How Do Electric Fields Affect Electron Motion Between Charged Plates?

[patrickmoloney]

Homework Statement

A uniform electric eld exists in a region between two oppositely charged parallel metal plates.
An electron is released from rest at the surface of the negatively charged plate and strikes the surface of the positively charged plate, 2.00 cm away, in a time 1.8 * 10^{-8} s

(i) What is the speed of the electron as it strikes the second plate?
(ii) What is the magnitude of the electricfield between the plates?

If the plates are circular with r = 15.0 cm find,

(i) The magnitude of the charge per unit area on the surface of either plate
(ii) the electrical force of attraction between two plates

Homework Equations

\Delta x = v_{avg} t = \frac{vt}{2}

\Delta x = \frac{1}{2} at^2

E = \frac{F}{q_e}

The Attempt at a Solution

B.

(i) \Delta x = v_{avg} t = \frac{vt}{2}

v = \frac{2\Delta x}{t} = \frac{2(2x10^{-2})}{1.8 *10^{-8}}\frac{m}{s} = 2.7*10^6 \frac{m}{s}

(ii) \Delta x = \frac{1}{2} at^2 and E = \frac{F}{q_e} = \frac {ma}{q_e}

E = \frac{ma}{q_e} = \frac{2 \Delta x m}{et^2} = \frac{2(2.0*10^{-2})(9.11*10^{-31})}{(1.6*10^{-19})(1.8*10^{-8})} = 1*10^3 N/C

C.

(i) \sigma = \frac{q}{2 \pi r^2} = \frac{1.6*10^{-19}}{2 \pi (15.0*10^{-2})^2} = 1.8*10^{-19} C/m^2

(ii) I'm sure the force equation is \vec{F} = k \frac{q_1q_2}{r^2} where,

k = \frac{1}{4 \pi \epsilon_0}

How do I find C. (ii) since I only have one charged particle $q_e$

 

[dauto]

The question is what's the force between the plates. First calculate the charge of the plates than calculate the force on that charge due to the field produced by the other plate (that's half of the total field between the plates).

 

[dauto]

Your question says time = 1.8∗10^-8s, you used time = 1.5∗10^-8s

 

[patrickmoloney]

Yeah sorry that was a typo. Could I say -

$E = \frac{\sigma}{\epsilon_0} = \frac{1.8*10^{-19}}{8.85*10^{-12}} = 2.03*10^{-8} NC^{-1}$

 

[dauto]

Yeah sorry that was a typo. Could I say -

$E = \frac{\sigma}{\epsilon_0} = \frac{1.8*10^{-19}}{8.85*10^{-12}} = 2.03*10^{-8} NC^{-1}$

No, that σ is wrong. Did you use the electric charge to find it? Why would you do that?