Final answer:
A) The electric field just inside the paint layer is -3.71 x 10^6 N/C. B) The electric field just outside the paint layer is also -3.71 x 10^6 N/C. C) The electric field 5.00 cm outside the surface of the paint layer is approximately -8.25 x 10^5 N/C.
Step-by-step explanation:
A) To find the electric field just inside the paint layer, we can use the formula for electric field due to a point charge:
E = k * q / r^2
where E is the electric field, k is the Coulomb's constant (9 x 10^9 Nm^2/C^2), q is the charge, and r is the distance from the charge.
In this case, the charge is -49.0 μC and the distance from the charge is the radius of the plastic sphere, which is 9.0 cm or 0.09 m. Plugging in these values, we get:
E = (9 x 10^9 Nm^2/C^2) * (-49.0 x 10^-6 C) / (0.09 m)^2 = -3.71 x 10^6 N/C
So, the electric field just inside the paint layer is -3.71 x 10^6 N/C.
B) To find the electric field just outside the paint layer, we can use the same formula. The distance from the charge is still the radius of the plastic sphere, so the electric field just outside the paint layer will be the same as just inside, -3.71 x 10^6 N/C.
C) To find the electric field 5.00 cm outside the surface of the paint layer, we just need to plug in the new distance from the charge, which is 14.0 cm or 0.14 m, into the formula. Using the same calculations, we find that the electric field 5.00 cm outside the surface of the paint layer is approximately -8.25 x 10^5 N/C.