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Electric Field and Work

  1. May 6, 2016 #1
    upload_2016-5-6_21-57-50.png
    Need help with Problem 1

    2. Relevant equations
    E = kq/r2
    V = -W/q
    V = kq/r

    3. The attempt at a solution
    1a. dE = kdq/rRSUP]2[/SUP]
    dq = p ds
    dE = kpds/R2
    Not sure if I am right up to this point and don't know how to proceed.

    1b. I think the electric field would be zero since the enclosed charge cancels out. Not sure on this either.

    1c. V = kq/r so V = kZe/a
    W = -kZ2e2/a

    1d. Not sure how to approach

    Any help would be appreciated! upload_2016-5-6_21-57-50.png
     
  2. jcsd
  3. May 6, 2016 #2

    Paul Colby

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    Assume the E field is radial and constant in magnitude at a given radius, ##r##, from the center. This really follows from the symmetry of the charge distribution. What is the total charge enclosed inside our sphere of radius ##r##? Try looking at Gauss's law.
     
  4. May 6, 2016 #3
    So then from Gauss's law, E = q/εA so E=Ze/επR2
    Would that be correct for R>r.
    And in the case of r>R would it be zero, since for q it would cancel out to zero?
     
  5. May 7, 2016 #4

    Paul Colby

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    Yes, so ##q(r)## is a function of ##r## which gives ##E(r)##. Just write out ##q(r)## and the rest follows.
     
  6. May 7, 2016 #5
    Awesome, so when r>R, its zero. Thanks! Can you assist me in telling me if I am right for part c please? I got V = kq/r so V = kZe/a and W = -kZ2e2/a and can you guide me through part d please?
     
  7. May 7, 2016 #6

    Paul Colby

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    Okay, the expression for the potential energy looks in order. Since the work needed has to equal the change in potential energy why does your expression for work have ##Z^2e^2## in it while the potential has just ##Ze##?

    On d I'd point out that the potential energy of the electron is the sum of two parts. One due to the nucleus and one due to the cloud. Look at the signs of these.
     
  8. May 7, 2016 #7
    So I got the electric potential, V, to be kZe/a and plugged that into the equation for work which is W = -Vq which is where I got the -kZ2e2/a. So you're saying this is not correct?
     
  9. May 7, 2016 #8

    Paul Colby

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    It boils down to what the symbols mean and what conventions one are using. Typically ##Z## is the integer number of charges in the nucleus and is the "Atomic Number". ##e## is the unit of charge or some ##1.6\times 10^{-19}## coulombs. ##k## is coulomb's constant. Okay, the electrostatic potential is ##V = kZe/a## and is in Joules/Coulomb (using MKS units). The potential energy is the ## V *(-e) ## so the answer should be ##-kZe^2/a## because the charge of the electron is ##-e##.
     
  10. May 7, 2016 #9
    But the q that is given is Q = Ze, so wouldn't you plug that in for W = -Vq which is where you would get -kZ2e2/a
     
  11. May 7, 2016 #10

    Paul Colby

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    I may have lost the bubble on which question is being discussed, if so sorry. Problem c is what being asked about, correct? If so q is the charge of a single electron and not the cloud. ##-Ze## is the total charge of the electron cloud which has ##Z## electrons. Problem c doesn't have an electron cloud by my reading.
     
  12. May 7, 2016 #11

    Paul Colby

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    Nope, the word "the" nucleus likely implies the cloud is there, my bad. There is a potential at each ##r## so you would need ##V(r)##. This would be multiplied by the electron charge which is ##-e## not ##-Ze##.
     
  13. May 7, 2016 #12
    Wouldn't it be multiplied by Ze since the electron cloud is there since Q is given to be +Ze?
     
  14. May 7, 2016 #13
    Additionally, the uniform charge density is also given and I haven't used it in any of the solutions thus far. Isn't the uniform charge density to be used in order to calculate the electric field?
     
  15. May 7, 2016 #14

    Paul Colby

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    It's really important to ask oneself what the symbols mean. ##V## in the problem is always the electrostatic potential. This is the amount of work that needs to be done to move a charge, ##q##, to infinity (which is V=0 by the usual convention). If one doubles the charge it takes twice the work. Your answer would require 4 times the work because ##(2Z)^2=4Z^2## so something has to be wrong with your answer.
     
  16. May 7, 2016 #15
    Ohhhh, ok I understand what you mean now. Thank you! But a last concern is that the uniform charge density is given in the problem but it's not used in any of the solutions. Is that fine?
     
  17. May 7, 2016 #16

    Paul Colby

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    On problem c if we assume the electron cloud then at ##r=a## where the electron starts, the electron is starting at a lower potential because the cloud charge between the nucleus and the electron orbit "shields" the nuclear charge making it look smaller by the amount of charge in between. An extreme case if ##a > R##, ##R## being the atomic radius, it would take no energy to release the electron. My guess is that the electron cloud should be included and that one should calculate ##V(r)## and set ##r=a##. Your answer (less the additional factor of ##Z##) is correct assuming no electron cloud.
     
  18. May 7, 2016 #17
    So what would you get for the work as a final answer in part c?
     
  19. May 7, 2016 #18

    Paul Colby

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    Well, The potential of the bare nucleus is,

    ##V_N(r) = k\frac{Ze}{r}##

    The potential of the cloud is harder. The ##V## field due to the uniform charge is,

    ##V_C(r) = k\frac{Q(r)}{r}##

    where ##Q(r)## is the total electron charge enclosed in the sphere of radius ##r##. That would be

    ##Q(r) = -Ze\frac{r^3}{R^3}## for ##r < R##

    and

    ##Q(r) = -Ze## for ## r \ge R##

    So adding these gives

    ##V(r) = V_N(r) + V_C(r) = kZe(1 - \frac{r^3}{R^3})\frac{1}{r}##

    for ##r < R## and 0 outside ##R##. This is if I haven't dropped a factor so I'd check my answer carefully. Hope this helps
     
  20. May 7, 2016 #19

    Paul Colby

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    Oh, and you'd have to evaluate at ##r=a##
     
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