Suppose we have an infinite cylinder of radius $R$ in vacuum with some surface charge density $\sigma=\sigma(\varphi)$, where $\varphi$ is azimuthal angle. We want to calculate the force acting on one half of the cylinder (per unit length).
Now I've done the dirty job and found the electric field inside ($\vec{E}_{in}$) and outside of the cylinder ($\vec{E}_{out}$). There is of course a discontinuity of the electric field due to the surface charge:
$$E_{out}^{\perp}-E_{in}^{\perp}=\frac{\sigma}{\varepsilon_0}$$
To calculate the force I can use Maxwell's tensor $\hat{T}$ and the momentum theorem.
$$\vec{F} = \oint_{\partial V} \hat{T}\vec{dS}$$
Given the condition that all this is happening in vacuum so the only physical thing here is the cylinder itself, I should be free to choose the volume $V$ from the theorem as I wish, as long as it only encloses the half of the cylinder (per unit length) stated in the problem. So I choose a half cylindrical surface with the radius $R+dR$. So when I calculate the tensor in the surface integral from the theorem I use $\vec{E}_{out}$ for the curved part of $\partial V$ and $\vec{E}_{in}$ otherwise.
The other way to find the force is to directly calculate it from the general formula:
$$\vec{F} = \int \rho(\vec{r})\vec{E}(\vec{r})dV$$
Here it becomes:
$$\vec{F} = \int_S \sigma(\vec{r})\vec{E}(\vec{r})dS$$
Where $S$ is the surface of the half cylinder on which the force acts. The problem is – which $\vec{E}$ should I use? The surface is just along the discontinuity line of $\vec{E}$.
Best Answer
Because of the discontinuity of $\vec E$, the solution isn't necessarily obvious. Surface charges are most easily understood as the limit of a volume charge density as the thickness of this volume goes to zero.
Consider a sheet of charge of density $\sigma$. Let us choose the $x$-axis to be normal to the sheet. To treat the surface charge as a thin sheet of volume charge, let $\rho(x)$ denote the volume charge density, confined to the region $0<x<\delta$. Let $E(x)$ denote the normal component of $\vec E$ (the tangential component is already continuous across the sheet of charge, so calculating the force due to this component is not problematic). We will let $E_+ = E(\delta)$ and $E_- = E(0)$ for convenience, where $E_+-E_- = \sigma/\epsilon_0$. Now, the normal force per unit area on the sheet of charge can be expressed as $$f=\int\limits_{0}^\delta\rho(x)E(x)dx.$$ Using Gauss' law $dE/dx = \rho/\epsilon_0$, $$f=\epsilon_0\int\limits_{0}^\delta\frac{dE}{dx}E\ dx = \left.\frac 1 2 \epsilon_0 E^2\right\rvert_0^\delta =\frac 1 2 \epsilon_0(E_+^2-E_-^2) = \frac 1 2 \epsilon_0(E_+-E_-)(E_+ + E_-)= \sigma \frac{E_+ + E_-}{2}.$$
So the short answer is: use the mean of the E-fields on either side of the surface charge for the purpose of calculating the force on this charge. Note that we didn't have to make assumptions about how exactly the charge is distributed along the thickness of the sheet.
We can quickly verify this result on a slightly simpler example. Consider a parallel-plate capacitor of area $A$, plate spacing $a$ and surface charge density $\sigma$, with vacuum or air as dielectric. Due to the opposite charges on the two plates, they attract each other, and we wish to calculate the force on either plate. Note that the $E = \sigma/\epsilon_0$ between the plates and $E = 0$ inside the plates (i.e. on the other side of the surface charge densities), so their mean is $\sigma/2\epsilon_0$, and the force per unit area on each plate is $\sigma^2/2\epsilon_0$.
We can also calculate the force per unit area using the method of virtual work. With $\mathscr E_{cap}$ as the energy stored in the capacitor,
$$ f = \frac{1}{A}\frac{d}{da}\mathscr E_{cap}=\frac{1}{A}\frac{d}{da}\frac{Q^2}{2C}=\frac{1}{A}\frac{d}{da}\frac{\sigma^2A^2}{2\epsilon_0A/a} = \frac{\sigma^2}{2\epsilon_0},$$ in agreement with the previous result.