CourseNotes.15 - Figure 9 One quadrant of a charged disk...

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Figure 9: One quadrant of a charged disk. This is clearly going to be an integration problem, and it will work out very similarly to the electric field problems from last chapter. We begin by writing down the differential bit of potential from a differential piece of charge on the disk. dV = k dq r 0 Note that this definition of the differential piece of potential already has the assumption that V = 0 at infinity built in. As usual, the challenge is to express the differential piece of charge dq in terms of the charge density and a variable of integration. From the definition of a surface charge density, we can see that dq = σ dA where dA is a differential piece of area on the disk. There are a number of good ways to express the differential piece of area on a circular object. The easiest in this case is to take a sliver of area at the same radius (like a very small piece of a donut). Expressing the area in this manner leads to: dq = σ dA = σ ( π 2 r ) dr We must be very careful at this point. The r
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