Spherical cavity electric field

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A positive charge of 2800.00 C is uniformly distributed over a sphere with a radius of 6.00 m, and a spherical cavity with a radius of 3.00 m is cut out, removing some charge. The electric field at point P, located 19.00 m from the center of the original sphere, is calculated using the principle of superposition, subtracting the electric field of the smaller sphere from that of the larger sphere. The charge of the smaller sphere is determined to be 3500 C through proportional calculations. The user is seeking assistance in resolving discrepancies in their calculations for the electric field magnitude.
GravityGirl
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A positive charge Q = 2800.00 C is uniformly distributed over the volume of a sphere of radius R = 6.00 m. Suppose a spherical cavity of radius R/2 is cut out of the solid sphere, the center of the cavity being a distance R/2 from the center of the original solid sphere (see figure). The cut-out material and its charge are discarded. What is the magnitude of the electric field produced by this new charge distribution at point P, a distance r = 19.00 m from the center of the original sphere?


so i am going to use the principle of superposition

E=Ebigsphere-Elittlesphere

inorder to find E i must find the charge and i can do so by using porportions

Qb/Vb=Ql/Vl

that is 28000/904.7787=Ql/113.0973 and Ql=3500

so i am going to use the formula E=KQ/R^2 for both big a little spheres

and i have E=(K28000/19^2)-(K3500/16^2)

with this i am not getting the right solution...please help me find where i am going wrong
 
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Looks OK to me. Where is point P with respect to the two spheres? (Since we don't have the diagram, it's up to you to describe it carefully.)
 
did you ever figure this out?
 

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