Help: Due Tonight Electric Field Question

In summary, the electron is deflected a distance of 1.5 meters when it leaves the plates. The final velocity of the electron is 9.8 kilometers per second.
  • #1
rbraunberger
10
0

Homework Statement


An electron is traveling through a region between two metal plates in which there is a constant electric field of magnitude E directed along the y direction as sketched in the figure below. This region has a total length of L, and the electron has an initial velocity of v0 along the x direction. (Use the following as necessary: e, E, L, m, and v0.)
17-p-044.gif


1. By what distance is the electron deflected when it leaves the plates?

2. What is the final velocity (x-direction) of the electron?

Homework Equations


F=qE
Motion equations

The Attempt at a Solution



The questions I have answered.

t= L/v0

F=eE

a=(eE)/m

(F=Electric Force, e=electron charge, E=electric field, m=mass of electron, L=length, t=time)

For the final velocity Vx, I tried to work with the motion equations. Vx=V0+at Vx being final velocity in the x direction, V0 being initial velocity, a is acceleration, and t is time. I plugged in the equations I worked out in the problem for t and a getting Vx=V0 + (eE/m)(L/V0), but the systems says this is an incorrect answer.

For the Distance deflected I also used a motion equation (delta)y=V0 t + 1/2a(y direction) t^2 I worked this down to deltaY= 1/2 (eE/m)(L/v0)^2 and am unsure where to go now.Thanks for any help!
 
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  • #2
welcome to pf!

hi rbraunberger! welcome to pf! :smile:

i'm not sure what your question is, but the important thing is to find the aceleration …

once you have that, you can treat it much as you would projectile motion under gravity :wink:
 
  • #3
Thanks, I found the acceleration and was trying to do that...look at my work now...I edited my post since you replied. I have this online homework system and it is not taking my answers...not sure if my logic is just way off.
 
  • #4
Here is a list of all the questions...I am only having trouble with d and first part of e

(a) How long does it take the electron to travel the length of the plates?
t =

(b) What are the magnitude and direction of the electric force on the electron while it is between the plates?
magnitude F =
direction

(c) What is the acceleration of the electron?
magnitude a =
direction

(d) By what distance is the electron deflected when it leaves the plates?
Δy =

(e) What is the final velocity of the electron?
vx =
vy =
 
  • #5
hi rbraunberger! :smile:

(have a delta: ∆ and try using the X2 and X2 icons just above the Reply box :wink:)
rbraunberger said:
For the final velocity Vx, I tried to work with the motion equations. Vx=V0+at Vx being final velocity in the x direction, V0 being initial velocity, a is acceleration, and t is time. I plugged in the equations I worked out in the problem for t and a getting Vx=V0 + (eE/m)(L/V0), but the systems says this is an incorrect answer.

For the Distance deflected I also used a motion equation (delta)y=V0 t + 1/2a(y direction) t^2 I worked this down to deltaY= 1/2 (eE/m)(L/v0)^2 and am unsure where to go now.

hmm … you're throwing v0 and a around like confetti :redface:

v0 is only in the x direction

a is only in the y direction

start again! :smile:
 
  • #6
I thought the motion equations were (the ones we are talking about)

Vx = V0 + at
Y=Y0 + V0t + 1/2 at^2
 
  • #7
ok...I removed the a from the X and Vo from the Y and got the answers correct...are the motion equations different in this way with electric fields? My book sucks!
 
  • #8
hi rbraunberger! :smile:

(just got up :zzz: …)
rbraunberger said:
ok...I removed the a from the X and Vo from the Y and got the answers correct...are the motion equations different in this way with electric fields? My book sucks!

your book does not suck! :rolleyes:

this is exactly the same as with projectiles in a gravitational field, isn't it?

you have to remember to deal with the x and y components separately, and that means using only the x or y component of each vector (velocity or acceleration or force) as the case may be …

in the wrong direction, that component is obviously zero :smile:

get some sleep! :zzz:​
 

Related to Help: Due Tonight Electric Field Question

1. What is an electric field?

An electric field is a physical field that surrounds a charged particle and exerts a force on other charged particles within its vicinity. It is represented by a vector quantity, with both magnitude and direction.

2. How is electric field strength calculated?

The electric field strength is calculated by dividing the force exerted on a test charge by the magnitude of the test charge itself. It is also affected by the distance between the charged particles, as the electric field strength decreases with increasing distance.

3. What is the unit of measurement for electric field?

The unit of measurement for electric field is Newtons per Coulomb (N/C) in the SI (International System of Units) system. In the CGS (centimeter-gram-second) system, the unit is dynes per electrostatic unit (dyn/esu).

4. How does an electric field affect charged particles?

An electric field exerts a force on charged particles, causing them to either attract or repel each other depending on their charges. The direction of the force is determined by the direction of the electric field.

5. Can electric fields be shielded?

Yes, electric fields can be shielded by placing a conductive material between the source of the electric field and the area to be shielded. The conductive material will absorb the electric field and prevent it from reaching the desired area.

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