distance from the center of the shell, select Aso the field does not depend on the distance. We use a Gaussian surface in the form of a sphere with radius rg, concentric with the spherical shell and within it (a < rg< b). Gauss’ law will be used to find the magnitude of the electric field a distance rgfrom the shell center. The charge that is both in the shell and within the Gaussian sphere is given by the integral qdVs=zρover the portion of the shell within the Gaussian surface. Since the charge distribution has spherical symmetry, we may take dVto be the volume of a spherical shell with radius rand infinitesimal thickness dr: dVr dr=42π. Thus, ()22224442.gggrrrsgaaaAqrdrrdrArdrArarπρπ====−³³³The total charge inside the Gaussian surface is
This is the end of the preview. Sign up
access the rest of the document.