Question

1. an electron enters two parallel plates moving horizontally with a speed of 1.8 x 10⁶ m/s as shown in the system diagram. the magnitude of the electric field between the plates is 120 n/c and the plates are 13 cm across. calculate the velocity of the electron as it emerges from the plates to show that the velocity is 2.4 x 10⁶ m/s 40° above the horizontal

Explanation:

Step1: Find time in horizontal motion

The horizontal speed \( v_x = 1.8\times 10^6\ \text{m/s} \), horizontal distance \( d = 13\ \text{cm}=0.13\ \text{m} \).
Time \( t=\frac{d}{v_x}=\frac{0.13}{1.8\times 10^6}\approx7.22\times 10^{-8}\ \text{s} \).

Step2: Find vertical acceleration

Electron charge \( q = -1.6\times 10^{-19}\ \text{C} \), mass \( m = 9.1\times 10^{-31}\ \text{kg} \), electric field \( E = 120\ \text{N/C} \).
Force \( F = |q|E \), acceleration \( a=\frac{F}{m}=\frac{|q|E}{m}=\frac{1.6\times 10^{-19}\times120}{9.1\times 10^{-31}}\approx2.11\times 10^{13}\ \text{m/s}^2 \).

Step3: Find vertical velocity

Vertical initial velocity \( v_{y0} = 0 \), so \( v_y = at = 2.11\times 10^{13}\times7.22\times 10^{-8}\approx1.52\times 10^6\ \text{m/s} \).

Step4: Find magnitude of velocity

Using \( v=\sqrt{v_x^2 + v_y^2}=\sqrt{(1.8\times 10^6)^2+(1.52\times 10^6)^2} \)
\( v=\sqrt{3.24\times 10^{12}+2.31\times 10^{12}}=\sqrt{5.55\times 10^{12}}\approx2.36\times 10^6\approx2.4\times 10^6\ \text{m/s} \).

Step5: Find angle

\( \tan\theta=\frac{v_y}{v_x}=\frac{1.52\times 10^6}{1.8\times 10^6}\approx0.844 \), so \( \theta\approx\arctan(0.844)\approx40^\circ \).

Answer:

The velocity of the electron as it emerges is \( \boldsymbol{2.4\times 10^6\ \text{m/s}} \) at \( \boldsymbol{40^\circ} \) above the horizontal, which matches the given result.