# Suppose that the particle in Exercise in 1.33 is an electron projected

Question:

Suppose that the particle in Exercise in $1.33$ is an electron projected with velocity $v_{x}=2.0 \times 10^{6} \mathrm{~m} \mathrm{~s}^{-1} .$ If $E$ between the plates separated by $0.5$ $\mathrm{cm}$ is $9.1 \times 10^{2} \mathrm{~N} / \mathrm{C}$, where will the electron strike the upper plate? $\left(|e|=1.6 \times 10^{-19} \mathrm{C}, m_{e}=9.1 \times 10^{-31} \mathrm{~kg} .\right)$

Solution:

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

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

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

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

Mass of an electron, $m_{\mathrm{e}}=9.1 \times 10^{-31} \mathrm{~kg}$

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

Therefore,

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

$L=\sqrt{\frac{2 d m v_{x}^{2}}{q E}}$

$=\sqrt{\frac{2 \times 0.005 \times 9.1 \times 10^{-31} \times\left(2.0 \times 10^{6}\right)^{2}}{1.6 \times 10^{-10} \times 9.1 \times 10^{2}}}$

$=\sqrt{0.025 \times 10^{-2}}=\sqrt{2.5 \times 10^{-4}}$

$=1.6 \times 10^{-2} \mathrm{~m}$

$=1.6 \mathrm{~cm}$

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