An insulator in the shape of a spherical shell is shown in cross-section above. The insulator is defined by an inner radius a = 4.0 cm and an outer radius b = 6.0 cm and carries a total charge of Q = + 9.0 μC (1 μC = 10-6 C). (You may assume that the charge is distributed uniformly throughout the volume of the insulator). What is Ey, the y-component of the electric field at point P which is located at (x,y) = (0, -5.0 cm) as shown in the diagram?

Question

An insulator in the shape of a spherical shell is shown in cross-section above. The insulator is defined by an inner radius a = 4.0 cm and an outer radius b = 6.0 cm and carries a total charge of Q = + 9.0 μC (1 μC = 10-6 C). (You may assume that the charge is distributed uniformly throughout the volume of the insulator).  What is Ey, the y-component of the electric field at point P which is located at (x,y) = (0, -5.0 cm) as shown in the diagram?

anonymous · Answer

I used the equation for surface area of a Gaussian surface

anonymous · Answer

with the integral of Electric field = charge /E_0 but still can't get it

anonymous · Answer

So you know that this is a question pertaining to Gauss' Law, which states: $$\phi =\int\limits_{}^{}EdA =\frac{ Q(enclosed) }{ \epsilon0 }$$

anonymous · Answer

sorry I'm new at this and that took me forever. lol

anonymous · Answer

so you know that E(4 pi P^3)= (Q enclosed)/E_0

anonymous · Answer

that would be E*(....) btw

anonymous · Answer

Q enclosed is 4/3 pi (p^3 - a^3) and Q total=4/3 pi (b^3-a^3)

Q enclosed = Qtotal(p^3-a^3)/(b^3-a^3)

so E=[Qtot/(4piE_0 p^2)]*[(p^3-a^3)/(b^3-a^3)]

anonymous · Answer

i really hope that makes sense

anonymous · Answer

Yeah i think I actually used that formula more derived and had r: ro/3E_0*(b^3-a^3)/r^2