# An electron is released from rest at the negative plate of a parallel plate capa

An electron is released from rest at the negative plate of a parallel plate capacitor. The charge per unit area on each plate is
$\sigma =1.8×{10}^{-7}\mathrm{C}/{\mathrm{m}}^{2}$
and the plates are separated by a distance of
$1.5×{10}^{-2}\mathrm{m}.$
How fast is the electron moving just before it reaches the positive plate?

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Crom1970
Step 1
Concept
The Electric field due to capacitor having surface charge density $\sigma$ is
$E=\frac{\sigma }{{ϵ}_{0}}$
The acceleration of electron is given by
$\stackrel{\to }{a}=\frac{\stackrel{\to }{F}}{m}$
$\stackrel{\to }{a}=\frac{q\stackrel{\to }{E}}{m}$
We have to find the speed of electron as it reaches the plate
Step 2
The acceleration
$\stackrel{\to }{a}=\frac{q\sigma }{m{ϵ}_{0}}$
$\stackrel{\to }{a}=\frac{\left(1.6×{10}^{-19}C\right)\left(1.8×{10}^{-7}\frac{C}{{m}^{2}}\right)}{\left(9.11×{10}^{-31}kg\right)\left(8.85×{10}^{-12}N\cdot \frac{{m}^{2}}{{C}^{2}}\right)}$
$\stackrel{\to }{a}=3.572×{10}^{15}\frac{m}{{s}^{2}}$
Step 3
The velocity
The velocity can be found from the equation of kinematics
${v}^{2}={v}_{0}^{2}+2\left(\stackrel{\to }{a}\right)s$
${v}^{2}=0+2\left(3.572×{10}^{15}\frac{m}{{s}^{2}}\right)\left(1.5×{10}^{-2}m\right)$
${v}^{2}=1.0716×{10}^{14}\frac{{m}^{2}}{{s}^{2}}$
$v=\sqrt{1.0716×{10}^{14}\frac{{m}^{2}}{{s}^{2}}}$
$\begin{array}{|c|}\hline v=1.035×{10}^{7}m/s\\ \hline\end{array}$
Result
$v=1.035×{10}^{7}\frac{m}{s}$