Nonconducting nonuniform non-nonsymmetric cylinder A long, nonconducting, solid cylinder of radius 4.0 cm has a nonuniform volume charge destiny $p$ that is a function of radial distance $r$ from the cylinder axis: $p=Ar^2$. For $A=2.5 \mu /m^5$, what is the magnitude of the electric field at (a) r = 3.0 cm and (b) r = 5.0 cm?

So, clearly, I overdid it. See #2 for a legible version of solution to (a). For me, apparently, that was step one. So now the question is, WHAT ABOUT REPULSION? See below, the concentric circles on top right: For every differential excerpt $dr$, the enclosed charge $q_enc$ generates an electric field perpendicular to the circumference. This field points both ways - outward and inward. Now, don't inward electric field lines of $r+dr$ cancel those of $r$ and form a certain $E_{net_{r}}$?

At first, I've tried to formulate an integral based on $E_r - E_{r+dr}$. You can imagine how that went ($\int ...(r+dr)dr$...). Then, I thought instead I should do $E_{net}=E_r-E_{R-r}$. It led to the circled big formula , derived after an hour of globbling - above.

Something was wrong. This equation is clearly not elegant. And when physics throws you an elegant problem, you must get an elegant answer. Then I realized something... do you know what it is? It made me mad. I'll leave it to you to figure out ;)

NOTE Spoilers ahead! If you don't know the answer to above, give it a thought!

So, what if we DID have to resort to such measures? What if there was no internal symmetric cancellation? How would we resolve that issue? Say, for example, we have a big rectangle - and its charge density obeys the same pattern as that cylinder. If we place the rectangle on origin with length parallel to y axis, let $r$ be $x$. What would be its electric field lines (assume it's infinitely long)? NOW my issue is real.  Note by John Muradeli
4 years, 6 months ago

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