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Final Velocity in parallel plate capicator?

  • Thread starter outxbreak
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  • #1
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1. The plates of a parallel plate capacitor are separated by a distance (4cm). A point charge of magnitude (q=2e) (3.2*10^-19 C) is fired horizontally through a small hole in the positive plate of the capacitor. The charge is fired with speed (8.5 *10^5 m/s) and, upon reaching point M (located @ 2cm between each plate), is observed to have speed (4*10^5 m/s). The point M is located from the negative plate. Assume the charge has mass 2proton (3.34 *10^-27 kg).



2. Will the charge reach the negative plate? If so, what will its speed be when it reaches
the negative plate? If not, what is its closest point of approach to the negative plate?




3. the answer is: The charge will be momentarily brought to rest at a point ~1.33 cm
from the right hand plate.

I do not understand how they came up with the above answer.
When I calculate the Vf using the Vi as the Velocity at point M I get a very large velocity (not zero). What equation should I use to get their answer?
I currently use:
Vf= sqrt (2q deltaV)/(m)

 

Answers and Replies

  • #2
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This formula is for a charge accelerated from rest.
You have a charge sowed down from some non-zero initial velocity.

You can use the work-energy theorem. The work is done by the electric field.
 
  • #3
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Okay I still get positive values..
 
  • #4
gneill
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Okay I still get positive values..
You'll have to show more of your work; Can't tell what's going wrong without seeing what you're actually doing :smile:
 
  • #5
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I just really don't understand the first persons comment. The only work energy formula I have ever been taught starts with a charged particle accelerating from zero.

I guess work energy is:
Change in potential energy=Change in kinetic energy

ΔU = −W = −Fd = −q Ed


So: −q Ed=1/2mvf2 OR

vf=sqrt (-2qed/m)

which is sqrt ( (-2)(-3.2*10^-19)(125,250 V/m)(0.04m)/(3.34*10^-27))

= 4.89 *10^6 which is not zero so I have no clue what he is doing..
 
  • #6
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The change in kinetic energy equals the work done by external forces.
This is the work-energy theorem.
So write the change or difference between the final and initial KE and equate it with the work done by the electric force. This will give you the velocity at any distance from the entry plate.
 
  • #7
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So change in kinetic is just Vf-Vi or 4*10^5 m/s
and work done by field is
W=qED

Setting these equal and solving for D

4*10^5=(3.2*10^-19)(125,250 V/m)x

x=9*10^18 meters doesn't really make any sense
 
  • #8
gneill
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How did you arrive at the value you did for the electric field (125,250 V/m)?

I'm not sure why you're looking for a final velocity at a distance of 4 cm since the particle may not reach that point (in fact we know it does not, so results would be nonsense). Instead, consider that the charge has a potential energy that is associated with its location in the field. It begins with some initial KE and as it travels forward it trades KE for PE in the field.

You can determine the amount of KE lost in going halfway across the capacitor from the given information. Given that the field is uniform between the plates, how much KE would be required to go the second half of the distance? Is there enough KE remaining?

Knowing your ΔKE per unit distance, at what distance will the initial KE be reduced to zero?
 
  • #9
gneill
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Energy is measured in Joules, not meters per second. Kinetic energy is not velocity (although it is certainly related to velocity). What's the formula for kinetic energy?
 
  • #10
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The E field was calculated by:


Change in potential energy= sqrt (mass (vi2 -vf2)/(2q)
which is 125,250 V/m

This is also in the answer key so I'll go with it.

I don't understand what formula to use. I can understand that in 0.02 meters the kinetic energy decreases 4*10^5 m/s so it would take another 0.02 cm to get the velocity down to zero because that would be a total decrease of 8*10^ m/s. This doesn't match up with his answer at all.

I would guess change in kinetic energy would be

4*10^5 m/s so
4.5*10^5=qEd

d=1.12*10^19 meters.
 
  • #11
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Speed is not kinetic energy. Do you understand this?
The kinetic energy does not decrease by 4*10^5 m/s. It does not even make sense to say it.

Do you know the expression (formula) for kinetic energy?
How much is the initial KE?
 
  • #12
gneill
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The E field was calculated by:


Change in potential energy= sqrt (mass (vi2 -vf2)/(2q)
which is 125,250 V/m
That doesn't look like a correct formula for change in potential energy (and it's not clear what the square root is supposed to include, since the parentheses are unbalanced).

What does unit analysis say the resulting units would be?

This is also in the answer key so I'll go with it.
I would be suspicious of the answer key values; they don't tally with what I'm seeing by my own calculations. It could be that the problem values have been changed and the answer key has not been updated.

I don't understand what formula to use. I can understand that in 0.02 meters the kinetic energy decreases 4*10^5 m/s so it would take another 0.02 cm to get the velocity down to zero because that would be a total decrease of 8*10^ m/s. This doesn't match up with his answer at all.

I would guess change in kinetic energy would be

4*10^5 m/s so
4.5*10^5=qEd

d=1.12*10^19 meters.
Velocity is not kinetic energy. What is the formula for kinetic energy? What is the initial kinetic energy of the particle? Hint: Kinetic energy depends only upon mass and velocity.
 
  • #13
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yes.. the E field formula I put was a little messed up.
but my initial formula took mass into account saying that Kinetic Energy is: 1/2mv2

I got the answer to E field with the formula and plugging in:

Change in potential energy= (mass (vi2 -vf2)/(2q)

(3.34*10^-27) ( (8.5 *10^5 m/s)2 -(4*10^5 m/s)2)/2(3.2*10^-19)
=2505V
so change in potential is 2505 V
and E field=ΔV/Δx
so 2505 V/0.02 meters=
which is 125,250 V/m

Anyway if the initial kinetic energy is 0.5mv2
that would be (0.5)(3.34*10^-27)(8*10^5 m/s)2=1.07*10^-15 kgm2/s2
K at 0.2 meters is (0.5)(3.34*10^-27)(4*10^5 m/s)2=2.67*10^-16 kgm2/s2
Change in Ke= 8.03 *10^-16 kg m2/s2

So the kinetic energy changes 4.01*10^-14 every meter

By dividing:
initial Kinetic by the
4.01*10^-14 every meter
I got 2.665 cm
so 4-2.665=1.333


Thanks for so much help!!!
 
  • #14
gneill
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yes.. the E field formula I put was a little messed up.
but my initial formula took mass into account saying that Kinetic Energy is: 1/2mv2

I got the answer to E field with the formula and plugging in:

Change in potential energy= (mass (vi2 -vf2)/(2q)

(3.34*10^-27) ( (8.5 *10^5 m/s)2 -(4*10^5 m/s)2)/2(3.2*10^-19)
=2505V
so change in potential is 2505 V
If I run those same numbers I obtain a value of 2963 V. So something's not right.

and E field=ΔV/Δx
so 2505 V/0.02 meters=
which is 125,250 V/m

Anyway if the initial kinetic energy is 0.5mv2
that would be (0.5)(3.34*10^-27)(8*10^5 m/s)2=1.07*10^-15 kgm2/s2
Should be 1.208 x 10-16 J since the problem statement sets the initial velocity at 8.5 *10^5 m/s. You've used 8.0 * 10^5 m/s. I think you might have used it in other calculations, too. Is the problem statement value correct?
 

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