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Electric potential energy problem

  • Thread starter mcano
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  • #1
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Homework Statement


"Two parallel plates having charges of equal magnitude but opposite sign are separated by 12cm. Each plate has a charge superficial density of 36nC/m2. A proton is launched from the negative plate to the positive plate. What is the minimum speed the proton should be launched to get to the positive plate?"

Well here's the problem, however, there are steps the teacher gave to solve, here are those

1. Express the electric field between the plates as a function of the charge superficial density (done)
2. Express the potential difference between the plates as a function of the charge superficial density(done)
3. Use the conservative nature of the electric fields and find an expression for the initial speed of the proton as a function of the charge superficial density, the distance between the plates and the speed when it gets to the negative plate.(Here is where i have problems)

Homework Equations



from 1. we have that
[tex]
E = {\frac{\sigma}{\epsilon_{0}}}
[/tex]
from 2 we have that

[tex]
\Delta V = E.d = {\frac{d\sigma}{\epsilon_{0}}}
[/tex]

Given that the field is uniform, well, i'm considering it uniform since the size of the plates is not stated

Now, we have the kinetic energy
[tex]
K = {\frac{1}{2}}mv^2
[/tex]

and also we have the Torriceli's equation

[tex]
v_{f}^2 =v_{i}^2 + 2ad
[/tex]

The Attempt at a Solution



Well, given the things before i know i can use the Torriceli's equation and get the expression, however i have no idea of how it would be possible to ge the relation between kinetic and electric potential energy, mostly because i'm not sure what is the equation for the electric potential energy in this exact problem, i have found many equations and my only guess is this:

Assuming that the potential energy at the positive plate (point B) is maximum and assuming that the kinetic energy in the negative plate is maximum (point A), thus at point A electric potential energy is zero

i have this,

[tex]
\Delta V =U_{B}-U_{A} = E.d = {\frac{d\sigma}{\epsilon_{0}}}
[/tex]

And given that at the point A:

[tex]
U_{A} = 0
\rightarrow U_{B} = {\frac{d\sigma}{\epsilon_{0}}}
[/tex]

Then i could do
[tex]
K = U
[/tex]

But i'm not so sure about this, so i was wondering if someone could give me a better explanation
Thanks beforehand
 

Answers and Replies

  • #2
vela
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You're headed in the right direction, but you're mixing up electric potential V with electric potential energy U.
 
  • #3
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You're headed in the right direction, but you're mixing up electric potential V with electric potential energy U.
So now, i only have to remember that
[tex]
W = -(U_{B}-U_{A}) = -\oint \vec{E}\cdot d\vec{r}
[/tex]

Well considering that in this case F = E

And use the principle of conservation to make K = U and then i can just get the final speed, and replace in the torricelli's equation and replace right?
right?
 
Last edited:
  • #4
vela
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No, you're still making the same mistake. Check the units of what you wrote. They don't work out.
 
  • #5
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No, you're still making the same mistake. Check the units of what you wrote. They don't work out.

Yeah, sorry, i forgot that the field is N/C and the Work should be J/m, so first i need to calculate the force that the field is exerting over the particle right? i know the field would be:
[tex]
E = \frac{\sigma}{\epsilon_{0}}
[/tex]

and thus
the force would be:

[tex]
F = q {\vect{E}}
[/tex]
where q would be the charge right?
that would be the proton's charge, now, as i obtain the force i should now integrate to obtain the work right?

[tex]
W = -(U_{B}-U_{A})=-\oint \vec{F}\cdot d\vec{r} = -\oint q\vec{E}\cdot d\vec{r}
[/tex]
 
Last edited:
  • #6
vela
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Work is in joules, and the integral symbol shouldn't have a circle, which denotes a closed path. The negative sign in front of the integral shouldn't be there either.
 
  • #7
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Work is in joules, and the integral symbol shouldn't have a circle, which denotes a closed path. The negative sign in front of the integral shouldn't be there either.
Teah, i already worked it out, thanks for your help...
 

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