- #1
Dell
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we have solid ball with a radius of R=5cm and a charge density of "rho"=-3C/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.
diagram below
http://picasaweb.google.com/devanlevin/DropBox?authkey=Gv1sRgCL_4l4PpvP_YsQE#5314956645280048530
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
==============================================================
these are all my calculations, but i am only stuck with C, so if you don't fell like going through the whole story,skip to the last step (4)
===============================================================
what i did was the following:
my workings:
http://picasaweb.google.com/devanlevin/DropBox?authkey=Gv1sRgCL_4l4PpvP_YsQE#5315181847071965298
i say that my field is the total of 2 fields, one of the original ball, and one of the "imaginary hollowed out ball" (which has a density equal to the negative of the density of the big ball)
1) i chose a gauss surface as a sphere, after my calculations i get : E=([tex]\rho[/tex]r/3[tex]E[/tex]0)
i will use this expression throughout my calculation changing the radius every time
2) for point A, i can say that the field in the y and z axis is 0 because of the symetry of the point.
as for the axis:
Ex=([tex]\rho[/tex]R/3[tex]E[/tex]0) - ([tex]\rho[/tex](R/3)/3[tex]E[/tex]0)
EAx=(2[tex]\rho[/tex]r/9[tex]E[/tex]0)
-----------------------------------------------------------------------------------
3) for point B, z axis is still 0 because of symetry
Ey=([tex]\rho[/tex](sqrt(5)R/3)/3[tex]E[/tex]0*sin(26.565)) -([tex]\rho[/tex](R/3)/3[tex]E[/tex]0) = 0
Ex=([tex]\rho[/tex](sqrt(5)R/3)/3[tex]E[/tex]0*cos(26.565)) - 0 =
EBx=(2[tex]\rho[/tex]r/9[tex]E[/tex]0)
------------------------------------------------------------------------------------
4) for point C i say that since it is exactly in the centre of the big ball, the big ball's field at C is equal to 0, because of perfect symetry and because R=0, so the total field at point C is only the field of my hollowed out imaginary ball,
because of the symetry, the imaginary ball too has 0 field on y and z axis so i only care about x
ECx=([tex]\rho[/tex](2R/3)/3[tex]E[/tex]0)
ECx=(2[tex]\rho[/tex]r/9[tex]E[/tex]0)
---------------------------------------------------------------------------------------
HOW CAN THIS BE??
for A, B i understand that it is possible that the field is the same, but for C i do not, the correct answer is supposed to be
ECx=(2[tex]\rho[/tex]r/36[tex]E[/tex]0)
but i cannot get to it, where am i going wrong??
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.
diagram below
http://picasaweb.google.com/devanlevin/DropBox?authkey=Gv1sRgCL_4l4PpvP_YsQE#5314956645280048530
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
==============================================================
these are all my calculations, but i am only stuck with C, so if you don't fell like going through the whole story,skip to the last step (4)
===============================================================
what i did was the following:
my workings:
http://picasaweb.google.com/devanlevin/DropBox?authkey=Gv1sRgCL_4l4PpvP_YsQE#5315181847071965298
i say that my field is the total of 2 fields, one of the original ball, and one of the "imaginary hollowed out ball" (which has a density equal to the negative of the density of the big ball)
1) i chose a gauss surface as a sphere, after my calculations i get : E=([tex]\rho[/tex]r/3[tex]E[/tex]0)
i will use this expression throughout my calculation changing the radius every time
2) for point A, i can say that the field in the y and z axis is 0 because of the symetry of the point.
as for the axis:
Ex=([tex]\rho[/tex]R/3[tex]E[/tex]0) - ([tex]\rho[/tex](R/3)/3[tex]E[/tex]0)
EAx=(2[tex]\rho[/tex]r/9[tex]E[/tex]0)
-----------------------------------------------------------------------------------
3) for point B, z axis is still 0 because of symetry
Ey=([tex]\rho[/tex](sqrt(5)R/3)/3[tex]E[/tex]0*sin(26.565)) -([tex]\rho[/tex](R/3)/3[tex]E[/tex]0) = 0
Ex=([tex]\rho[/tex](sqrt(5)R/3)/3[tex]E[/tex]0*cos(26.565)) - 0 =
EBx=(2[tex]\rho[/tex]r/9[tex]E[/tex]0)
------------------------------------------------------------------------------------
4) for point C i say that since it is exactly in the centre of the big ball, the big ball's field at C is equal to 0, because of perfect symetry and because R=0, so the total field at point C is only the field of my hollowed out imaginary ball,
because of the symetry, the imaginary ball too has 0 field on y and z axis so i only care about x
ECx=([tex]\rho[/tex](2R/3)/3[tex]E[/tex]0)
ECx=(2[tex]\rho[/tex]r/9[tex]E[/tex]0)
---------------------------------------------------------------------------------------
HOW CAN THIS BE??
for A, B i understand that it is possible that the field is the same, but for C i do not, the correct answer is supposed to be
ECx=(2[tex]\rho[/tex]r/36[tex]E[/tex]0)
but i cannot get to it, where am i going wrong??
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