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/s. If the magnitude of the electric between the plates is 9.1 V/cm, then the vertical component of velocity of electron is (mass of electron =9.1×10−31kg and charge of electron =1.6×10−19C

To calculate the vertical component of the velocity of the electron when it exits the electric field, we use the following steps:Step 1: Force acting on the electronThe force FF on the electron due to the electric field EE is given by:F=qEF = qEwhere:q=1.6×10−19 Cq = 1.6 \times 10^{-19} \, \text{C} (charge of the electron),E=9.1 V/cm=9.1×102 V/mE = 9.1 \, \text{V/cm} = 9.1 \times 10^2 \, \text{V/m}.F=(1.6×10−19)⋅(9.1×102)=1.456×10−16 N.F = (1.6 \times 10^{-19}) \cdot (9.1 \times 10^2) = 1.456 \times 10^{-16} \, \text{N}.Step 2: Acceleration of the electronThe vertical acceleration aa of the electron is given by Newton's second law:a=Fma = \frac{F}{m}where:m=9.1×10−31 kgm = 9.1 \times 10^{-31} \, \text{kg} (mass of the electron).a=1.456×10−169.1×10−31=1.6×1014 m/s2.a = \frac{1.456 \times 10^{-16}}{9.1 \times 10^{-31}} = 1.6 \times 10^{14} \, \text{m/s}^2.Step 3: Time spent in the electric fieldThe time tt the electron spends in the field is determined by the horizontal motion. Since the electron moves with a horizontal velocity vx=106 m/sv_x = 10^6 \, \text{m/s} and the length of the plates is L=10 cm=0.1 mL = 10 \, \text{cm} = 0.1 \, \text{m}, the time is:t=Lvxt = \frac{L}{v_x} t=0.1106=10−7 s.t = \frac{0.1}{10^6} = 10^{-7} \, \text{s}. Step 4: Vertical velocityThe vertical velocity vyv_y acquired by the electron due to acceleration in the electric field is:vy=a⋅tv_y = a \cdot tSubstitute a=1.6×1014 m/s2a = 1.6 \times 10^{14} \, \text{m/s}^2 and t=10−7 st = 10^{-7} \, \text{s}:vy=(1.6×1014)⋅(10−7)=16×106 m/s.

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/s. If the magnitude of the electric between the plates is 9.1 V/cm, then the vertical component of velocity of electron is (mass of electron =9.1×10−31kg and charge of electron =1.6×10−19C

To calculate the vertical component of the velocity of the electron when it exits the electric field, we use the following steps:Step 1: Force acting on the electronThe force FF on the electron due to the electric field EE is given by:F=qEF = qEwhere:q=1.6×10−19 Cq = 1.6 \times 10^{-19} \, \text{C} (charge of the electron),E=9.1 V/cm=9.1×102 V/mE = 9.1 \, \text{V/cm} = 9.1 \times 10^2 \, \text{V/m}.F=(1.6×10−19)⋅(9.1×102)=1.456×10−16 N.F = (1.6 \times 10^{-19}) \cdot (9.1 \times 10^2) = 1.456 \times 10^{-16} \, \text{N}.Step 2: Acceleration of the electronThe vertical acceleration aa of the electron is given by Newton's second law:a=Fma = \frac{F}{m}where:m=9.1×10−31 kgm = 9.1 \times 10^{-31} \, \text{kg} (mass of the electron).a=1.456×10−169.1×10−31=1.6×1014 m/s2.a = \frac{1.456 \times 10^{-16}}{9.1 \times 10^{-31}} = 1.6 \times 10^{14} \, \text{m/s}^2.Step 3: Time spent in the electric fieldThe time tt the electron spends in the field is determined by the horizontal motion. Since the electron moves with a horizontal velocity vx=106 m/sv_x = 10^6 \, \text{m/s} and the length of the plates is L=10 cm=0.1 mL = 10 \, \text{cm} = 0.1 \, \text{m}, the time is:t=Lvxt = \frac{L}{v_x} t=0.1106=10−7 s.t = \frac{0.1}{10^6} = 10^{-7} \, \text{s}. Step 4: Vertical velocityThe vertical velocity vyv_y acquired by the electron due to acceleration in the electric field is:vy=a⋅tv_y = a \cdot tSubstitute a=1.6×1014 m/s2a = 1.6 \times 10^{14} \, \text{m/s}^2 and t=10−7 st = 10^{-7} \, \text{s}:vy=(1.6×1014)⋅(10−7)=16×106 m/s.

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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/s. If the magnitude of the electric between the plates is 9.1 V/cm, then the vertical component of velocity of electron is
$($ mass of electron $= 9.1 \times 10^{- 31}$ kg and charge of electron $= 1.6 \times 10^{- 19} C)$

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$0$

$16 \times 10^{6} m / s$

$16 \times 10^{4} m / s$

$1 \times 10^{6} m / s$

answer is C.

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Detailed Solution

To calculate the vertical component of the velocity of the electron when it exits the electric field, we use the following steps:

Step 1: Force acting on the electron

The force $F$ on the electron due to the electric field $E$ is given by:

$F = qE$

where:

$q = 1.6 \times 10^{-19} \, \text{C}$ (charge of the electron),
$E = 9.1 \, \text{V/cm} = 9.1 \times 10^2 \, \text{V/m}$ .

$F = (1.6 \times 10^{-19}) \cdot (9.1 \times 10^2) = 1.456 \times 10^{-16} \, \text{N}.$

Step 2: Acceleration of the electron

The vertical acceleration $a$ of the electron is given by Newton's second law:

$a = \frac{F}{m}$

where:

$m = 9.1 \times 10^{-31} \, \text{kg}$ (mass of the electron).

$a = \frac{1.456 \times 10^{-16}}{9.1 \times 10^{-31}} = 1.6 \times 10^{14} \, \text{m/s}^2.$

Step 3: Time spent in the electric field

The time $t$ the electron spends in the field is determined by the horizontal motion. Since the electron moves with a horizontal velocity $v_x = 10^6 \, \text{m/s}$ and the length of the plates is $L = 10 \, \text{cm} = 0.1 \, \text{m}$ , the time is: