# What does Ohm's law mean in this context?

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?

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 $$\phi$$.