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Q)

Suppose the particle in $Q_{33}$ is an electron projected with velocity $V_x =2.0 \times 10^6 m/s$. 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 ? $(|e|=1.6 \times 10^{-19}\;C,m_e=9.1 \times 10^{-3}\;kg)$

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A)
Answer :
Given :$V_x=2 \times 10^6 m/s$
$E= 9.1 \times 10^2 N/c$
$q= e= 1.6 \times 10^{-19}\;C$
$me= 9.1 \times 10^{-31}\;kg$
$d=0.5 \;cm=0.5 \times 10^{-2}\;m$
$\qquad= 5 \times 10^{-3}\;m$
Let the electron strike the upper plate at $x=L$ as it gets deflected
$Y= \large\frac{d}{2}=\frac{5 \times 10^{-3}}{2}$$=2.5 \times 10^{-3}\;m$
$y= \large\frac{qEL^2}{2m V_x^2}$
$L^2 =\large\frac{2 mvx^2y}{qE}$
$\therefore L= \sqrt{\large\frac{2mY}{qE}}$$ \times V_x$
Substituting the values ,
$L= \sqrt{ \large\frac{2 \times 9.1 \times 10^{-31} \times 2.5 \times 10^{-3}}{1.6 \times 10^{-19}\times 9.1 \times 10^2}}$$ \times 2 \times 10^{6}$
$L= 1.12 \times 10^{-2}\;m$
$\qquad= 1.12\;cm$
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