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 σ = 1.69E-7 C/m2, and the plates are separated by a distance of 1.75E-2 m. How fast is the electron moving just before it reaches the positive plate?

Answer 1

Answer:

$v=1.08\times 10^7\ m/s$

Explanation:

Initial speed of the electron, u = 0

The charge per unit area of each plate, $\dfrac{Q}{A}=1.69\times 10^{-7}\ C/m^2$

Separation between the plates, $d=1.75\times 10^{-2}\ m$

An electron is released from rest, u = 0

Using equation of kinematics,

$v^2-u^2=2ad$..........(1)

Acceleration of the electron in electric field, $a=\dfrac{qE}{m}$............(2)

Electric field, $E=\dfrac{\sigma}{\epsilon_o}$............(3)

From equation (1), (2) and (3) :

$v=\sqrt{\dfrac{2q\sigma d}{m\epsilon_o}}$

$v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\times 1.69\times 10^{-7}\times 1.75\times 10^{-2}}{9.1\times 10^{-31}\times 8.85\times 10^{-12}}}$

v = 10840393.1799 m/s

or

$v=1.08\times 10^7\ m/s$

So, the electron is moving with a speed of $1.08\times 10^7\ m/s$ before it reaches the positive plate. Hence, this is the required solution.