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In this problem Griffiths states that the potential at radius $a$ and angle $\phi$ is $V(a,\phi) = \frac {V_0 \phi}{2 \pi}$

And yes that satisfies the boundary conditions, that at $ \phi = \pi$ , $ V=\frac{V_0}{2}$ , at $ \phi = - \pi$ , $ V = - \frac {V_0}{2}$

But how are we sure that at any other angle between $ - \pi $ and $ \pi$, that the potential obeys the equation? He says 'According to Ohm's law, $V(a,\phi) = \frac {V_0 \phi}{2 \pi}$ '

Where from Ohm's law did that come from?

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The current $I$ is constant, and resistivity $r$ is constant so the resistance is linear with arc length around the cylinder. You know the potential drop around the whole circle, so you can calculate the potential drop around a given angle $\phi$.

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