Question
A disk of radius $R$ (Fig. P25. 71$)$ has a nonuniform surface charge density $\sigma=C r,$ where $C$ is a constant and $r$ is measured from the center of the disk. Find (by direct integration) the potential at $P .$
Step 1
The charge on this ring, $dq$, is given by the surface charge density $\sigma$ times the area of the ring. The area of the ring is $2\pi r dr$, so we have \[dq = \sigma dA = C r \cdot 2\pi r dr = 2\pi C r^2 dr.\] Show more…
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