An electron is made to enters symmetrically between two paralel 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 m/s. If the magnitude of the elect

Question

An electron is made to enters 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 $10^{6}$ m/s. If the magnitude of the electric field 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 = $1.6 \times 10^{- 19}$ ) [2025]

Smart Achievers · Accepted Answer

(3)

Horizontal component of velocity will remain same.

$\Rightarrow t = \frac{l}{V_{x}} = \frac{10 \times 10^{- 2}}{10^{6}} = 10^{- 7}$

$\Rightarrow V_{y} = u_{y} + a_{y} t \Rightarrow V_{y} = 0 + \frac{e E}{m} \times 10^{- 7}$

$\Rightarrow V_{y} = \frac{1.6 \times 10^{- 19}}{9.1 \times 10^{- 31}} \times 9.1 \times 10^{+ 2} \times 10^{- 7}$

$\Rightarrow V_{y} = 16 \times 10^{6} m / s$

Solution

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

2 0

3 $16 \times 10^{6} m$

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

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Solution

1 1×106 m/s

2 0

3 16×106 m

4 16×104 m/s

Ans. (3) Horizontal component of velocity will remain same. ⇒ t=lVx=10×10–2106=10–7 ⇒ Vy=uy+ayt ⇒ Vy=0+eEm×10–7 ⇒ Vy=1.6×10–199.1×10–31×9.1×10+2×10–7 ⇒ Vy=16×106 m/s

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

3 $16 \times 10^{6} m$

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