Figure 8: Charged shell with a positive point charge at its center.by our Gaussian surface of radiusrincludes the central point charge as well as the charge fromthe enclosed portion of the shell.Lets first calculate the charge from the shell enclosed by theGaussian sphere. This is not a simple as multiplying the charge density times the volume becausethe charge density is a funtion ofr.We must therefore sum up all of the contributions at eachvalue of r (i.e. integrate).Qenc,shell=Zρ dV=ZraAr04πr02dr0= 4πAZrar0dr0= 2πA(r2-a2)Plugging this into Gauss’s law and exploiting the symmetry gives the electric field with in the shell.E=Qenc0A=104πr2q+ 2πA(r2-a2)=A20+q-2πAa24π0r2The question at hand is what does the value ofAneed to be such thatEis independent ofr?Clearly, the first term is a constant, but the second term has somerdependence. In order forEto be a constant, the numerator in the second term must therefore vanish.
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