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

Question

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

protisvitfc 2021-11-16 Answered

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

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Expert Answer

Crom1970

Answered 2021-11-17 Author has 6 answers

Step 1
Concept
The Electric field due to capacitor having surface charge density

σ

is

E = \frac{σ}{ϵ_{0}}

The acceleration of electron is given by

\vec{a} = \frac{\vec{F}}{m}

\vec{a} = \frac{q \vec{E}}{m}

We have to find the speed of electron as it reaches the plate
Step 2
The acceleration

\vec{a} = \frac{q σ}{m ϵ_{0}}

\vec{a} = \frac{(1.6 \times 10^{- 19} C) (1.8 \times 10^{- 7} \frac{C}{m^{2}})}{(9.11 \times 10^{- 31} k g) (8.85 \times 10^{- 12} N \cdot \frac{m^{2}}{C^{2}})}

\vec{a} = 3.572 \times 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 (\vec{a}) s

v^{2} = 0 + 2 (3.572 \times 10^{15} \frac{m}{s^{2}}) (1.5 \times 10^{- 2} m)

v^{2} = 1.0716 \times 10^{14} \frac{m^{2}}{s^{2}}

v = \sqrt{1.0716 \times 10^{14} \frac{m^{2}}{s^{2}}}

\begin{matrix} v = 1.035 \times 10^{7} m / s \end{matrix}

Result

v = 1.035 \times 10^{7} \frac{m}{s}

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