Unit 6 - Problem 2 ==> Parallel Plates

Given:

• $$ E_{field} = 1000 $$
• $$ d = 1 \; \mbox{cm} $$
• $$ m_{proton} = 1.673 * 10^{-27} $$
• $$ q = 1.6024 * 10^{-19} $$ 
• $$ \mbox{Initial velocity of charge}\; = 0 $$

Question:

• $$\mbox{How long, in nanoseconds, does it take the proton to cross between the two plates?} $$

Overview of Approach:

1. Convert centimeters to meters for final calculations
2. Determine force on charge due to electrical field
3. Determine acceleration on charge due to force on charge
4. Determine time to transit using the acceleration

Rationale:

Convert centimeters to meters

• $$ 1 \; \mbox{cm} = 1 * 10^{-2} \; \mbox{meters} $$

Determine force on charge due to electrical field

• $$ E = \frac{F_e}{q} $$
• $$ F_e = E * q $$

Determine acceleration on charge due to force on charge

• $$ F = m * a $$
• $$ a = \frac{F}{m} $$
• $$ a_p = \frac{F_e}{m} = \frac{E * q}{m} $$

Determine time to transit using the acceleration

• $$ \Delta{x} = (V_0 * t)  + (\frac{1}{2} * a * t^2) $$
• $$ t^2 = \frac{2\Delta{x}}{a} $$
• $$ t^2 = \frac{(2 * d)}{ \frac{E * q}{m}}$$
• $$ t^2 = \frac{(2 * d * m)}{(E * q)} $$
• $$ t = \sqrt \frac{(2 * d * m)}{(E * q)} $$

Calculate

• $$ t = \sqrt \frac{(2 * (10^{-2}) * ( 1.673 * 10^{-27}))}{(1000 * (1.6024 * 10^{-19} ))} $$
• $$ t = \sqrt\frac{(3.346 * 10^{-29})}{(1.6024 * 10^{-16})}$$
• $$ t = \sqrt{(2.088 * 10{-13})} $$
• $$ t = .4569 * 10^{-6} =  456.9 * 10^{-9}$$