Proton's Final Speed in Double-Charged Capacitor

In summary, doubling the charge on a capacitor will increase the voltage across the capacitor and increase the potential energy of the proton. This will cause the proton's final speed to double.
  • #1
Xaspire88
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0
A proton is released from rest at the positive plate of a parallel-plate capacitor. It crosses the capacitor and reaches the negative plate with a speed of 50,000 m/s. What will be the proton's final speed if the experiment is repeated with double the amount of charge on each capacitor?

I seem to be stuck on this one. I initially tried to use
1/2 m[tex]^{}_{i}[/tex]v[tex]^{2}[/tex] + V[tex]^{}_{i}[/tex]q = 1/2 m[tex]^{}_{f}[/tex]v[tex]^{2}[/tex] + V[tex]^{}_{f}[/tex]q
but i ran into a wall. Help please.
 
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  • #2
That's the wrong way to think of it.

What is the same about the capacitor? (Hint, it's the same capacitor)
What does doubling the charge of a capacitor do to the voltage across the capacitor?
What effect does this voltage increase have on the electric potential energy of the electron?
 
  • #3
the capacitor plates are still the same distance apart. same size. same signs(one is positive and one is still negative)

by doubling the the charge of the plates it would seem that the voltage would increase by a factor of 2 as well.

increasing the voltage would increase the electrical potential of the proton.

Are these asumtions correct?
 
  • #4
Xaspire88 said:
the capacitor plates are still the same distance apart. same size. same signs(one is positive and one is still negative)

by doubling the the charge of the plates it would seem that the voltage would increase by a factor of 2 as well.

increasing the voltage would increase the electrical potential of the proton.

Are these asumtions correct?
Doubling the voltage will increase the potential energy of the proton by how much?

And then, how much more KE will the proton have when it reaches the other plate?
 
  • #5
the potential energy = Vq so if the voltage increases by a factor of two then the potential energy will increase by a factor of two. Which would then make sense to me that the kenetic energy would have to double to keep conservation of energy?
 
  • #6
Xaspire88 said:
the potential energy = Vq so if the voltage increases by a factor of two then the potential energy will increase by a factor of two. Which would then make sense to me that the kenetic energy would have to double to keep conservation of energy?
Correct. The proton (initially at rest, so KE = 0) is accelerated across the potential difference. The electric potential energy is converted to the proton's KE.
 
  • #7
So if initially the proton had a final velocity of 50,000m/s, once the charge on the plates was doubled and the potential energy was doubled and the kinetic was therefor doulbed would the resulting final velocity be double that of the original , 100,000m/s? if Ke= 1/2mv^2. the mass is the same but since the kinetic energy doubled the velocity would as well? Correct line of thought?
 
  • #8
Careful. What is the relationship between magnitude of velocity and kinetic energy.

Take V - and the kinetic energy, KE = 1/2 mV2,

now let V become 2V and KE = 1/2 m (2V)2.


Find the expression for V in terms of KE from the first expression.
 
  • #9
If we solve for v in terms of KE from the first expresion

v= [tex]\sqrt{(2KE)/m}[/tex]

now that is without using 2v

using KE= 1/2m2v^2v= [tex]\sqrt{KE/m}[/tex]

or since you have it as KE = 1/2 m (2V)^2 when you solve for V in terms of KE would

v= 1/2 [tex]\sqrt{(2KE)/m}[/tex]

my work:

KE = 1/2 m (2V)^2

2KE=m(2V)^2

(2KE)/m = (2V)^2

squareroot((2KE)/m)= 2V

1/2 squareroot((2KE)/m) = V
 
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  • #10
So if the energy doubles... how does the speed change?
 
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  • #11
rather than have (2V) wouldn't i have (2KE) since we have determined that we have double the PE which become double the KE? so the expression would then be

2KE = 1/2 mv^2

which would mean that V in terms of KE would equal

v = [tex]\sqrt{(4KE)/m}[/tex]

which simplified would be v = 2 [tex]\sqrt{KE/m}[/tex]?

would make sense to me that the increase in v due to the increase in KE would be a factor of 2? or should it only be an increase of [tex]\sqrt{2}[/tex]?

without the increase v= [tex]\sqrt{(2KE)/m}[/tex]
after v = [tex]\sqrt{(4KE)/m}[/tex] since the only difference is from 2 to 4... the increase would be by a factor of [tex]\sqrt{2}[/tex]? which would then make the velocity after doubling the charge on the plates v= 70,710.7m/s.
 
