Question

Take the particle in with an electron 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 plate if the length of plate is (Take me $=9.1\times 10^{-31}kg$)

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

Answer 1

Answer: (D)

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 $=\frac{d}{2}=\frac{10^{-2}}{2}m=5\times 10^{-3}m$
From eqn. $\left(iii\right)$,
$L=\sqrt{\frac{2m_e{v}_x^{2}y}{qE}}$
$=\sqrt{\frac{2\times 9.1\times 10^{-31}\times {\left(4\times 10^6\right)}^2\times 5\times 10^{-3}}{1.6\times 10^{-19}\times 8.2\times 10^2}}$
$L=3.3\times 10^{-2}m=3.3cm$
The electrons will strike the upper plate at its other end, if length of plate is $3.3cm$