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Conducting shell and proton with KE

  • Thread starter origamipro
  • Start date
  • #1
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Homework Statement



http://imageshack.us/photo/my-images/543/15jd.jpg/

[IMG=[PLAIN]http://img543.imageshack.us/img543/8674/15jd.jpg][/PLAIN] [Broken]

Homework Equations



Qnet outside sphere = Q1+Q2
KEi + PEi = KEf + PEf

q = 1.6E-19
mass of q = 1.672E-27

The Attempt at a Solution



+(q ∫ KQ1 / R^2 ] 3R to R) + (q ∫ K(-Q2) / R^2 ] 3R to R) - 1/2 m v^2
 
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Answers and Replies

  • #2
74
6
I don't know if it's just me, but I cannot see the picture that you posted.
 
  • #4
14
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Bump. Q net is +.04E-16C. Both charge sources are modeled from the very center. How does the proton come to rest?
 
  • #5
gneill
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2,773
Your Q net does not look right to me. Show your work.

Presumably the proton comes to rest only momentarily before being accelerated away from the net positive charge on the shell.
 
  • #6
14
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Q1 = 4E-16 Q2 = ?
q = 1.6E-19 qm = 1.67E-27

given answer Q2 = -3.96E-16

at the surface of the shell kinetic energy must equal zero. The potential energy at the surface does not need to be accounted for since it only wants value of Q2 when proton gets to the surface. Is this assumption correct, otherwise calculating E field at its surface becomes an approximation since the integral does not converge.

so proton traveling in negative x direction, with two electric potentials both modeled from the same source point.

(1/2)m(v^2) -(q ∫ kQ1/r^2 ] 3R to R) + (q ∫ KQ2/r^2 ] 3R to R) = 0
(4.0915x10^-24) -2.74286E-24 + (1.6E-19)Q2(4.28571E10) = 0
solving for Q2 = -1.9667E-16
 
  • #7
gneill
Mentor
20,793
2,773
Q1 = 4E-16 Q2 = ?
q = 1.6E-19 qm = 1.67E-27

given answer Q2 = -3.96E-16

at the surface of the shell kinetic energy must equal zero. The potential energy at the surface does not need to be accounted for since it only wants value of Q2 when proton gets to the surface. Is this assumption correct, otherwise calculating E field at its surface becomes an approximation since the integral does not converge.
The potential energy is key. The change in potential energy between the staring and ending locations will equal the magnitude of the change in kinetic energy of the particle. The potential for a charge q very near the surface of a spherical conductor with charge Q and radius r is kQq/r. The spherical charged surface behaves like a point charge located at its center.

so proton traveling in negative x direction, with two electric potentials both modeled from the same source point.

(1/2)m(v^2) -(q ∫ kQ1/r^2 ] 3R to R) + (q ∫ KQ2/r^2 ] 3R to R) = 0
(4.0915x10^-24) -2.74286E-24 + (1.6E-19)Q2(4.28571E10) = 0
solving for Q2 = -1.9667E-16
No need to integrate. Use conservation of energy via initial KE and change in PE for the two locations.
 
  • #8
14
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Right that is what i showed.

Ke = 1/2 m v^2

the integration is the delta V(voltage) = q(kQ1/3R - KQ2/R)
 
  • #9
gneill
Mentor
20,793
2,773
Ah. My mistake. When I read the problem I misread the key phrase "...from a distance of 2R from the surface of the sphere." That puts the initial distance at 3R from the center. I suspect that whoever provided the answer for the problem did the same thing.

So I agree with your result that Q2 should be ~ -1.97 x 10-16 C and not the given value.
 

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