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MCQEasyJEE 2025Electric Field & Field Lines

JEE Physics 2025 Question with Solution

An electron is made to enters symmetrically between two parallel and equally but oppositely charged metal plates, each of 10 cm10 \, \text{cm} length. The electron emerges out of the field region with a horizontal component of velocity 106 m/s10^6 \, \text{m/s}. If the magnitude of the electric field between the plates is 9.1 V/cm9.1 \, \text{V/cm}, then the vertical component of velocity of electron is (mass of electron = 9.1×10−31 kg9.1 \times 10^{-31} \, \text{kg} and charge of electron = 1.6×10−19 C1.6 \times 10^{-19} \, \text{C}):

  • A

    1×106 m/s1 \times 10^6 \, \text{m/s}

  • B

    00

  • C

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

  • D

    16×104 m/s16 \times 10^4 \, \text{m/s}

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: Length of plates, L=10 cm=0.1 mL = 10 \, \text{cm} = 0.1 \, \text{m}; horizontal velocity, vx=106 m/sv_x = 10^6 \, \text{m/s}; electric field, E=9.1 V/cm=9.1×102 V/mE = 9.1 \, \text{V/cm} = 9.1 \times 10^2 \, \text{V/m}; charge of electron, e=1.6×10−19 Ce = 1.6 \times 10^{-19} \, \text{C}; mass of electron, m=9.1×10−31 kgm = 9.1 \times 10^{-31} \, \text{kg}.

Find: The vertical component of velocity of the electron on emerging from the plates.

The motion is like projectile motion in an electric field: the horizontal velocity remains constant, while the electron acquires vertical acceleration due to electric force.

Time spent between the plates:

t=Lvx=0.1106=1×10−7 st = \frac{L}{v_x} = \frac{0.1}{10^6} = 1 \times 10^{-7} \, \text{s}

Force on the electron:

F=eE=(1.6×10−19)(9.1×102)=1.456×10−16 NF = eE = (1.6 \times 10^{-19})(9.1 \times 10^2) = 1.456 \times 10^{-16} \, \text{N}

Acceleration of the electron:

a=Fm=1.456×10−169.1×10−31=1.6×1014 m/s2a = \frac{F}{m} = \frac{1.456 \times 10^{-16}}{9.1 \times 10^{-31}} = 1.6 \times 10^{14} \, \text{m/s}^2

Vertical component of velocity gained in time tt:

vy=at=(1.6×1014)(1×10−7)=1.6×107 m/sv_y = at = (1.6 \times 10^{14})(1 \times 10^{-7}) = 1.6 \times 10^7 \, \text{m/s}

Thus,

vy=16×106 m/sv_y = 16 \times 10^6 \, \text{m/s}

Therefore, the vertical component of velocity of the electron is 16×106 m/s16 \times 10^6 \, \text{m/s} and the correct option is C.

Using Electric Force and Transit Time

Given: Convert all quantities to SI units before calculation.

The electric force on the electron is

F=e⋅EF = e \cdot E

where e=1.6×10−19 Ce = 1.6 \times 10^{-19} \, \text{C} and E=9.1 V/cm=910 V/mE = 9.1 \, \text{V/cm} = 910 \, \text{V/m}. So,

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

Using F=maF = ma,

a=Fm=1.456×10−169.1×10−31=1.6×1014 m/s2a = \frac{F}{m} = \frac{1.456 \times 10^{-16}}{9.1 \times 10^{-31}} = 1.6 \times 10^{14} \, \text{m/s}^2

The electron travels horizontally through plate length 0.1 m0.1 \, \text{m} with constant horizontal speed 106 m/s10^6 \, \text{m/s}, so the time is

t=0.1106=10−7 st = \frac{0.1}{10^6} = 10^{-7} \, \text{s}

Now the vertical velocity is

vy=a⋅t=1.6×1014×10−7=16×106 m/sv_y = a \cdot t = 1.6 \times 10^{14} \times 10^{-7} = 16 \times 10^6 \, \text{m/s}

Therefore, the correct option is C.

Common mistakes

  • Using 9.1 V/cm9.1 \, \text{V/cm} directly as SI electric field is incorrect because acceleration must be calculated in SI units. First convert it to 910 V/m910 \, \text{V/m}, then use F=eEF = eE.

  • Assuming the horizontal velocity also changes is incorrect. The electric field gives acceleration only in the vertical direction here, so the horizontal component remains constant at 106 m/s10^6 \, \text{m/s} while computing transit time.

  • Using the plate length as vertical displacement or substituting it directly into a kinematics equation for vertical motion is incorrect. The plate length determines only the time of flight between the plates through t=Lvxt = \frac{L}{v_x}.

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