Unit 6 - Problem 4 ==> Cathode Ray Tube

Given:

• $$ d = 0.5 \;\mbox{meters} $$
• $$  V_0 = 1 * 10^7 \;\mbox{meters/second}   $$
• $$ q = 1.602 * 10^{-19} \;\mbox{coulombs} $$
• $$ m_e = 9.11 * 10^{-31} \;\mbox{kg} $$

Question:

• $$\mbox{What electrical field is required to cause a displacement} = \Delta{y} \;\mbox{?} $$

Overview of Approach:

1. Determine horizontal time
2. Determine the vertical acceleration
3. Determine electrical field necessary for vertical displacement
4. Finish by substituting for the time to transit

Rationale:

Determine horizontal time

• $$ d = V_0 * t_x $$
• $$ t _x = \frac{d}{V_0} $$

Determine the vertical acceleration

• $$ F_{movement}  = F_e $$
• $$ F_{movement} = m * a $$
• $$ F_e = E * q $$
• $$ F_{movement} = F_e $$
• $$ E * q = m * a $$
• $$ a_y = \frac{(E * q)}{m} $$

Determine electrical field necessary for vertical displacement

• $$ t_x = t_y $$


• $$ \Delta{y} = \frac{1}{2}  * a_y * t_y^2$$
• $$ \frac{(2 * \Delta{y})}{t_x^2} = a_y = \frac{(E * q)}{m} $$
• $$ E = \frac{(m * 2 * \Delta{y})}{(q * t_x^2)}$$

Finish by substituting for the time to transit

• $$ t_x^2 = \frac{d^2}{V_0^2}  $$
• $$ E = \frac{(V_0^2 * m * 2 * \Delta{y})}{(q * d^2)} $$

Calculate

• $$ E = \frac{(10^{14} * 9.11 * 10^{-31} * 2 * 0.1)}{(1.602 * 10^{-19} * 0.25)}  $$
• $$ E = \frac{(10^{-17} * 1.822)}{(10^{-19} * 0.4005)}  $$
• $$ E = 10^2 * 4.54931 = 454.931 = 455 $$