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 [Broken]

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 dont 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 [Broken]

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]

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:

E

E

-----------------------------------------------------------------------------------

3) for point B, z axis is still 0 because of symetry

E

E

E

------------------------------------------------------------------------------------

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

E

E

---------------------------------------------------------------------------------------

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

E

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 [Broken]

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 dont 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 [Broken]

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:

E

_{x}=([tex]\rho[/tex]R/3[tex]E[/tex]_{0}) - ([tex]\rho[/tex](R/3)/3[tex]E[/tex]_{0})E

_{Ax}=(2[tex]\rho[/tex]r/9[tex]E[/tex]_{0})-----------------------------------------------------------------------------------

3) for point B, z axis is still 0 because of symetry

E

_{y}=([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}) = 0E

_{x}=([tex]\rho[/tex](sqrt(5)R/3)/3[tex]E[/tex]_{0}*cos(26.565)) - 0 =E

_{Bx}=(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

E

_{Cx}=([tex]\rho[/tex](2R/3)/3[tex]E[/tex]_{0})E

_{Cx}=(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

E

_{Cx}=(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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