A nonconducting spherical shell of inner radius R1 and outer radius R2 contains a uniform volume charge density rho throughout the shell. Use Gauss's law to derive an equation for the magnitude of the electric field at the following radial distances rrr from the center of the sphere. Your answers should be in terms of rho, R1, R2, r, ε0, and π.

(a) r < R1
(b) R1 (c) r>R2

a) 0; b) 0; c) E= ρ(R2^3-R1^3)/(3ε0r^2)

Explanation:

To solve this problem, we should carefully analyse Gauss's law. Equation of the gauss's law says, that the flux of the electric field for the given enclosed surface is equal to the total charge, which is enclosed by this surface, divided by ε0.

Note, that the flux is a multiplication of the electric field and the area of the given enclosed surface.

From this idea, we can get the following answers without mathematical analysis:

a) when the distance is less then R1, there are no charges stored inside the area, so the flux will be equal to 0:

E*A=Σq/ε0, as q=0, then E=0 V/m;

b) At the distance R1, we have similar situation- on R1 there are no charges being stored inside the surface. As a result, we can repeat the calculations from above: E*A=Σq/ε0

c) For the case, when r>R2, we have the entire charge of the sphere being stored inside the enclosed surface. In this case, the enclosed surface is a sphere, with radius r.

For such sphere, the area of it can be calculated as: A=4πr^2

To calculate electric field, we should also analyse the total charge inside the given surface. As the charge is stored in the volume, we should multiply charge's density with the given volume. Then, the total charge can be found as: Q=ρ* 4π(R2^3-R1^3)/3. Note, that the total volume of the charged material is equal to the subtraction of the volume of the sphere with the radius R1 and radius R2.

Using Gauss's Law, we can derive the following:

E*A=Q/ε0

E*4πr^2=ρ* 4π(R2^3-R1^3)/(3ε0)

From where, E=(ρ(R2^3-R1^3))/(3ε0r^2)

I would like to add, that we can also find the filed in the area between R1 and R2 (R1<r<R2). In this case, the area of the enclosed surface will be the same, as above, bu the total charge will be limited by the volume inside sphere with the radius r. Then, the field will be calculated as:

E=(ρ(r^3-R1^3))/(3ε0r^2)

energy can cross the boundaries of a closed system in the form of heat and mass

explanation:

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