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Discussion Quiz 3 Solution

Since this is only part of charge you need to

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Unformatted text preview: the volume into a small geometry where the density is nearly constant. For spherical geometries, you break it up into spherical shells, and for cylindrical geometries, you break it up into cylindrical shells. For planer, you break it up into thin sheets of width dx (or whatever variable that describes the variation of the density). The following is useful in these problems. dV= 4πr2 dr volume of thin spherical shell of width dr dV= 2πrL dr volume of thin cylindrical shell of length L dV = Adx volume of thin sheet of width dx Within these shells, you can figure out how much charge there is by using the simple rule: dq= ρ*dV, where p is the density at the location of the shell or sheet, and dV is chosen from the paragraph above. Since this is only part of charge, you need to integrate to get the total. The solutions below show you examples of how to figure out the limits of integration. When the point of interest is outside of the contained charge in the actual object, then you just need to figure out...
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