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Charged particle entering a uniform field between plates: deflection, angle or exit velocity

This type of question, asked 4 times: 28 Jul 2022, 24 Jan 2023, 25 Jan 2023, 22 Jan 2025. All 4 questions below, oldest first.

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  1. Question 1

    28 Jul 2022

    100
    A uniform electric field E=(8 m/e)V/mE = (8\text{ }m/e)V/m is created between two parallel plates of length 1 m1\text{ }m as shown in figure, (where m=m = mass of electron and e=e = charge of electron). An electron enters the field symmetrically between the plates with a speed of 2 m/s2\text{ }m/s. The angle of the deviation (θ)(\theta) of the path of the electron as it comes out of the field will be

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  2. Question 2

    24 Jan 2023

    24
    A stream of a positively charged particles having q m=2×1011Ckg\frac{q}{\text{ }m}= 2 \times10^{11}\frac{C}{kg} and velocity v→0=3×107im^s{\overrightarrow{v}}_{0}= 3 \times\frac{10^{7}\widehat{im}}{s} is deflected by an electric field 1.8ȷ^kVm\frac{1.8\widehat{\jmath}kV}{m}. The electric field exists in a region of 10 cm10\text{ }cm along xx direction. Due to the electric field, the deflection of the charge particles in the yy direction___ is mmmm

  3. Question 3

    25 Jan 2023

    21
    A uniform electric field of 10 N/C10\text{ }N/C is created between two parallel charged plates (as shown in figure). An electron enters the field symmetrically between the plates with a kinetic energy 0.5eV0.5eV. The length of each plate is 10 cm10\text{ }cm. The angle (θ)(\theta) of deviation of the path of electron as it comes out of the field is ___(in degree).

  4. Question 4

    22 Jan 2025

    9
    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 106 m/s10^{6}\text{ }m/s. If the magnitude of the electric field between the plates is 9.1 V/cmV/cm, then the vertical component of velocity of electron is (mass of electron =9.1×10−31 kg= 9.1 \times10^{- 31}\text{ }kg and charge of electron =1.6×10−19C= 1.6 \times10^{- 19}C )

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