Final answer:
After removing a core of radius 0.266R from a uniformly charged sphere, the electric field magnitude at point P will be 98.12% of the original electric field E1, using the concept of superposition and Gauss's law for a sphere.
Step-by-step explanation:
To determine the electric field at a point P after a core of radius 0.266R is removed from a uniformly charged sphere, we can use the principle of superposition. Initially, with the solid ball of radius R, the electric field at point P (which is 2.56R away from the ball's center) is E1. We know from Gauss's law for a sphere with uniform volume charge density, the electric field outside the sphere is proportional to the total charge and inversely proportional to the square of the distance from the center of the sphere.
When a spherical core of radius 0.266R is removed, the remaining shell contributes to the electric field at point P. The core that was removed would have contributed to the original electric field E1 by an amount proportional to its volume, hence its charge. Since the core removed is smaller than the original sphere (0.266R compared to R), it has a volume that is (0.266)^3 = 0.0188 times the volume of the larger sphere. Assuming uniform charge density, this core would contribute the same fraction to the electric field intensity. Thus, we subtract this fraction from 1 to find the remaining fraction of E1 due to the outer shell.
The fraction of E1 that remains is 1 - 0.0188, which equals 0.9812 (or 98.12%). Therefore, the electric field magnitude at point P after the core is removed will be 98.12% of E1.