A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter 12.0 cm, giving it a charge of - 49.0 mC. Find the electric field (a) just inside the paint layer; (b) just outside the paint layer; (c) 5.00 cm outside the surface of the paint layer.

A) E = 0 N/C

B) E = 3.06 x 10^ (7) N/C

C) E = 1.52 x 10^ (7) N/C

Explanation:

A) Electric Flux = EA = Q / εo.

However, the "enclosed charge" Q "just under the paint layer" is 0, so

the field E = 0 N/C (everywhere inside the paint layer).

B) Q = - 49 mC

Diameter = 12cm = 0.12m

εo = (8.854x10^ (-12))

EA=Q/εo

A = Area

E x 4π (0.12²) = (-49x10^ (-3)) / (8.854x10^ (-12))

E = - 3.06 x 10^ (7) N/C

Electric field can never be negative, thus, we take the absolute value as;

E = 3.06 x 10^ (7) N/C

C) Now, Area (A) = 4π (0.12 + 0.05) ²

4π (0.17) ²

E x 4π (0.17) ² = (-49x10^ (-3)) / (8.854x10^ (-12))

E=-0.015238 x 10^ (9) = - 1.52 x 10^ (7) N/C

Again, Electric field can never be negative, thus, we take the absolute value as;

E = 1.52 x 10^ (7) N/C