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

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

(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 ##

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