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A pair of parallel metal plates are kept with a separation 'd'. One plate is at a potential $+\mathrm{V}$ and the other is at ground potential. A narrow beam of electrons enters the space between the plates with a velocity $\mathrm{v}_{0}$ 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?
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Verified Answer
The correct answer is:
$\tan ^{-1}\left(\frac{\mathrm{eVL}}{\mathrm{mdv}_{0}^{2}}\right)$
Hint:
$t=\frac{L}{v_{0}}$
$v_{y}=\frac{e}{m} \cdot \frac{v}{d} \times \frac{L}{v_{0}}=\frac{e v L}{m d v_{0}}$
$\tan \theta=\frac{v_{y}}{v_{x}}=\frac{e v L}{m d v_{0} \cdot v_{0}}$
$\Rightarrow \theta=\tan ^{-1}\left(\frac{e v L}{m d v_{0}^{2}}\right)$
$t=\frac{L}{v_{0}}$
$v_{y}=\frac{e}{m} \cdot \frac{v}{d} \times \frac{L}{v_{0}}=\frac{e v L}{m d v_{0}}$
$\tan \theta=\frac{v_{y}}{v_{x}}=\frac{e v L}{m d v_{0} \cdot v_{0}}$
$\Rightarrow \theta=\tan ^{-1}\left(\frac{e v L}{m d v_{0}^{2}}\right)$
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