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Electric field of ball

  1. Mar 19, 2009 #1
    as seen in the diagram below, ->
    http://picasaweb.google.com/devanlevin/DropBox?authkey=Gv1sRgCL_4l4PpvP_YsQE#5314956645280048530 [Broken]

    is a solid ball with a radius of R=5cm and a charge density of [tex]\rho[/tex]=-3[tex]\mu[/tex]C/m3,
    inside this ball, we make a hollow ball shaped space with a radius of R/3 with its centre at 2R/3 from the centre of the big ball.

    what is the Electric field at point:

    A-on the leftmost point of the hollow
    B-on the top point of the hollow
    C-at the centre of the big ball

    ---------------------------------------------------
    how do i do this?

    i think what i need to do is say that the field is equal to (the field of the original ball) + ( the field of a ball the size of the hollow, with a charge density of -[tex]\rho[/tex] )??

    for C i know that the field before is 0 since it is at the centre, how do i continue from there?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 19, 2009 #2
    what i did so far is: [tex]E[/tex]=epsilon0
    i took a surface at the radius of the ball and said

    [tex]\varphi[/tex]=[tex]\oint[/tex]EdA=EA=E(4[tex]\pi[/tex]R2)

    [tex]\varphi[/tex]=Q/[tex]E[/tex]=(V[tex]\rho[/tex])/[tex]E[/tex]=(0.75[tex]\pi[/tex]R3/[tex]E[/tex])

    E(4[tex]\pi[/tex]R2)=(0.75[tex]\pi[/tex]R3/[tex]E[/tex])
    E=([tex]\rho[/tex]R)/(3[tex]E[/tex])

    now what i will do is subtract the "field" of the imaginary ball from the field of the big ball to get the total

    E=E1-E2
    E=([tex]\rho[/tex]R)/(3[tex]E[/tex])-([tex]\rho[/tex]R)/(9[tex]E[/tex])
    and i get

    E=(2[tex]\rho[/tex]R)/(9[tex]E[/tex])
    but where is this answer valid for? A,B or C?? is this the field at A since i took the radius of the big ball and found the flux according to that? for the others do i need to do the same using the radius 2R/3 for point B and C and saying the field of the big ball alone at C is 0?
     
    Last edited: Mar 19, 2009
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