An electron enters the region between the plates of a parallel-plate capacitor with same initial velocity at an angle $θ$ to the plants. The plate width is l and the plate separation is d. The electron follows the path shown, just missing the upper plate. Neglect gravity. Then,

t a n θ = \frac{2 d}{l}

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t a n θ = \frac{4 d}{l}

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t a n θ = \frac{6 d}{l}

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t a n θ = \frac{8 d}{l}

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Solution

The correct option is B $t a n θ = \frac{4 d}{l}$
$d = \frac{u^{2} s i n^{2} θ}{2 a}$ and $l = \frac{2 u^{2} s i n θ c o s θ}{a}$
$\Rightarrow \frac{d}{l} = \frac{t a n θ}{4}$
$\Rightarrow t a n θ = \frac{4 d}{l}$

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An electron enters the region between the plates of a parallel-plate capacitor with same initial velocity at an angle θ to the plants. The plate width is l and the plate separation is d. The electron follows the path shown, just missing the upper plate. Neglect gravity. Then,

The correct option is B tan θ=4dld=u2 sin2 θ2a and l=2u2 sin θ cos θa ⇒dl=tan θ4 ⇒tan θ=4dl

An electron enters the region between the plates of a parallel-plate capacitor with same initial velocity at an angle $θ$ to the plants. The plate width is l and the plate separation is d. The electron follows the path shown, just missing the upper plate. Neglect gravity. Then,

The correct option is B $t a n θ = \frac{4 d}{l}$
$d = \frac{u^{2} s i n^{2} θ}{2 a}$ and $l = \frac{2 u^{2} s i n θ c o s θ}{a}$
$\Rightarrow \frac{d}{l} = \frac{t a n θ}{4}$
$\Rightarrow t a n θ = \frac{4 d}{l}$