Question

The electric field strength is $20,000\mathrm{N}/\mathrm{C}$inside a parallel-plate capacitor with a $1.0mm$ spacing. An electron is released from rest at the negative plate. What is the electron’s speed when it reaches the positive plate?

Short answer

Electron speed is$2.7\cdot {10}^6\mathrm{m}/\mathrm{s}$when it reaches the positive speed.

Step-by-step solution

Step :1 Introduction 

According to the law of conservation of momentum, all energy generated from deformation in an electrostatic potential is transitioned to angular momentum. This enables us to compose.

$q\mathrm{\Delta }V=\frac{m{v}^2}{2}$

where can we discover out what the speed will be

$v=\sqrt{\frac{2q\mathrm{\Delta }V}{m}}$

Step :2  Theoretical difference 

Everything we really need to do is determine to see what the alternative potential is. Realizing that the electromagnetic field within a resistor has a magnitude of

$E=\frac{\mathrm{\Delta }V}{d}$

The theoretical discrepancy can be identified as follows:

$\mathrm{\Delta }V=Ed$

Step :3  Substitution 

By swapping this, we can reach at our finalized performance articulation:

$v=\sqrt{\frac{2qEd}{m}}$

Consequently, when we apply proportionally, we get

$$\begin{gathered} v=\sqrt{\frac{2\cdot 1.6\cdot {10}^{-19\cdot 2\cdot {10}^4\cdot 1\cdot {10}^{-3}}}{9.1\cdot {10}^{-31}}} \\ =2.7\cdot {10}^6\mathrm{m}/\mathrm{s} \end{gathered}$$