in 3D, electric field of a piont charge is inversely proportional to the square of distance while the potential is inversely proportional to distance. We can derive it from Coulomb's law.
however, I don't known how to derive the formula in 2D and 1D. I read in a book that electric potential of a point charge in 2D is proportional to the logarithm of distance.
How to prove it?
[Physics] Electric field and electric potential of a point charge in 2D and 1D
coulombs-lawelectrostaticsgauss-lawhomework-and-exercisesspacetime-dimensions
Best Answer
The Coulomb potential has the following forms for a positive charge in each dimensionality: \begin{align} \Phi_{\operatorname{1-d}}(r) &= -\frac{\sigma}{2\epsilon_0} r, \\ \Phi_{\operatorname{2-d}}(r) &= -\frac{\lambda}{2\pi \epsilon_0} \ln(r),\ \mathrm{and} \\ \Phi_{\operatorname{3-d}}(r) &= \frac{q}{4\pi\epsilon_0} \left(\frac{1}{r}\right). \end{align} The reason for this is that the electric field, defined as $-\nabla\Phi$ in general and $-\dfrac{\partial\Phi}{\partial r} \hat{r}$ in this case, times the measure of the boundary of a "ball" has to be a constant. In 1-d, a ball is a line and the measure of its boundary is just the number of points on its ends (i.e. 2). In 2-d the ball is a circle and the measure of its boundary is the circumference (i.e. $2\pi r$). In 3-d the ball is a sphere and the measure of its boundary is the sphere's surface area ($4\pi r^2$). Notice that those are exactly the quantities in the denominator when we calculate the electric field from the potentials: \begin{align} \vec{E}_{\operatorname{1-d}}(r) &= \frac{\sigma}{\epsilon_0} \left(\frac{\hat{r}}{2}\right), \\ \vec{E}_{\operatorname{2-d}}(r) &= \frac{\lambda}{\epsilon_0} \left(\frac{\hat{r}}{2\pi r}\right),\ \mathrm{and} \\ \vec{E}_{\operatorname{3-d}}(r) &= \frac{q}{\epsilon_0} \left(\frac{\hat{r}}{4\pi r^2}\right). \end{align}
The name of the law that implies this is known as Gauss's law.