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Physics 2: Electromagnetic Theory (Amperè's Law, Biot-Savart, etc)

  1. Dec 12, 2013 #1
    1. The problem statement, all variables and given/known data
    First off: these are not homework problems technically, I'm doing these for my own benefit...they're not being graded or collected. Just trying to study for my final.

    Problem 1
    A thin spherical, conducting metal shell has a radius of 15.0cm. A positive charge of 1.30µC is placed on this shell. What is the electrical potential at a distance of 7.50cm from the center of the shell?
    Answer: 160.0kV

    Problem 2
    Consider a very thin, and infinite in extent sheet of metal which is carrying an evenly distributed electrical current leading to a current density of 10A/m. What is the magnetic field strength a distance of 1.6m above the sheet?



    2. Relevant equations

    [itex]\vec{B}[/itex] = [itex]\frac{µ0}{4*∏}[/itex] * [itex]\frac{I*Δ \vec{S} X \widehat{r} }{r^2}[/itex]

    B = [itex]\frac{µ0*I*b}{2}[/itex]

    [itex]\oint[/itex][itex]\vec{B}[/itex][itex]\bullet[/itex] d[itex]\vec{s}[/itex] = µ0Ithrough

    3. The attempt at a solution

    Problem 1
    My query: I'm looking how to solve this. I'm not sure if you can use an assumption that the charge placed on the shell is now evenly distributed around the shell or not. In addition, I'm also not sure how to get down to electrical potential distance away.
    I am assuming that I'll be using the electric field created by the charge placed on the sphere in association with some distance relation, but I'm not sure where to go further than that.

    Problem 2
    The answer is 0, which means that this is more of a theoretical question than computational, but I'm not sure why.
    I tried reading through several sources, and they're effectively resonating with the equation B=[(μ0)(J0)(b)]/2 which seems to be the magnetic field created by the sheet, but void of a distancing element.
    Thanks in advance to any and all that help me with this, it's much appreciated and I've spent quite a long time reading my physics textbook trying to figure these out!
     
    Last edited: Dec 12, 2013
  2. jcsd
  3. Dec 13, 2013 #2
    Problem1​


    If the charge isn't distributed uniformly on its surface and we don't need to consider the electrostatic induction on the surface, then this question would be easy. It's like a point charge in the space. However, this leads to next problem: where do we need to calculate the electrical potential? The sets of all points whose distance from the center of the shell is 7.5cm forms a spherical surface. I think it's not what the problem means. In addition, if we consider the electrostatic induction, then it's very difficult to calculate the potential. So, as far as I'm concerned, I suggest we regard this "charge" as being uniformly distributed on the surface and everything will be ok!

    By Gauss' Law

    [itex]\oint Eda\:=\:\frac{Q_{enclosed}}{\epsilon_{0}}[/itex]

    [itex]E*4π*(0.075)^{2}=\frac{0}{\epsilon_{0}}[/itex]

    [itex]∴E\:=\:0[/itex]

    [itex]∴V(7.5cm\:from\:the\:center) = V_{surface}[/itex]

    [itex]V_{surface}=kQ/r[/itex]

    [itex]∴V_{surface}=\frac{9*10^{9}*1.3*10^{-6}}{0.15}[/itex]

    [itex]∴V_{surface}\:=\:7.8*10^{4}[/itex]

    Problem 2​


    Your third relevant equation might be wrong.

    [itex]\oint Bds\:=\:0[/itex]

    [itex]\oint Bdl\:=\:\mu_{0}I_{enclosed}[/itex]

    I'll use [itex]\oint Bdl\:=\:\mu_{0}I_{enclosed}[/itex] to calculate the magnetic field in your problem 2.

    [itex]\oint Bdl\:=\:B*2L[/itex]

    [itex]\mu_{0}*I_{enclosed}=\mu_{0}*10L[/itex]

    [itex]∴B*2L\:=\:\mu_{0}*10L[/itex]

    [itex]∴B\:=\:5\mu_{0}[/itex]
     
    Last edited: Dec 13, 2013
  4. Dec 13, 2013 #3

    collinsmark

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    Hello PropulsionMan,

    Welcome to Physics Forums! :smile:

    No, the answer is not 160.0 kV. I'm not sure where you got that answer, but something isn't right.

    No, the answer is not zero.

    Where are you getting these questions and answers?
     
  5. Dec 13, 2013 #4

    collinsmark

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    Expanding on my previous post...

    It's a conducting shell. That means the charge can move freely around the shell. Given that the shell is spherically symmetric, and given that there are no other charges nearby, the charge will naturally distribute itself uniformly over the surface of the shell.

    You need to integrate the electric field over a path. Traditionally, the "reference" position is infinity. So you must integrate the electric field from infinity to 7.5 cm.

    Given that the charge distribution is spherically symmetric, the electric field anywhere inside the shell's surface is zero. That means that the electric potential inside the surface of the shell is constant (but not necessarily zero -- in this particular problem it certainly is not zero). The point is, that integrating from infinity to 7.5 cm is the same as integrating from infinity to 15 cm, since the electric field from 15.0 cm to 7.5 cm is zero.

    But I don't know where you got the answer of 160.0 kV. It's not right. I would question the source of that answer.

    Once again, I question the source of that answer. Use Ampère's law to get the correct answer (It's not zero). By the way, in the configuration described, with an infinitely wide sheet, the magnitude of the magnetic field is uniform. It does not depend on the height above the sheet.
     
  6. Dec 13, 2013 #5

    rude man

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    Do in 2 steps:
    1. work needed to drag a unit positive test charge from infinity to the shell.
    2. work needed to drag the test charge from the shell to the position 7.5 cm from the center.
    If the current density is 10A/m that means that the current is finite. E.g if your sheet is 1 mm thick the current is 10A/m * 0.001m = 0.01A.

    So what happens with Ampere's law if a finite current pierces an infinite contour?

    BTW both your given answers are correct by me.
     
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