Question 86 of 311

A uniform electric field of $10N{C}^{-1}$ is created between two parallel charged plates (as shown in figure). An electron enters the field symmetrically between the plates with a kinetic energy $0.5eV.$ The length of each plate is $10cm$. The angle $(\theta )$ of deviation of the path of electron as it comes out of the field is ______(in degree).

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Answer: 45

Given here, kinetic energy $\dfrac{1}{2}e=\dfrac{1}{2}m{u}_{x}^{2}\Rightarrow {u}_{x}=\sqrt{\dfrac{e}{m}}$
Acceleration of electron due to electric field is $a=\dfrac{eE}{m}$, where, $e$ is charge on electron.
Using equation of motion, $s={u}_{x}t+\dfrac{1}{2}{a}_{x}{t}^{2}$ along $x$ direction, we get
$L={u}_{x}t+0\left(∵{a}_{x}=0\right) \Rightarrow 10\times {10}^{-2}={u}_{x}t \Rightarrow t=\dfrac{{10}^{-1}}{{u}_{x}}....\left(1\right)$
Using equation of motion, ${v}_{y}={u}_{y}+{a}_{y}t$ along y direction, we get
${v}_{y}=0+\dfrac{eE}{m}t=\dfrac{eE}{m}t...\left(2\right)$
From equation $\left(1\right)and\left(2\right)$, we have
${v}_{y}=\dfrac{eE}{m}\left(\dfrac{{10}^{-1}}{{u}_{x}}\right)$
Now, $\tan \theta =\dfrac{{v}_{y}}{{v}_{x}}=\dfrac{eE}{m}\left(\dfrac{{10}^{-1}}{{u}_{x}^{2}}\right)=\dfrac{e\times 10\times {10}^{-1}}{m\times \dfrac{e}{m}}=1$
The angle of deviation of the path of electron as it comes out of the field is $\theta =45^{\circ}$.