Answered

Discover a wealth of knowledge at Westonci.ca, where experts provide answers to your most pressing questions. Join our platform to get reliable answers to your questions from a knowledgeable community of experts. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.

Consider the parallel-plate capacitor shown in the figure. The plate separation is 1.8 mm and the the electric field inside is 23 N/C. An electron is positioned halfway between the plates and is given some initial velocity, vi.

(a) What speed, in meters per second, must the electron have in order to make it to the negatively charged plate?

(b) If the electron has half the speed needed to reach the negative plate, it will turn around and go towards the positive plate. What will its speed be, in meters per second, when it reaches the positive plate in this case?

Sagot :

okpalawalter8

We have that the Velocities is mathematically given as

a)v=2.8e5m/s

b)v_2=3.11e5m/s

From the question we are told

  • Consider the parallel-plate capacitor shown in the figure.
  • The plate separation is 1.8 mm and the the electric field inside is 23 N/C.
  • An electron is positioned halfway between the plates and is given some initial velocity, vi.

Velocity

Generally the equation for the workdone   is mathematically given as

a)

[tex]-qEr=0.5m(u/2)^2\\\\Therefore\\\\1.6e-19*19*2.9e-3=0.5*9.1e-31*v^2/4[/tex]

v=2.8e5m/s

b)

applying work

[tex]-qEr=0.5m(u/2)^2\\\\1.6e-19*19*2.9e-3=0.5*9.1e-31(v_2^2-7.75e10)[/tex]

v_2=3.11e5m/s

For more information on Velocity visit

https://brainly.com/question/7359669

We appreciate your time. Please revisit us for more reliable answers to any questions you may have. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.