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Electric field of ball

  • Thread starter Dell
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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:
590
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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:

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