What Is the Electric Field at Various Distances From a Charged Spherical Shell?

[22steve]

Homework Statement

The figure below shows a spherical shell with uniform volume charge density ρ = 1.63 nC/m3, inner radius a = 6.0 cm, and outer radius b = 2.70a. What is the magnitude of the electric field at the following radial distances:
r=1.5, r=b, r=3b

Homework Equations

Q= sigma dV
flux = E x A
flux = charge enclosed/permittivity constant (8.85e-12)
k x q/r^2 = E (electric field outside a spherical shell of charge)

E = k x r(q/R^3) R= radius r= distance from center of shell to point where E is measured.

The Attempt at a Solution

I've tried integrating, and substituting different values of q into the Electric Field equations. I've gotten 18094, 82, and 1813 as answers. I would show my integration techniques, but it's kinda hard to do on the keyboard. haha

 

[rootX]

Part I:
When r = 1.5, what's charge inside the sphere of r = 1.5 m?

When r = b or r =3b, that sphere encloses all the charge. So, Q is same but Area changes for part II and part III

 

[nlightner]

I would first like to commend you for attempting to solve the problem using equations and calculations. However, there are a few points that need to be addressed.

Firstly, in your equations, you have used the symbol "sigma" which typically represents surface charge density, but the problem states that the charge density is uniform throughout the volume of the spherical shell. Therefore, we should use the symbol "rho" instead.

Secondly, it seems that you have tried to use different equations for finding the electric field at different radial distances. However, for a spherical shell with uniform charge density, the electric field at any point outside the shell is given by E = kq/r^2, where q is the total charge enclosed within the Gaussian surface of radius r.

Finally, when solving for the electric field, it is important to consider the direction as well. Since the problem does not specify the sign of the charge, we cannot determine the direction of the electric field. Therefore, we can only solve for the magnitude of the electric field at each radial distance.

To solve this problem, we can use Gauss's Law, which states that the electric flux through a closed surface is equal to the charge enclosed divided by the permittivity constant. In this case, the closed surface can be taken as a spherical Gaussian surface with radius r.

At r = 1.5, the charge enclosed within the Gaussian surface is 0 since it is outside the spherical shell. Therefore, the electric field is 0.

At r = b, the charge enclosed within the Gaussian surface is the total charge of the spherical shell, which is given by q = rho x (4/3)π(b^3 - a^3). Substituting the values given in the problem, we get q = 1.63 x (4/3)π[(2.70a)^3 - (6.0 cm)^3] = 1.63 x 0.001 x 4π(2.70^3 - 6.0^3) = 1.63 x 0.001 x 4π(19.683) = 0.1285 nC. Therefore, the electric field at r = b is given by E = kq/r^2 = (8.99 x 10^9)(0.1285 x 10^-9)/(2.70 x 0.06)^2 = 0.