Finding an kinetic energy of an electron moving to a positive plate

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Homework Help Overview

The problem involves calculating the kinetic energy of an electron moving between two parallel plates held at different electric potentials. The first part of the problem has been solved, which involved finding the electric field between the plates. The second part requires determining the kinetic energy of the electron when it reaches the positive plate.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the application of conservation of energy principles and the relationship between potential difference and kinetic energy. There are attempts to calculate the kinetic energy using different methods, including direct multiplication of voltage and charge, as well as considering the distance the electron falls.

Discussion Status

There is ongoing exploration of the calculations involved, with participants questioning the accuracy of their results and discussing the significance of units and significant figures. Some participants suggest alternative approaches and clarify the relationship between different units of energy.

Contextual Notes

Participants note the importance of significant figures based on the precision of the given voltages. There is also a mention of potential discrepancies in the answers provided by practice problems, indicating a need for careful consideration of the problem setup.

Patdon10
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Homework Statement


2 part problem. I've already solved the first part.

Two parallel plates are held at potentials of +380.0 V and -150.0 V. The plates are separated by 3.00 cm.
a. Find the electric field between the plates.
b. An electron is initially placed halfway between the plates. Find its kinetic energy when it hits the positive plate.

Homework Equations


E = deltaV/d
W = delta U = qEd

The Attempt at a Solution



For the first one I plugged into delta V, which was simply 380 - -150 = 530 V.
then took 530V/(0.03m) = 17,666 V/m
the 2nd part isn't working.
q(E)d = (1.6e-19 C)(17,666 V/m)(0.015 m)
= 4.2398 e-17 CV (needs to be in units of eV?)

This answer is wrong. Anyone know what I'm doing it wrong?

by the way. I think it is obvious I'm using the conservation of energy. All potential energy at the start, converts to all kinetic energy at the end.
 
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You might just need to do ΔW = ΔV*q, since the electric field is uniform.
 
I tried that way too. Didn't work. deltV = 530 and q= 1.60e-19. Multiply the two and it's equal to 8.49e10-17. Still saying it's wrong.
 
Patdon10 said:
8.49e10-17
Is that how you wrote the answer? You really mean 8.49e-17 here. Also, do they require you to include units in your answer? And how many significant figures should your answer have?

You basically have it right.
 
Sorry. I wrote it exactly as 8.49e-17 eV. Should have 3 sig figs. The practice problems which are the same problem but slightly different numbers have answers like 205 eV when the numbers change to "Two parallel plates are held at potentials of +230.0 V and -180.0 V. The plates are separated by 8.00 cm. "
 
Patdon10 said:
Sorry. I wrote it exactly as 8.49e-17 eV.
Okay.

Should have 3 sig figs.
But the voltages are given to an accuracy of 0.1V. I.e., how many sig figs are in "380.0 V"?

The practice problems which are the same problem but slightly different numbers have answers like 205 eV ...
Oh. In that case, you might also try answering in units of eV.
 
8.49e-17 is in units of CV? How do I convert that to eV?
 
CV units are equivalent to Joules (J). Can you convert between Joules and eV?
 
Patdon10 said:
8.49e-17 is in units of CV? How do I convert that to eV?

Firstly when writing the units of energy (any kind) it's best to write as Joule and not CV. It is correct yes but less confusing that just J.

Second thing, 1 eV = 1.6(10-19) C * 1 V = 1.6(10-19) J

so just divide your answer by 1.6(10-19) and it will be in eV.
 
  • #10
ahhh. I had no idea CV = J. Yeah, I knew I could simply divide J/1.6e-19 to get eV. That would make the answer 530.62 eV. This makes sense, but the answer was also marked as wrong : /

I'm not really sure what I'm doing wrong. Anymore ideas?
 
  • #11
530.62 ... that's remarkably close to the number you got for question (a).

Think about it: you had 530 (really 530.0) for the voltage between the plates. You multiplied that by 1.6e-19, and then later you divided it by the same number, 1.6e-19.

Again, you mainly have it right. The problem is with sig figs, or units, or it's always possible the website has it wrong too.
 
  • #12
Thanks. I will email my teacher and report back as to what's going on.
 
  • #13
The electron is placed halfway between the plates. So the ΔV that it falls through is half of 530V. An electron falling through 1V is attributed 1eV. So an electron falling through 530/2 Volts must acquire..._____ eV ?
 
  • #14
Is it really that easy? I hope that wasn't my only mistake. I've since emailed my teacher and asked for another submission. I'll see if that's right. I imagine it would be.
 
  • #15
gneill said:
The electron is placed halfway between the plates.
Ah, can't believe I missed that. :redface:
 

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