An e- of mass m is released from rest under gravity- An electric field E - mg-e is applied upwards- find the accleration of electron-

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Electric Field

An e- of mass...

Question

An $e^{-}$ of mass $m$ is released from rest under gravity. An electric field $E = \frac{m g}{e}$ is applied upwards, find the accleration of electron.

g

down

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zero

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g

down

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Solution

The correct option is D 2 $g$ down
$F_{n e t} = q E + m g = e E + m g$ (since electron has negative charge, therefore accleration will be acting in downward direction)
Substituting E,
$F_{n e t} = m g + m g = 2 m g$ (downwards)

$\to a = \frac{F_{n e t}}{m} = 2 g ↓$

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An e− of mass m is released from rest under gravity. An electric field E=mge is applied upwards, find the accleration of electron.

The correct option is D 2g downFnet=qE+mg=eE+mg (since electron has negative charge, therefore accleration will be acting in downward direction) Substituting E, Fnet=mg+mg=2mg (downwards) →a=Fnetm=2g↓

An $e^{-}$ of mass $m$ is released from rest under gravity. An electric field $E = \frac{m g}{e}$ is applied upwards, find the accleration of electron.

The correct option is D 2 $g$ down
$F_{n e t} = q E + m g = e E + m g$ (since electron has negative charge, therefore accleration will be acting in downward direction)
Substituting E,
$F_{n e t} = m g + m g = 2 m g$ (downwards)

$\to a = \frac{F_{n e t}}{m} = 2 g ↓$