A pair of parallel metal plates are kept with a seperation ' $d$ '. One plate is at a potential $+ V$ and the other is at ground potential. A narrow beam of electrons enters the space between the plates with a velocity $v_{0}$ and in a direction parallel to the plates. What will be the angle of the beam with the plates after it travels an axial distance $L$ ?

t a n^{- 1} [\frac{e V L}{m d v_{0}}]

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t a n^{- 1} [\frac{e V L}{m d v_{0}^{2}}]

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s i n^{- 1} [\frac{e V L}{m d v_{0}}]

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c o s^{- 1} [\frac{e V L}{m d v_{0}^{2}}]

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Solution

The correct option is B $t a n^{- 1} [\frac{e V L}{m d v_{0}^{2}}]$
Assume the electron beam are travelling close to one of the plate.

$t a n θ = \frac{V_{y}}{V_{x}} = \frac{a t}{V_{x}} = \frac{q E t}{m V_{x}}$
Also, $E = \frac{V}{d}$
$\Rightarrow tan θ = (\frac{q}{m}) (\frac{V}{d}) \frac{t}{V_{x}}$ .
Now, $L = V_{x} t$
Where, $V_{x} = v_{0}$
$\Rightarrow tan θ = (\frac{q}{m}) (\frac{V}{d}) \frac{L}{v_{0}^{2}}$
Where, $q = e$
$∴ θ = t a n^{- 1} [\frac{e V L}{m d v_{0}^{2}}]$

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The correct option is B tan−1[eVLmdv20]Assume the electron beam are travelling close to one of the plate. tan θ=VyVx=atVx=qEtmVx Also, E=Vd ⇒tanθ=(qm)(Vd)tVx. Now, L=Vxt Where, Vx=v0 ⇒tanθ=(qm)(Vd)Lv20 Where, q=e ∴θ=tan−1[eVLmdv20]