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  • #12
Xaspire88 said:
[tex]\sqrt{2}[/tex]?

Yes, that's the answer... if you are unsure, just take the ratio of the new velocity to the old:

[tex]\sqrt{(2KE)/m}[/tex]

This is correct for v... so v1 = [tex]\sqrt{(2KE_1)/m}[/tex]

v2 = [tex]\sqrt{(2KE_2)/m}[/tex]

What does v2/v1 come out to?
 
  • #13
Xaspire88 said:
which would then make the velocity after doubling the charge on the plates v= 70,710.7m/s.

yes.
 
  • #14
well the ratio of the new velocity to the old is going to be [tex]\sqrt{2}[/tex]

(70,710.7m/s)/(50,000m/s) = [tex]\sqrt{2}[/tex]

which doesn't really confirm anything for me because i came up with Vfinal by multiplying
50,000m/s by [tex]\sqrt{2}[/tex] so it only make since that working backwards would give me that? Maybe i missed the point of your question. I'm sorry if this is frustrating. I really appreciate the time and effort from all of you.
 
  • #15
Xaspire88 said:
well the ratio of the new velocity to the old is going to be [tex]\sqrt{2}[/tex]

(70,710.7m/s)/(50,000m/s) = [tex]\sqrt{2}[/tex]

which doesn't really confirm anything for me because i came up with Vfinal by multiplying
50,000m/s by [tex]\sqrt{2}[/tex] so it only make since that working backwards would give me that? Maybe i missed the point of your question. I'm sorry if this is frustrating. I really appreciate the time and effort from all of you.

Oops... sorry. I understand now... I didn't mean take the ratio of the numbers... I gave expressions for v2 and v1 in terms of E2 and E1... take the ratio of v2/v1 using those two expressions...
 
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  • #16
oh ok now i see that by simplifying

[tex]\sqrt{(2(2KE))/m}[/tex] / [tex]\sqrt{(2KE)/m}[/tex] = [tex]\sqrt{2}[/tex]

which confirms that it would increase by a factor of [tex]\sqrt{2}[/tex]. Thank you guys.
 
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  • #17
Xaspire88 said:
oh ok now i see that by simplifying

[tex]\sqrt{(2(2KE))/m}[/tex]/[tex]\sqrt{(2KE)/m}[/tex] = [tex]\sqrt{2}[/tex]

which confirms that it would increase by a factor of [tex]\sqrt{2}[/tex]. Thank you guys.

yup. :smile: you're welcome.
 

1. What is a double-charged capacitor?

A double-charged capacitor is a type of capacitor that has two layers of insulating material, known as dielectrics, between its two metal plates. This design allows for a larger amount of charge to be stored compared to a single-charged capacitor.

2. How does a double-charged capacitor affect a proton's final speed?

A double-charged capacitor has a stronger electric field between its plates compared to a single-charged capacitor. This stronger field can accelerate a proton to a higher final speed.

3. How is the final speed of a proton calculated in a double-charged capacitor?

The final speed of a proton in a double-charged capacitor can be calculated using the equation v = √(2qV/m), where v is the final speed, q is the charge of the proton, V is the voltage of the capacitor, and m is the mass of the proton.

4. How does the voltage affect the final speed of a proton in a double-charged capacitor?

The voltage of a double-charged capacitor directly affects the final speed of a proton. A higher voltage means a stronger electric field, which can accelerate the proton to a higher final speed.

5. Are there any other factors that can affect the final speed of a proton in a double-charged capacitor?

Yes, the distance between the plates of the capacitor and the strength of the magnetic field in the vicinity can also affect the final speed of the proton. In general, a shorter distance and a stronger magnetic field can result in a higher final speed.

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