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

(C) $16 \times 10^{6} \mathrm{~m} / \mathrm{s}$

(A) $1 \times 10^{6} \mathrm{~m} / \mathrm{s}$

(D) $16 \times 10^{4} \mathrm{~m} / \mathrm{s}$

JEE Main · 2025 · 22 Jan 2025 · Shift 1 · Electrostatics

An electron is made to enters symmetrically between two parallel and equally but oppositely charged metal plates, each of 10 cm length. The electron emerges out of the field region with a horizontal component of velocity

1 0^{6} m / s

. If the magnitude of the electric between the plates is 9.1

V / cm

, then the vertical component of velocity of electron is

(mass of electron

= 9.1 \times 1 0^{- 31} kg

and charge of electron

= 1.6 \times 1 0^{- 19} C

)

(A) $1 \times 1 0^{6} m / s$
(B)0
(C) $16 \times 1 0^{6} m / s$ ✓ Correct
(D) $16 \times 1 0^{4} m / s$

Answer: (C)

Solution

The time spent between the plates is

t = \frac{ℓ}{v _{x}} = \frac{10 \times 1 0 ^{- 2}}{1 0 ^{6}} = 1 0^{- 7} s .

The transverse acceleration due to the field

E = 9.1 V/cm = 910 V/m

a_{y} = \frac{e E}{m} = \frac{1.6 \times 1 0 ^{- 19} \times 910}{9.1 \times 1 0 ^{- 31}} = 1.6 \times 1 0^{14} m/s^{2} .

Starting from rest in the vertical direction, the vertical velocity on leaving the field is

v_{y} = a_{y} t = 1.6 \times 1 0^{14} \times 1 0^{- 7} = 1.6 \times 1 0^{7} m/s = 16 \times 1 0^{6} m/s .

JEE Main · 2025 · 22 Jan 2025 · Shift 1 · Single correct option

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JEE Main · 2025 · 22 Jan 2025 · Shift 1 · Electrostatics

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