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Integral help

  1. Jun 6, 2009 #1
    a question and a solution is here

    http://i42.tinypic.com/2qch7vo.gif

    J=I/A
    V=qE(\tao)/me

    by the intervals iknow that they are scanning the whole sphere
    I=J*S (the double integral is S)
    why
    they are doing Q_0 -Q(t) ??
     
    Last edited: Jun 6, 2009
  2. jcsd
  3. Jun 6, 2009 #2

    Cyosis

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    Homework Helper

    I=JA is only valid if the current density is constant along the surface. If it is not, as always, you have to integrate.

    [itex]
    dI=\vec{J}\cdot d\vec{a} \Rightarrow I=\int \vec{J}\cdot d\vec{a}
    [/itex]

    In this case the current density is different along the surface of the sphere, but luckily the surface vector points in the same direction as the current density, so the integral reduces to [itex]I=\int J da[/itex].

    You know two boundary conditions [itex]Q(0)=Q_0[/itex], [itex]I(0)=0[/itex] and [itex]Q_{tot}(t)=Q_0+Q(t)[/itex]. Evaluate the integrals and solve the integration constant using the boundary conditions.
     
    Last edited: Jun 6, 2009
  4. Jun 6, 2009 #3
    on what basis you say that I(0)=0
    on what basis you say Q_total(t)=Q_0+Q(t)
    how do you know that surface vector points in the same direction as the current density
    ??
     
  5. Jun 6, 2009 #4

    Cyosis

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    Homework Helper

    The total charge is the sum of the separate charges. What are the separate charges? Q_0, and the charge the current causes the sphere to have over time. At t=0 it is given that the sphere only possesses a charge Q_0, so there needs to be zero contribution from the current.
    J points in the rhat direction. Any normal vector, nhat, on a sphere is a vector perpendicular to the surface of the sphere. Therefore nhat=rhat and rhat.rhat=1.
     
    Last edited: Jun 6, 2009
  6. Jun 6, 2009 #5
    whats rhat and nhat ??
     
  7. Jun 6, 2009 #6

    Cyosis

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    They are the unit vectors [itex]\hat{r}, \hat{n}[/itex].
     
  8. Jun 6, 2009 #7
    we have already done
    [tex]
    I=\int J da
    [/tex]
    we have some expresion
    and i know the law of I=dq/dt (derivative of Q)
    but why they say that its -I=dq/dt (minus derivative of Q)
    ??
     
    Last edited: Jun 6, 2009
  9. Jun 8, 2009 #8
    why minus
    ??
     
  10. Jun 8, 2009 #9

    Mark44

    Staff: Mentor

    They are keeping track of directions. From the page you posted, "The current is directed along the positive [itex]\hat{r}[/itex] axis (away from the charge inside the sphere).

    I read this as meaning that the current is moving outward, but Q increases in the opposite direction; that is Q decreases (so dQ/dt is negative) in the direction I is flowing. I and dQ/dt have the same magnitudes, but opposite signes.
     
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