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Vlad1618 [11]
3 years ago
14

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?
Physics
1 answer:
Readme [11.4K]3 years ago
4 0

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.

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