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JEE Main 202522 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
22 January
Shift
Morning Shift
Subject
Physics
Chapter
Electrostatics

JEE Main 2025 · 22 January · Morning Shift · Physics · Electrostatics

An electron is made to enter symmetrically between two parallel and...

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/s10^6 \mathrm{~m} / \mathrm{s}. If the magnitude of the electric field between the plates is 9.1 V/cm9.1 \mathrm{~V} / \mathrm{cm}, then the vertical component of velocity of electron is (mass of electron =9.1×10−31 kg=9.1 \times 10^{-31} \mathrm{~kg} and charge of electron =1.6×10−19C=1.6 \times 10^{-19} \mathrm{C} )An electron is made to enter symmetrically between two parallel and equally but oppositely charged metal plates, each

Options

  1. A0
  2. B1×106 m/s1 \times 10^6 \mathrm{~m} / \mathrm{s}
  3. C16×106 m/s16 \times 10^6 \mathrm{~m} / \mathrm{s}
  4. D16×104 m/s16 \times 10^4 \mathrm{~m} / \mathrm{s}

Correct answer

C. 16×106 m/s16 \times 10^6 \mathrm{~m} / \mathrm{s}

Step-by-step solution

An electron is made to enter symmetrically between two parallel and equally but oppositely charged metal plates, each
⇒t=ℓVx=10×10−2106=10−7⇒ Vy=uy+ayt⇒Vy=0+eEm×10−7⇒ Vy=1.6×10−199.1×10−31×9.1×10−2×10−7⇒ Vy=16×106\begin{aligned} & \Rightarrow \mathrm{t}=\frac{\ell}{\mathrm{V}_{\mathrm{x}}}=\frac{10 \times 10^{-2}}{10^6}=10^{-7} \\ & \Rightarrow \mathrm{~V}_{\mathrm{y}}=\mathrm{u}_{\mathrm{y}}+\mathrm{a}_{\mathrm{y}} \mathrm{t} \\ & \Rightarrow \mathrm{V}_{\mathrm{y}}=0+\frac{\mathrm{eE}}{\mathrm{m}} \times 10^{-7} \\ & \Rightarrow \mathrm{~V}_{\mathrm{y}}=\frac{1.6 \times 10^{-19}}{9.1 \times 10^{-31}} \times 9.1 \times 10^{-2} \times 10^{-7} \\ & \Rightarrow \mathrm{~V}_{\mathrm{y}}=16 \times 10^6\end{aligned}
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