Respuesta :
Explanation:
Formula to calculate electric field because of the plate is as follows.
E = [tex]\frac{\sigma}{2 \times \epsilon_{o}}[/tex]
= [tex]\frac{2.10 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}}[/tex]
= [tex]1.18 \times 10^{5} N/C[/tex]
Now, we will consider that equilibrium of forces are present there. So,
ma = qE
a = [tex]\frac{1.6 \times 10^{-19} \times 1.18 \times 10^{5}}{1.67 \times 10^{-27}}[/tex]
= [tex]1.13 \times 10^{13} m/s^2[/tex]
According to the third equation of motion,
[tex]v^{2} = 2 \times a \times d[/tex]
or, d = [tex]\frac{v^{2}}{2d}[/tex]
= [tex]\frac{(2.4 \times 10^{6})^{2}}{2 \times 1.13 \times 10^{13}}[/tex]
= 0.254 m
Thus, we can conclude that the proton will travel 0.254 m before reaching its turning point.
The turning point of the particle is the maximum distance to which the
particle travels before being reversed by the electric field.
Response:
- The distance traveled by the particle before reaching its turning point is approximately 0.253 m
What is the electric field strength?
The electric field strength indicates the strength of the electric field in a region.
Given;
Charge density, σ = -2.10 × 10⁻⁶ C/m²
Speed of the proton, v = 2.40 × 10⁶ m/s
Required:
The distance the proton travels before reaching its turning point
Solution:
The electric field strength is given by the following formula;
[tex]E = \mathbf{ \dfrac{\sigma}{2 \cdot \epsilon_0}}[/tex]
Which gives;
ε₀ = The permittivity of free space = 8.85 × 10⁻¹² Farad
Which gives;
- [tex]E = \dfrac{-2.10 \times 10^{-6} \, C/m^2}{2 \times 8.85 \times 10^{-12} \, F} \approx \mathbf{118,644.07 \, N/C}[/tex]
According to Newton's third law of motion, we have;
Force due to the acceleration of the particle = Force due to the electric field
Which gives;
m·a = q·E
Which gives;
[tex]a = \mathbf{ \dfrac{q \cdot E}{m}}[/tex]
Where;
m = Mass of a proton = 1.67262 × 10⁻²⁷ kg
a = Acceleration of the proton
q = The charge of the proton = +1 e ≈ 1.602 × 10⁻¹⁹ C
Therefore;
- [tex]a = \dfrac{1.602 \times 10^{-19} \, C\times 118,644.07 \, N/C}{1.67262 \times 10^{-27} \. Kg } \approx \mathbf{ 1.14 \times 10^{13} \, m/s^2}[/tex]
From the kinematic equation of motion, we have;
v² = u² + 2·a·s
Given that the initial velocity, u, is 0, we have;
v² = 2·a·s
[tex]s = \mathbf{\dfrac{v^2}{2 \cdot a}}[/tex]
Which gives;
[tex]s = \dfrac{\left(2.40 \times 10^6 \right)^2}{2 \times 1.14 \times 10^{13}} \approx \mathbf{0.253}[/tex]
- The distance the particle travels before coming to rest is approximately 0.253 m.
Learn more about electric field strength here:
https://brainly.com/question/9027809
