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Motion between two charged surfaces, find initial speed

  • #1
rlc
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1

Homework Statement


Two charged, parallel, flat conducting surfaces are spaced d = 1.3 cm apart and produce a potential difference ΔV = 625 V between them. An electron is projected from one surface directly toward the second. What is the initial speed of the electron if its comes to rest just at the second surface?

Homework Equations


Velocity=sqrt(2qV/m)
U=sqrt(2V(e/m))

The Attempt at a Solution


Are those equations the right ones to use for this problem? Usually, I try to find my physics homework on yahoo answers or even on this website, but I haven't had much luck in finding it or relevant equations. How do I start on this problem?
 

Answers and Replies

  • #2
DEvens
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Usually, I try to find my physics homework on yahoo answers or even on this website, but I haven't had much luck in finding it or relevant equations. How do I start on this problem?
I'm only guessing, but what I'm guessing is, people are responding poorly to this because it looks like you are trying to get the 'net to do your homework for you. I know I had to "bite my tongue" a couple times to avoid saying nasty things.

"I looked on yahoo" isn't really an attempt to answer the question. Can you at least make some attempt to solve it? You have a charged particle moving in an electric field. What will happen to it?
 
  • #3
rlc
128
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Ok.
Charged particles move in circles at a constant speed if projected into a magnetic field at right angles to the field.
Charged particles move in straight lines at a constant speed if projected into a magnetic field along the direction of the field.

From all my attempts, the equations have all been pretty similar. It's just that whenever I try to make sense of how they used the equations with their problems (different problems, but same concepts), I get lost. The equations I'm referring to are the ones I posted above, but most especially velocity=sqrt((2q)(volt)/m))

Is this the wrong equation and I just need to drop it, or am I following the correct lead and should push on?
 
  • #4
rude man
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Magnetic field? Didn't someone say a couple of hundred years ago that to get a mag field you need a CHANGING E field?
 
  • #5
rlc
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I'm guessing that this isn't a magnetic field, then :)
 
  • #6
rude man
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I'm guessing that this isn't a magnetic field, then :)
Good guess.
So, what kind do you think it is?
 
  • #7
rlc
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Would it be a constant electric field?
 
  • #8
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I'm just going to take a stab at this question. I'm assuming that the electric field is slowing down the electron since the velocity is zero when it reaches the second conducting surfaces. Since you know the initial potential energy (voltage) and initial speed, and you need the final speed, you can use the algebraic equation relating final and initial kinetic energy and potential energy. Not sure if this helps.
 
  • #9
rude man
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Would it be a constant electric field?
Sure would!
 
  • #10
rlc
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acceleration=(change in velocity)/time=force/mass=absolutevalue(qE)/m
The final velocity of the electron is zero. I'm to solve for the initial velocity.
I've been given that the charged surfaces are 1.3 cm apart, and that the potential difference between them is 625 V.
Potential difference=Work/Charge=(change in potential energy)/Charge

Can someone please give me a hint as what to do next? I haven't found an example in my textbook that is similar to this problem, other than that first equation about acceleration. But is that even applicable to this problem?
 
  • #11
rude man
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acceleration=(change in velocity)/time=force/mass=absolutevalue(qE)/m
The final velocity of the electron is zero. I'm to solve for the initial velocity.
I've been given that the charged surfaces are 1.3 cm apart, and that the potential difference between them is 625 V.
Potential difference=Work/Charge=(change in potential energy)/Charge

Can someone please give me a hint as what to do next? I haven't found an example in my textbook that is similar to this problem, other than that first equation about acceleration. But is that even applicable to this problem?
How about energy conservation? You can call p.e. = 0 at any one point, so I suggest making p.e. at the first surface = 0. Then, at the first surface you have k.e. only and at thesecond surface you have p.e. only.
"p.e." = "potential energy"
"k.e." = "kinetic energy"
 
  • #12
rlc
128
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(1/2)mV0^2=q(deltaV)
So: VO=SQRT( 2*q*deltaV / m )
where q=1.60E-19 C and m=9.11E-31 kg

V0=SQRT[ [(2)(1.60E-19 C)(625 V)] / (9.11E-31kg)]
V0=1.48E7 m/s

I haven't checked this yet because I'm on my last attempt to get this right on my online homework, but does this look like the correct formula? I did some searching for examples involving kinetic energy and work, and this one popped up. The only thing is that the distance between the two charged surfaces isn't included in the calculations. Is that just an extra number that doesn't "matter" for the velocity? Is that because the surfaces are charged?
 
  • #13
rude man
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(1/2)mV0^2=q(deltaV)
So: VO=SQRT( 2*q*deltaV / m )
where q=1.60E-19 C and m=9.11E-31 kg

V0=SQRT[ [(2)(1.60E-19 C)(625 V)] / (9.11E-31kg)]
V0=1.48E7 m/s

I haven't checked this yet because I'm on my last attempt to get this right on my online homework, but does this look like the correct formula? I did some searching for examples involving kinetic energy and work, and this one popped up. The only thing is that the distance between the two charged surfaces isn't included in the calculations. Is that just an extra number that doesn't "matter" for the velocity? Is that because the surfaces are charged?
The formula is right. But you need to understand why.
For example: work W = force F x distance d.
F = qE, q = charge, E = E field
V = Ed
So W = qEd = qV.
Then you equate work done on the charge to slow it down to zero once it reaches V.
 
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  • #14
rlc
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Thank you for helping me!
 

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