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

work: w = fd (given)

work done by chet   =50*1 = 50 j

work done by mika =40*2 = 80 j

work done by sara =30*1.5 = 45 j

work done by bill =60*0.5 = 30 j

students in order from the greatest amount of work done to the least is mika> chet> sara> bill

hence option (b) is correct.