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Charge density

  1. Oct 13, 2009 #1
    1. The problem statement, all variables and given/known data
    i have:
    a nonconducting sphere of radius 25 cm. the electric field points radially away from the center of the sphere, and increase linearly from 0 to 3700V/m at the surface of the sphere.
    determine charge density?

    2. Relevant equations

    i think the E=E0+r*148

    3. The attempt at a solution

    so i did [tex]\int E *da = \int p(r) d \tau / \epsilon0 [/tex]

    then i substitute E

    [tex]da = 4* pi * r^{2}

    [/tex]

    [tex]
    d \tau = 4* pi * r^{2} dr
    [/tex]


    then i simplify for p(r)


    is that right????
     
  2. jcsd
  3. Oct 13, 2009 #2

    rl.bhat

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    Yes. It is right.
     
  4. Oct 13, 2009 #3
    thank you
     
  5. Oct 13, 2009 #4
    i got


    p(r) = [tex]\epsilon 0 *(E_{}o 2/r +3)[/tex]

    can that be???

    and now the problem is asking for the potential difference between the center and the surface of the sphere

    i know that

    [tex]\Delta V = -\int E dr[/tex]

    will that do it??
     
  6. Oct 13, 2009 #5

    rl.bhat

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    p(r) = [tex]\epsilon 0 *(E_{}o 2/r +3)[/tex]
    How did you get this one?
    Your formula for dV is correct.
     
  7. Oct 13, 2009 #6
    [tex]
    (Eo+r*148)*( 4* pi * r^{2})* \epsilon 0 = \int p(r) (4*pi*r^{2}) dr


    [/tex]


    then i take the derivative on both sides and isolate p(r)

    i obtain =

    [tex]
    \epsilon 0*(E0*2/r +396) = p(r)

    [/tex]
     
  8. Oct 13, 2009 #7
    i got -15000 for the potential. I consider Eo to be zero is that a good assumption?
     
  9. Oct 13, 2009 #8

    rl.bhat

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    think the E=E0+r*148
    Is it given in the problem?
     
  10. Oct 14, 2009 #9
    no. I thought since the problem said that it is linearly that might work?
     
  11. Oct 14, 2009 #10

    rl.bhat

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    No. It is not correct.
    Electric field is given.
    You have found E = ρ*r/εο. Find ρ.
    Now dV = - dE*dr To find the potential V on the surface find the integration.
     
  12. Oct 14, 2009 #11

    ehild

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    ..."the electric field points radially away from the center of the sphere, and increase linearly from 0 to 3700V/m at the surface of the sphere."

    - this means that E=0 at the centre of the sphere and E=Es=3700 V/m at the surface, and increases linearly inside the sphere from r=0 to r=R, the radius of the sphere.
    The enclosed charge in the sphere is
    [tex]Q=4/3 R^3 \pi \rho [/tex] and by Gauss' Law, [tex] Q=4R^2 \pi \epsilon_0 E_s [/tex]. Comparing the equations for the total charge, you get the charge density.
    As for the potential at the surface of the sphere: remember that we usually choose zero potential at infinity, and outside a homogeneously charged sphere the electric field is the same as that of a point charge....

    ehild
     
  13. Oct 14, 2009 #12
    [tex]
    4/3 R^3 \pi \rho = 4R^2 \pi \epsilon_0 E_s
    [/tex]

    so i just set it equal to each other and solve for p


    [tex]
    \rho = 3 \epsilon_0 E_s / R
    [/tex]

    is that it?
     
  14. Oct 14, 2009 #13

    ehild

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    Yes. Now calculate it with the given data.

    ehild
     
  15. Oct 14, 2009 #14
    ok i got the charge density


    how do i do the potential


    so how do i do this integral

    dV = - dE*dr
     
  16. Oct 15, 2009 #15

    ehild

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    Previously you have written the integral correctly:


    [tex]
    \Delta V = -\int E dr
    [/tex]

    You want the potential at the surface of the sphere and you know that it is nought at infinity:

    [tex]
    \Delta V= V(\infty)-V(R) = -\int_R^\infty{ E dr}
    [/tex]

    You certainly have learnt that the electric field outside a homogeneously charged sphere is the same as if its total charge were concentrated in the centre. What is the field around a point charge?

    [tex]

    E=\frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}

    [/tex],

    is it right? You know Q, the total charge of the sphere.

    Now integrate this formula for E.

    ehild
     
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