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

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