As shown in the following figure,an electron | vedclass

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(B) The force $F$ experienced by an electron of charge $e$ in an electric field $E$ is given by $F = eE$.According to Newton's second law of motion,$F

Vedclass · Accepted Answer

$3.52 	imes 10^{15} \ ms^{-2}$

Vedclass · Answer

(B) The force $F$ experienced by an electron of charge $e$ in an electric field $E$ is given by $F = eE$.According to Newton's second law of motion,$F = ma$,where $m_e$ is the mass of the electron and $a$ is its acceleration.Equating the two expressions for force: $ma = eE$.Therefore,the acceleration $a$ is given by $a = \frac{eE}{m_e}$.Given:Charge of electron $e = 1.6 	imes 10^{-19} \ C$Electric field $E = 2.0 	imes 10^4 \ NC^{-1}$Mass of electron $m_e = 9.1 	imes 10^{-31} \ kg$Substituting the values:$a = \frac{(1.6 	imes 10^{-19} \ C) 	imes (2.0 	imes 10^4 \ NC^{-1})}{9.1 	imes 10^{-31} \ kg}$$a = \frac{3.2 	imes 10^{-15}}{9.1 	imes 10^{-31}} \ ms^{-2}$$a \approx 0.3516 	imes 10^{16} \ ms^{-2}$$a \approx 3.52 	imes 10^{15} \ ms^{-2}$.

As shown in the following figure,an electron falls through a distance of $1.5 \ cm$ in a uniform electric field of magnitude $2.0 \times 10^4 \ NC^{-1}$. Find the acceleration of the electron due to the electric field.

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