Electron Deflection in Parallel Plate Capacitor

[TheLegace]

Homework Statement

A Problem of Practical interest is to make a beam of electrons turn a 90 degree corner. This can be done wit ha parallel plate capacitor shown. An electric with kinetic energy of 3.0x10^-17J moves up through a small holes in the bottom plate of capacitor.

What is field strength needed if electron is to emerge from an exit hole of 1.0cm away from entrance hole traveling at right angles to its original direction.

http://img4.imageshack.us/img4/6061/platecapacitor.th.jpg

Homework Equations

KE = 1/2mv^2
v1^ = v0^2 +2as

The Attempt at a Solution

I know this somewhat similar to parabola example.
There is constant electric field acceleration caused by plate capacitor.
Initially KE = 1/mv^2 or split with x and y. KE = 1/2mv0x^2 + 1/2mv0y^2 and the y and x-axis will be tilted to adjust to angle of plate capacitor, which I am assuming is 45 degrees.
I know I need to find time with the constant velocity but I am having a bit trouble getting that far. I am not sure if my statements about KE are even correct so if someone could help me through that process and once I derive initial velocities in y and x this problem should be a piece of cake.

Thank You Very Much.

 

[dslowik]

Yes, 1/2mv^2 = 1/2 m (vx^2 + vy^2)
And I agree it's same as a parabola in a constant force field like gravity.
I would orient x-axis along the capacitor plate, and the y-axis perpendicular to the plates, so then is like height in gravity. So then the x and y motions decouple since the force is only along y.
Try it from there using uniform motion along x,
and for y, by the symmetry of the problem, you have to reverse the motion in the alloted time, some kinematics...

 

[nlightner]

I would like to clarify that the problem statement is not entirely clear. It is not specified whether the electron is initially moving parallel or perpendicular to the plates of the capacitor. Assuming that the electron is initially moving parallel to the plates, I will provide a solution based on that assumption.

First, we can use the equation KE = 1/2mv^2 to find the initial velocity of the electron. Given that the kinetic energy is 3.0x10^-17J and the mass of an electron is approximately 9.11x10^-31 kg, we can calculate the initial velocity to be approximately 2.46x10^7 m/s.

Next, we can use the equation v1^2 = v0^2 +2as to find the final velocity of the electron after it has traveled a distance of 1 cm (0.01 m) perpendicular to its original direction. Since the initial velocity in the y-direction is 0, we can rewrite the equation as v1^2 = v0x^2 +2ax. Solving for v1, we get v1 = √(v0x^2 + 2ax). Plugging in the known values, we get v1 = √(2.46x10^7 m/s)^2 + (2*9.0x10^-17 m/s^2)*0.01 m) = 2.46x10^7 m/s.

Now, we can use the equation F = ma to find the force required to accelerate the electron in the y-direction. Since the mass of an electron is 9.11x10^-31 kg and the final velocity in the y-direction is 2.46x10^7 m/s, we get F = (9.11x10^-31 kg)*(2.46x10^7 m/s^2) = 2.24x10^-23 N.

Finally, we can use the equation E = F/q to find the electric field strength required to produce this force on the electron. Since the charge of an electron is 1.6x10^-19 C, we get E = (2.24x10^-23 N)/(1.6x10^-19 C) = 1.4x10^-4 N/C.

In conclusion, the field strength needed for the electron to emerge from the exit hole of the parallel plate capacitor