Question

An electron is released from rest at the negative plate of a parallel plate capacitor. The charge per unit area on each plate is = 2.1 × 10^-7 C/m^2, and the plates are separated by a distance of 1.2 × 10^-2 m. How fast is the electron moving just before it reaches the positive plate?

Answer 1

Explanation:

An electron is released from rest, u = 0

We know that charge per unit area is called the surface charge density i.e. $\sigma=\dfrac{q}{A}=2.1\times 10^{-7}\ C/m^2$

Distance between the plates, $d=1.2\times 10^{-2}\ m$

Let E is the electric field,

$E=\dfrac{\sigma}{\epsilon_o}$

$E=\dfrac{2.1\times 10^{-7}}{8.85\times 10^{-12}}$

E = 23728.81 N/C

Now, $ma=qE$

$a=\dfrac{qE}{m}$

$a=\dfrac{1.6\times 10^{-19}\times 23728.81}{9.1\times 10^{-31}}$

$a=4.17\times 10^{15}\ m/s^2$

Let v is the speed of the electron just before it reaches the positive plate. So, third equation of motion becomes :

$v^2=2ad$

$v^2=2\times 4.17\times 10^{15}\times 1.2\times 10^{-2}$

$v=10.003\times 10^6\ m/s$

Hence, this is the required solution.