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Capacitor question

2025 · 22 Jan · Shift 1 · Q64
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Capacitor question

2025 · 22 Jan · Shift 1 · Q64

JEE MainPhysicsCapacitorMCQ+4 / −1
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}106 m/s. If the magnitude of the electric field between the plates is 9.1 V/cm9.1 \mathrm{~V} / \mathrm{cm}9.1 V/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}=9.1×10−31 kg and charge of electron =1.6×10−19C=1.6 \times 10^{-19} \mathrm{C}=1.6×10−19C )
  1. A
    1×106 m/s1 \times 10^6 \mathrm{~m} / \mathrm{s}1×106 m/s
  2. B
    16×106 m/s16 \times 10^6 \mathrm{~m} / \mathrm{s}16×106 m/s
  3. C
    16×104 m/s16 \times 10^4 \mathrm{~m} / \mathrm{s}16×104 m/s
  4. D
    0
View written solutionFree

Correct answer: B

  1. Given data
  • Length of plates: L=10 cm=0.1 mL = 10\text{ cm} = 0.1\text{ m}L=10 cm=0.1 m
  • Horizontal component of velocity: vx=106 m/sv_x = 10^6\text{ m/s}vx​=106 m/s
  • Electric field: E=9.1 V/cm=910 V/mE = 9.1\text{ V/cm} = 910\text{ V/m}E=9.1 V/cm=910 V/m
  • Charge of electron: e=1.6×10−19 Ce = 1.6 \times 10^{-19}\text{ C}e=1.6×10−19 C
  • Mass of electron: m=9.1×10−31 kgm = 9.1 \times 10^{-31}\text{ kg}m=9.1×10−31 kg

Since the electron enters horizontally and the electric field is vertical, the horizontal velocity remains constant.

  1. Time spent between the plates
t=Lvx=0.1106=10−7 st = \frac{L}{v_x} = \frac{0.1}{10^6} = 10^{-7}\text{ s}t=vx​L​=1060.1​=10−7 s
  1. Vertical acceleration of the electron

Magnitude of electric force:

F=eE=(1.6×10−19)(910)F = eE = (1.6 \times 10^{-19})(910)F=eE=(1.6×10−19)(910) F=1.456×10−16 NF = 1.456 \times 10^{-16}\text{ N}F=1.456×10−16 N

Hence acceleration:

a=Fm=1.456×10−169.1×10−31a = \frac{F}{m} = \frac{1.456 \times 10^{-16}}{9.1 \times 10^{-31}}a=mF​=9.1×10−311.456×10−16​ a=1.6×1014 m/s2a = 1.6 \times 10^{14}\text{ m/s}^2a=1.6×1014 m/s2
  1. Vertical component of velocity at exit

Initially, vertical velocity is zero. So,

vy=at=(1.6×1014)(10−7)v_y = at = (1.6 \times 10^{14})(10^{-7})vy​=at=(1.6×1014)(10−7) vy=1.6×107 m/sv_y = 1.6 \times 10^7\text{ m/s}vy​=1.6×107 m/s

This can be written as

vy=16×106 m/sv_y = 16 \times 10^6\text{ m/s}vy​=16×106 m/s
  1. Match with options

The correct option is:

B: 16×106 m/s16 \times 10^6\text{ m/s}16×106 m/s

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