Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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?

I got my questions answered at in under 10 minutes. Go to now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly


Get your free account and access expert answers to this and thousands of other questions

The deflection angle is required?
deflection angle is not mentioned in the book
then what is asked? what you mean find deflection??

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

|dw:1361792988295:dw| Y is the deflection that has to be calculated
the angle should be 37
|dw:1361855358005:dw|for angles 180=x+y+z
oh 45 is the other number it could be
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?
Cartesian coordinate system + newton law+Gauss's Law+Coulomb's Law=electric field

Not the answer you are looking for?

Search for more explanations.

Ask your own question