Electric Fields Question

1. Apr 9, 2014

patrickmoloney

1. The problem statement, all variables and given/known data

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

2. Relevant equations

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

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

$E = \frac{F}{q_e}$

3. 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 $q_e$

Last edited: Apr 9, 2014
2. Apr 9, 2014

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).

3. Apr 9, 2014

dauto

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

4. Apr 9, 2014

patrickmoloney

Yeah sorry that was a typo. Could I say -

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

5. Apr 9, 2014

dauto

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