An electron is made to enter symmetrically between two parallel and equally but oppositely charged metal plates, each of 10 cm length. The electron emerges out of the electric field region with a horizontal component of velocity 10⁶ m/s. If the magnitude of the electric field between the plates is 9.1 V/cm, then the vertical component of velocity of electron is (mass of electron = 9.1 × 10⁻³¹ kg and charge of electron = 1.6 × 10⁻¹⁹ C) (a) 1 × 10⁶ m/s (b) 16 × 10⁶ m/s (c) 16 × 10⁴ m/s (d) 0

Answer: (b). Acceleration: a = eE/m = 1.6×10⁻¹⁹×9109.1×10⁻³¹ = 1.6×10¹⁴ m/s² ( 9.1 V/cm = 910 V/m). Time between plates: t = 0.110⁶ = 10⁻⁷ s vy = at = 1.6×10¹⁴×10⁻⁷ = 1.6×10⁷ m/s So the answer is (b) 16×10⁶ m/s.

An electron is made to enter symmetrically between two parallel and… | JEE Main 2025 (22 Jan, Shift 1)

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Parallel Plate Capacitor

An electron is made to enter symmetrically between two parallel and equally but oppositely charged metal plates, each of 10 cm length. The electron emerges out of the electric field region with a horizontal component of velocity $10^{6}\text{ m/s}$ . If the magnitude of the electric field between the plates is $9.1\text{ V/cm}$ , then the vertical component of velocity of electron is (mass of electron $= 9.1 \times 10^{-31}\text{ kg}$ and charge of electron $= 1.6 \times 10^{-19}\text{ C}$ )

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Answer: (b)

Acceleration: $a = \frac{eE}{m} = \frac{1.6\times10^{-19}\times910}{9.1\times10^{-31}} = 1.6\times10^{14}\text{ m/s}^2$ ( $9.1\text{ V/cm} = 910\text{ V/m}$ ).
Time between plates: $t = \frac{0.1}{10^{6}} = 10^{-7}\text{ s}$
$v_y = at = 1.6\times10^{14}\times10^{-7} = 1.6\times10^{7}\text{ m/s}$
So the answer is (b) $16\times10^{6}\text{ m/s}$ .

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