# How do I go from the vectoral form of Ohm's law to the scalar form of Ohm's law?

The vectoral form of Ohm's law is $$\boldsymbol{J}=\sigma\boldsymbol{E}\text{ }\left[\dfrac{\text{A}}{\text{m}^{2}}\right]$$ I want to shown that $$I=\sigma V_{ab}$$ In other words, I want to show that $${\displaystyle \int_{S}}\boldsymbol{J}\cdot\vec{ds}=\sigma\cdot-{\displaystyle \int_{b}^{a}}\boldsymbol{E}\cdot\vec{d\ell}$$ I find myself stuck at this point. On the left hand side of the equation is the surface integral of $$\boldsymbol{J}$$ over $$\boldsymbol{S}$$, where $$\boldsymbol{S}$$ is the surface through which charge is propogating. On the right hand side is the integral of an electron moving from point b to point a. The movement of the electron is perpendicularly through the surface area that is being integrated, so the integrals have no relationship apparent to me. So how do you go from $$\boldsymbol{J}=\sigma\boldsymbol{E}\text{ }\left[\dfrac{\text{A}}{\text{m}^{2}}\right]$$ to $${\displaystyle \int_{S}}\boldsymbol{J}\cdot\vec{ds}=\sigma\cdot-{\displaystyle \int_{b}^{a}}\boldsymbol{E}\cdot\vec{d\ell}\text{ ?}$$

• Why do you think that $I = \sigma V$ is true? As your third equation shows, it's not even dimensionally consistent with $J = \sigma E$. Commented Feb 13, 2021 at 5:48

If $$V$$ is the electric voltage $$\Delta\phi$$ then ohms law for a cable is not what you have written: $$V=\frac{L}{\sigma A}I\equiv RI,$$ where $$L$$ is the length of the cable (the total resistance $$R$$ increases if you look at the path of the current thrue more cable), $$A$$ is the cross section area of the cable and $$\sigma(=nq^2\tau/m)$$ is the electrical conductivity
As you said we have $$\vec j=\sigma \vec E. \,\,\,\,\,\, (1)$$ Moreover we have $$I=\int_A \vec j \cdot d\vec A=|\vec j|A, \,\,\,\,\,\, (2)$$ because $$\vec j\parallel d\vec A$$. Furthermore we have $$V=-\int_a^b \vec E \cdot d\vec l=EL, \,\,\,\,\,\, (3)$$ because $$\vec E\parallel d\vec l$$ ($$\vec j, \vec E, d\vec A, d\vec l$$ are all in direction of the cable). Finally we get $$\frac{I}{A}\stackrel{(2)}{=}|\vec j|\stackrel{(1)}{=}\sigma E\stackrel{(3)}{=}\sigma \frac{V}{L}\,\,\,\,\,\,\,\,\text{or}\,\,\,\,\,\,\,\,V=\frac{L}{\sigma A}I,$$ which is the expression we were looking for.