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 $1 \,cm$ 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. $2.14 \,cm$
• B. $3.9 \,cm$
• C. $1.23 \,cm$
• D. $3.3 \,cm$

Answer 1

Answer: (D)

Given $v_x = 4 \times 10^6\, m\, s^{-1}$, $d = 1\,c\, m = 1 \times 10^{-2} \,m$
$E=8.2\times10^{2}\,N\,C^{-1}$, $q=e=1.6\times10^{-19}\,C$, $m_{e}=9.1\times10^{-31}\,kg$
The electron will strike the upper plate at its other end of $x = L$ as soon as its deflection.
And $=\frac{d}{2}=\frac{10^{-2}}{2}m=5\times10^{-3}\,m$
From eqn. $\left(iii\right)$,
$L=\sqrt{\frac{2m_{e}v^{2}_{x}y}{qE}}$
$=\sqrt{\frac{2\times9.1\times10^{-31}\times\left(4\times10^{6}\right)^{2}\times5\times10^{-3}}{1.6\times10^{-19}\times8.2\times10^{2}}}$
$L=3.3\times10^{-2}\,m=3.3\,cm$
The electrons will strike the upper plate at its other end, if length of plate is $3.3 \,cm$