Question

A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter 10.0cm , giving it a charge of -15.0μC .

I have figured out the answers for part A and B, but I am having trouble with solving C.

A: Find the electric field just inside the paint layer. E = 0 N/C
B: Find the electric field just outside the paint layer. E= 5.39×10^7 N/C
C: Find the electric field 6.00cm outside the surface of the paint layer.

Thanks in advance!

Answer 1

Do you know Gauss's law?

Answer 2

\[(1/4\pi \epsilon0) q/r^2 = (9*10^9) * (-15)/(0.06*0.06) =3.75*10^13 newton/coloumb\]

Answer 3

@prasadrajupvmv I also came up with that answer, but it's incorrect.

Answer 4

\[\huge E dA = {q_{enclosed}\over \epsilon_0}\]
\[\huge EA = {q_{enclosed}\over \epsilon_0}\]
\[\huge E(4\pi r^2)={q_{enclosed}\over \epsilon_0}\]
\[\huge E={q_{enclosed}\over (4\pi r^2)\epsilon_0}\]
\[\huge E={-15 X 10^{-6} C\over (4\pi {(5X 10^-2+6X10^-2)}^2)\epsilon_0}\]

Answer 5

I converted the radius to meters
You want the electric field 6cm outside of the sphere so the radius is 5 cm (half of the diameter of the sphere) plus the 6cm so the net radius is actually 11cm.

Answer 6

@roadjester I tried that as well, then realized I have been using the wrong numbers because I forgot to convert. Thank you!