Question

A particle of mass $m$ and charge $-q$ enters the region between the two charged plates initially moving along x-axis with speed $v_x$ as shown in figure. The length of the plate is $L$ and a uniform electric field $E$ is maintained between the plates. An electron is projected with velocity $v_x=4\times {10}^{-6}m{s}^{-1}$. If electric field between the plates separated by $1cm$ is $8.2{\times 10}^{2}N{C}^{-1}$, then the electron will strike the upper pate if the length of the plate is (Take $m_e=9.1{\times 10}^{-31}$kg

• A. $2.14cm$
• B. $3.9cm$
• C. $1.23cm$
• D. $3.3cm$

Answer 1

Answer: (D)

$3.3cm$

Given $v_x=4\times {10}^{6}m{s}^{-1},d=1cm=1\times {10}^{-2}m,E=8.2\times {10}^{2}N{C}^{-1},q=e=1.6\times {10}^{-19}C,m_e=9.1\times {10}^{-31}kg$
The electron will strike the upper plate at its other end of $x=L$ as soon as its deflection
And $y=\frac{d}{2}=\frac{{10}^{-2}}{2}m=5\times {10}^{-3}m$
From eqn. (iii)
$L=\sqrt{\frac{2m_e{v}_x^{2}y}{qE}}=\sqrt{\frac{2\times 9.1\times {10}^{-31}\times {(4\times {10}^6)}^2\times 5\times {10}^{-3}}{1.6\times {10}^{-19}\times 8.2\times {10}^2}}$
$L=3.3\times {10}^{-2}m=3.3cm$
The electron will strike the upper plate at its other end, if length of plate is $3.3cm$