$\tan \theta=\frac{ v _{ y }}{ v _{ x }}=\frac{ qEt _{0}}{ m \cdot v _{0}},\left( t _{0}=\frac{ d }{ v _{0}}\right)$
$\tan \theta=\frac{q E d}{m \cdot v_{0}^{2}}$
slope $=\frac{- qEd }{ mv _{0}^{2}}$
Now we have to find eq $^{n}$ of straight line
whose slope is $\frac{- qEd }{ mv _{0}^{2}}$ and it pass through
point $\rightarrow\left(d,-y_{0}\right)$
Because after $x>d$
No electric field $\Rightarrow F _{ net }=0, \vec{ v }= const.$
$y=m x+c, \{ m = \frac{qEd}{mv_0^2}, (d, -y_0)\}$
$- y _{0}=\frac{- qEd }{ mv _{0}^{2}} \cdot d + c $
$\Rightarrow c =- y _{0}+\frac{ qEd ^{2}}{ mv _{0}^{2}}$
Put the value
$y =\frac{- qEd }{ mv _{0}^{2}} x - y _{0}+\frac{ qEd ^{2}}{ mv _{0}^{2}}$
$y _{0}=\frac{1}{2} \cdot \frac{ qE }{ m }\left(\frac{ d }{ v _{0}}\right)^{2}=\frac{1}{2} \frac{ qEd ^{2}}{ mv _{0}^{2}}$
$y =\frac{- qEdx }{ mv _{0}^{2}}-\frac{1}{2} \frac{ qEd ^{2}}{ mv _{0}^{2}}+\frac{ qEd ^{2}}{ mv _{0}^{2}}$
$y =\frac{- qEd }{ mv _{0}^{2}} x +\frac{1}{2} \frac{ qEd ^{2}}{ mv _{0}^{2}}$
$y =\frac{ qEd }{ mv _{0}^{2}}\left(\frac{ d }{2}- x \right)$