Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
SI123
Group Title
The electric field between the 2 parallel plates of an oscilloscope is 1.2 x 10^5 V/m. If an electron of energy 2 keV enters at right angles to the field, what will be its deflection if the plates are 1.5 cm long?
 one year ago
 one year ago
SI123 Group Title
The electric field between the 2 parallel plates of an oscilloscope is 1.2 x 10^5 V/m. If an electron of energy 2 keV enters at right angles to the field, what will be its deflection if the plates are 1.5 cm long?
 one year ago
 one year ago

This Question is Closed

Mashy Group TitleBest ResponseYou've already chosen the best response.0
The deflection angle is required?
 one year ago

SI123 Group TitleBest ResponseYou've already chosen the best response.0
deflection angle is not mentioned in the book
 one year ago

Mashy Group TitleBest ResponseYou've already chosen the best response.0
then what is asked? what you mean find deflection??
 one year ago

SI123 Group TitleBest ResponseYou've already chosen the best response.0
dw:1361792988295:dw Y is the deflection that has to be calculated
 one year ago

DLBlast Group TitleBest ResponseYou've already chosen the best response.0
the angle should be 37
 one year ago

DLBlast Group TitleBest ResponseYou've already chosen the best response.0
dw:1361855358005:dwfor angles 180=x+y+z
 one year ago

DLBlast Group TitleBest ResponseYou've already chosen the best response.0
oh 45 is the other number it could be
 one year ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.1
in this case, since it is constant, uniform acceleration due to the electric field, by Newton's second law, \(\Sigma \vec F= q \vec E = m \vec a\) where q and m of the electron are known constants. so, \(\vec a=\frac{q \vec E}{m}\), which will be different in direction of the electric field, as expected, now, the original speed of the electron can be calculated from: \(\frac{1}{2}mu^2= 2KeV\), so \(u=\sqrt{ \frac{4KeV}{m}}\) by the equations of motion, \(\vec S_x=\vec u_x t + \frac{1}{2} \vec a_x t^2\), but \(\vec u_x=\vec u\) and \(\vec a_x =0\) and \(\vec S_x =1.5cm\) so, \(t=\frac{1.5}{\sqrt{ \frac{4KeV}{m}}}\) now, for \(\vec S_y=\vec u_y t + \frac{1}{2} \vec a_y t^2\), but this time,\(\vec u_y=0\), \(\vec a_y = \vec a = \frac{q \vec E}{m}\) so, \(\vec S_y= \frac{1}{2} \vec a_y t^2\) sub it all in to get the displacement along the y axis. Do you need further help?
 one year ago

DLBlast Group TitleBest ResponseYou've already chosen the best response.0
Cartesian coordinate system + newton law+Gauss's Law+Coulomb's Law=electric field
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.