A particle of mass m and charge (-q) enters region between two charged plates initially moving along

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1. Electric Charges and Fields

medium

A particle of mass $m$ and charge $(-q)$ enters the region between the two charged plates initially moving along $x$ -axis with speed $v_{x}=2.0 \times 10^{6} \;m\, s ^{-1} .$ If $E$ between the plates separated by $0.5 \;cm$ is $9.1 \times 10^{2} \;N / C ,$ where will the electron strike the upper plate in $cm$?

$\left(|e|=1.6 \times 10^{-19} \;C , m_{e}=9.1 \times 10^{-31}\; kg .\right)$

$4.6$

$8.4$

$1.6$

$5.2$

Solution

Velocity of the particle, $v_{x}=2.0 \times 10^{6} \,m / s$

Separation of the two plates, $d =0.5\, cm =0.005\, m$

Electric field between the two plates, $E =9.1 \times 10^{2} \,N / C$

Charge on an electron, $q=1.6 \times 10^{-19} \,C$

Mass of an electron, $m _{ e }=9.1 \times 10^{-31} \,kg$

Let the electron strike the upper plate at the end of plate $L$, when deflection is s. Therefore,

$s=\frac{q E L^{2}}{2 m V_{x}^{2}}$

$\Rightarrow L=\sqrt{\frac{2 s m V_{x}^{2}}{q E}}$$=\sqrt{\frac{2 \times 0.005 \times 9.1 \times 10^{-31}}{1.6 \times 10^{-19} \times 9.1 \times 10^{2}}}$

$=\sqrt{0.00025}=0.016 \,m=1.6 \,cm$

Therefore, the electron will strike the upper plate after travelling $1.6 \;cm .$

Standard 12

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