JEE Mains 2025 Physics Electron in Electric Field Question | 22 Jan Shift 1
In this post, we solve a question from the JEE Mains 2025 Physics paper (22nd January, Shift 1) based on the motion of an electron in a uniform electric field. This question tests your understanding of force, acceleration, and kinematics of charged particles.
📘 Question
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 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 the electron is:
Given:
- Length of plates, \( L = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Electric field, \( E = 9.1 \, \text{V/cm} = 910 \, \text{V/m} \)
- Electron charge, \( e = 1.6 \times 10^{-19} \, \text{C} \)
- Electron mass, \( m = 9.1 \times 10^{-31} \, \text{kg} \)
- Horizontal velocity, \( v_x = 10^6 \, \text{m/s} \)
🧮 Step-by-Step Solution
Time spent in the electric field:
\[
t = \frac{L}{v_x} = \frac{0.1}{10^6} = 1 \times 10^{-7} \, \text{s}
\]
Force on the electron due to electric field:
\[
F = eE = 1.6 \times 10^{-19} \times 910 = 1.456 \times 10^{-16} \, \text{N}
\]
Acceleration of the electron:
\[
a = \frac{F}{m} = \frac{1.456 \times 10^{-16}}{9.1 \times 10^{-31}} = 1.6 \times 10^{14} \, \text{m/s}^2
\]
Vertical velocity gained by the electron:
\[
v_y = at = 1.6 \times 10^{14} \times 1 \times 10^{-7} = 1.6 \times 10^7 \, \text{m/s}
\]
✅ Final Answer: Option (3) → \( 16 \times 10^6 \, \text{m/s} \)
🎥 Solution Video
Watch the detailed explanation here:
🔖 Tags
JEE Mains 2025, JEE Mains Physics 2025, JEE Mains 2025 PYQ, JEE Mains 2025 Physics PYQ, Electron in electric field, JEE Mains 2025 Electrostatics, 22 Jan Shift 1 Physics, JEE Previous Year Question
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