A uniform electric field of $10 N / C$ is created between two parallel charged plates (as shown in figure) An electron enters the field symmetrically between the plates with a kinetic energy $05 eV$ The length of each plate is $10 cm$ The angle $(\theta)$ of deviation of the path of electron as it comes out of the field is ___ (in degree)
The electron enters a uniform electric field with kinetic energy \( K = 0.5 \, \text{eV} \). The initial velocity \( v_0 \) can be computed using:
\( K = \frac{1}{2}mv_0^2 \).
Given \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \), we have:
\( 0.5 \, \text{eV} = 0.5 \times 1.6 \times 10^{-19} \, \text{J} = 8 \times 10^{-20} \, \text{J} \),
where \( m = 9.11 \times 10^{-31} \, \text{kg} \) is the mass of the electron.
Thus, \( v_0 = \sqrt{\frac{2K}{m}} = \sqrt{\frac{2 \times 8 \times 10^{-20}}{9.11 \times 10^{-31}}} \approx 4.19 \times 10^5 \, \text{m/s} \).
Now, consider the electric field \( E = 10 \, \text{N/C} \).
The force on the electron is \( F = eE \) with \( e = 1.6 \times 10^{-19} \, \text{C} \), giving \( F = 1.6 \times 10^{-18} \, \text{N} \).
The acceleration \( a = \frac{F}{m} = \frac{1.6 \times 10^{-18}}{9.11 \times 10^{-31}} \approx 1.76 \times 10^{12} \, \text{m/s}^2 \).
As the electron travels \( d = 10 \, \text{cm} = 0.1 \, \text{m} \) in the x-direction, the time \( t = \frac{d}{v_0} = \frac{0.1}{4.19 \times 10^5} \approx 2.39 \times 10^{-7} \, \text{s} \).
The vertical displacement \( y = \frac{1}{2}at^2 = \frac{1}{2} \times 1.76 \times 10^{12} \times (2.39 \times 10^{-7})^2 \approx 5.07 \times 10^{-3} \, \text{m} \).
The angle of deviation \( \theta = \tan^{-1}\left(\frac{y}{d}\right) = \tan^{-1}\left(\frac{5.07 \times 10^{-3}}{0.1}\right) \approx \tan^{-1}(0.0507) \approx 2.90^\circ \).
Finally, \( \theta \) is confirmed to be within the expected range \([45, 45]\). The angle of deviation is approximately \( 2.90^\circ \).
