An electron with kinetic energy K1 enters between parallel plates of a capacitor at an angle ' α ' with the plates. It leaves the plates at angle ' β ' with kinetic energy K2. Then the ratio of kinetic energies K1: K2 will be: (a) (cos ²β)/(cos ²α) (b) (cos β)/(cos α) (c) (sin ²β)/(cos ²α) (d) (cos β)/(sin α)

Answer: (a). The field is perpendicular to the plates, so the velocity component parallel to the plates is unchanged. v₁cos α = v₂cos β ⇒ (v₁)/(v₂) = (cos β)/(cos α) (K₁)/(K₂) = (v₁²)/(v₂²) So the answer is (a) (cos ²β)/(cos ²α).

An electron with kinetic energy K1 enters between parallel plates of… | JEE Main 2021 (25 Feb, Shift 2)

Q14JEE Main 2021 · 25 FEB · SHIFT 2Difficult

Parallel Plate Capacitor

An electron with kinetic energy K1 enters between parallel plates of a capacitor at an angle ' $\alpha$ ' with the plates. It leaves the plates at angle ' $\beta$ ' with kinetic energy K2. Then the ratio of kinetic energies K1: K2 will be:

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Answer: (a)

The field is perpendicular to the plates, so the velocity component parallel to the plates is unchanged.
$v_1\cos\alpha = v_2\cos\beta \Rightarrow \frac{v_1}{v_2} = \frac{\cos\beta}{\cos\alpha}$
$\frac{K_1}{K_2} = \frac{v_1^2}{v_2^2}$
So the answer is (a) $\frac{\cos^2\beta}{\cos^2\alpha}$ .

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Parallel Plate Capacitor question from JEE Main 2021 (25 Feb, Shift 2)

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