A thick sperical shell with inner radios Ri and outer radious Ro has uniform den

Question

A thick sperical shell with inner radios Ri and outer radious Ro has uniform density p.What is the total charge on the shell Q? What is the electric field magnitude E in terms of Ri,Ro,p, and distance r from the center of the shell?graph the electric field magnitude versus distance from the center of the shell for Ri=1.0cm, Ro=2.0 cm and p=10^-3C/m^3.The graph may be drawn,but use a ruler and correctly measure the distance between the points along both the E and r axes.Include the points r=0.0cm ,0.5cm, 1.0cm,1.5cm,2.0cm,2.5cm, and 3.0 cm on the graph.

Explanation / Answer

Let a = inner radius, b = outer radius, p = volume charge density, Q = total charge.

Consider three regions:

I - 0 < r < a
II - a <= r < b
III - r >= b

In region I, there is no charge so by Gauss's Law the E field is 0 ---> E = 0

In region II, the field is dependent on how large the raidus is and we need to do some math. For a <= r< b, the charge enclosed in a sphere of radius r is:

q(r) = p*4*pi/3*(r^3 - a^3)

Define Q = p*4*pi/3*(b^3 - a^3) ---> p = Q/{4*pi/3*(b^3 - a^3)}

So q(r) = Q*(r^3 - a^3)/(b^3 - a^3)

Now apply Gauss's Law :

Integral (E*dA) = q(r)/e0 ---> E*4*pi*r^2 = Q*(r^3 - a^3)/{e0*(b^3 - a^3)}

E = Q*(r^3 - a^3)/{4*pi*e0*r^2*(b^3 - a^3)} for a<= r < b

In region III, a Gaussian sphere encloses the entire charge distribution so:

Integral(E*da) = Q/e0 ---> E = Q/(4*pi*e0*r^2) r >= b

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