Electric field at points A, B, and C inside a hollow ball?

[Dell]

as seen in the diagram below, ->
http://picasaweb.google.com/devanlevin/DropBox?authkey=Gv1sRgCL_4l4PpvP_YsQE#5314956645280048530

is a solid ball with a radius of R=5cm and a charge density of $$\rho$$=-3$$\mu$$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

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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 -$$\rho$$ )??

for C i know that the field before is 0 since it is at the centre, how do i continue from there?

 

[Dell]

what i did so far is: $$E$$=epsilon0
i took a surface at the radius of the ball and said

$$\varphi$$=$$\oint$$EdA=EA=E(4$$\pi$$R²)

$$\varphi$$=Q/$$E$$=(V$$\rho$$)/$$E$$=(0.75$$\pi$$R³/$$E$$)

E(4$$\pi$$R²)=(0.75$$\pi$$R³/$$E$$)
E=($$\rho$$R)/(3$$E$$)

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=($$\rho$$R)/(3$$E$$)-($$\rho$$R)/(9$$E$$)
and i get

E=(2$$\rho$$R)/(9$$E$$)
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?

 

[nlightner]

I would first clarify the problem by asking for more information about the material of the hollow ball and the units for the charge density. Assuming that the charge density is in coulombs per cubic meter and the material of the hollow ball is non-conducting, I would approach the problem by using Gauss's Law.

First, I would find the total charge enclosed by the hollow ball by multiplying the charge density by the volume of the hollow ball (4/3π(R/3)^3). This would give me a total charge of -0.08μC enclosed by the hollow ball.

Next, I would use Gauss's Law to calculate the electric field at points A and B. Since the Gaussian surface for both points is a sphere with a radius of R/3, the electric field at both points would be equal to the total charge enclosed by the hollow ball divided by the surface area of the Gaussian sphere (4π(R/3)^2). This would give me an electric field of -60N/C at points A and B.

For point C, since it is located at the center of the big ball, the electric field would be zero. This is because the total charge enclosed by the Gaussian surface would be zero since the charge is distributed uniformly throughout the big ball.

In summary, the electric field at point A and B would be -60N/C and the electric field at point C would be 0N/C.