Determining the electron speed in parallel plates

  • #1

Homework Statement



Two parallel plates labelled W (negative) and X (positive) are separated by 5.2 cm. The electric potential between the plates is 150 V. An electron starts from rest at time tw and reaches plate X at time tx. The electron continues through an opening in plate X and reaches point P at time tP. Determine the speed at tx and tP

e = -1.6 x 10-19 C
me = 9.1 x 10-31

Homework Equations



V = kq1 / r
ΔV = ΔE/q
ε = ΔV / r
Ek = 0.5mv2

The Attempt at a Solution



I'm not sure how to determine v at tw and tx, but I did find the field strength: ε = 2884.6. We also know the electron starts from rest, so we could use the conservation of energy law to generate the following eq'n.

v2 = 2m-1kq1q2(ra-2 - rb2). But we don't have a "q2" so I'm quite lost. Any help is appreciated!

God bless and thanks!
 

Answers and Replies

  • #2
gneill
Mentor
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The Volt is a name given to the unit combination Joules/Coulomb :wink:

You should be able to find (in your notes or text) an expression for the work done on a charge falling through a potential difference. The work done shows up as kinetic energy...
 
  • #3
You can start by finding the acceleration of the electron.
F = ma = qE
a = qE/m
Then it's a kinematics problem
vf^2 = 2*a*x Thats good for finding Vx
 
  • #4
I used your equation (not on my eq'n sheet), a = qE/m, substituted Vq for E, leaving me with Vq˄2 / m. I found acc. to be 3.376 x 10^-6.

Plugged that into eq'n v22 = v12 + 2ad,

which reduces to v2 = sqr(2ad), and I got a velocity (vx) of 5.93 x 10^-4 m/s which seems really slow to me.
 
  • #5
gneill
Mentor
20,906
2,857
I used your equation (not on my eq'n sheet), a = qE/m, substituted Vq for E, leaving me with Vq˄2 / m. I found acc. to be 3.376 x 10^-6.
Why would you substitute Vq for E? E here would be the electric field strength (units of [N]/[C] or [V]/[m]). You calculated the electric field strength previously, calling it ε.

However, a more slick approach is to use q*ΔV, with q being the charge on the electron and ΔV the change in electric potential (the 150V). The units are then [C][J]/[C] → [J], which is energy. That's the work done on the electron as it falls through the field...
 
  • #6
Ok that makes more sense. When you wrote E, I figured you meant energy, not ε. Tak.
 

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