Respuesta :
Answer:
it take for the proton to reach the negative plate is 0.1089 ns
Explanation:
given data
area = 21 cm² = 21 × [tex]10^{-4}[/tex] m²
separated d = 3 mm = 3 × [tex]10^{-3}[/tex] m
charge Q = 9.8 nC = 9.8 × [tex]10^{-9}[/tex] C
to find out
How long does it take for the proton to reach the negative plate
solution
we know that electric field between the plate is express as
electric field E = [tex]\frac{Q}{A*\epsilon}[/tex] ..................1
here Q is charge and A is area and ∈ is permittivity of free space
put here value
electric field E = [tex]\frac{9.8 ×10^{-9}}{21×10^{-4}*8.85×10^{-12}}[/tex]
electric filed E = 527306.968 N/C
and
we know force on proton by electric field is
Force = electric filed × charge on proton
and force according to newton law
force = mass × acceleration
so
mass × acceleration = electric filed × charge on proton
here put all value and find acceleration
1.67 × [tex]10^{-27}[/tex] × acceleration = 527306.968 × 1.6 × [tex]10^{-19}[/tex]
acceleration = 5.05 × [tex]10^{13}[/tex] m/s²
and
final velocity of proton by equation of motion
v² - u² = 2as ..........2
here u is zero and v is final velocity and s is distance and a is acceleration
v² - 0 = 2(5.05 × [tex]10^{13}[/tex]) (3 × [tex]10^{-3}[/tex] )
velocity = 5.50 × [tex]10^{5}[/tex] m/s
so time is
time = [tex]\frac{velocity}{acceleration}[/tex]
time = [tex]\frac{5.50*10^{5}}{5.05*10^{13}}[/tex]
time = 0.1089 × [tex]10^{-9}[/tex] s
so it take for the proton to reach the negative plate is 0.1089 ns
In a parallel plate, the capacitor plate is parallel while the plates are separated by distance. The time required for the proton to reach the negative plate will be 0.1089×10⁻⁹ second.
What is a parallel plate capacitor?
It is a type of capacitor is in which two metal plates arranged in such a way so that they are connected in parallel and have some distance between them.
A dielectric medium is a must in between these plates helps to stop the flow of electric current through it due to its non-conductive nature.
The following date given in the problem,
A id the area of plates = 21 cm² = 21 × 10m²
d is the distance separated= 3 mm = 3 ×10⁻³m
q is the charge on capicitor = 9.8 nC = 9.8 ×10⁻⁹C
The formula of the electric field is as
[tex]\rm E = \frac{q}{A\varepsilon } \\\\ \rm E = \frac{9.8\times10^{-9}}{ 21\times10^{-4}\times8.85\times10^{-12}} \\\\\rm E = 527306\; N/C}[/tex]
The electric force is equal to the product of the charge and the electric field. Here the electric force is balanced by mechanical force.
[tex]\rm F= qE\\\\\rm{ F= ma}\\\\qE =ma\\\\ \rm{ a= \frac{qE}{m} }\\\\\rm{ a= \frac{1.6\times10^{-9}\times527306}{1.2\times10^{-10}} }\\\\ \rm a = 5.05\times10^{-31}\;m/sec^2[/tex]
From newton's third equation of motion
As the initial velocity is zero u=0
[tex]\rm{v^2= u^2-2as} \\\\ \rm{v^2= 2\times5.05\times10^-3\times10^{-3}} \\\\ \rm{v= \sqrt{2\times5.05\times10^-3\times10^{-3}}\\\\[/tex]
[tex]\rm v = 5.50\times10^5 \;m/sec.[/tex]
Time is defined as the ratio of velocity and acceleration.
[tex]\rm t= \frac{v}{a} \\\\ \rm t= \frac{5.50\times10^5}{5.05\times10^{-3}} \\\\ \rm t= 0.1089\times 10^{-9}}\;sec[/tex]
Hence the time required for the proton to reach the negative plate will be 0.1089×10⁻⁹ second.
To learn more about the parallel plate capacitor refer to the link;
https://brainly.com/question/12883102
