3. Electric force on surface charge: Consider a surface charge of local surface density 0' on an area element da = da ii of a smooth thin surface, where ii is the local unit vector normal to the surface (left gure of Figure 2). There can be an arbitrary charge distribution elsewhere in space. We would write down the electric force acting on this surface charge element as dF = O'Eda. (1) However, we have seen in class that the electric eld has a discontinuity across the charged surface, so it is unclear what value of E should be used in the above equation. The correct answer is that we should used the \"average\" value: 1 dF=0(E++EJd (m where E, and E- are the local electric eld immediately above and below the surface, respectively. As we showed in class, the discontinuity is mm=3 (m m Z , local normal direction FlED: volumetric charge density BLUE: normal component of the electric field Figure 2: Left: An (idealized) surface charge element. Right: A simple model for a surface charged layer with a nite thickness. Justify this conclusion in the following two ways: (a) (1 point) Use the intuition that according to Newton's 3rd law, the infinitesimal surface charge element cannot exert any force on itself - the force must come from electric field induced by all charge in space excluding this surface charge element. Further use the result we derived in class for the electric field around an infinitely large charged surface. [Hint: Think about which part of the charge distribution is responsible for the discontinuity in Eq. (3).] (b) (2 points) An infinitely thin surface charge does not exist in the real world. It can be considered as an idealization of a charge distribution within a layer of small but finite thickness. Let us consider a simple model of this situation: as shown in the right figure of Figure 2, a volumetric charge density p(z) is stratified along the normal direction (n or the z axis) and is localized within a small range of z. p(z) is otherwise arbitrary. Far away from this surface layer, the normal component of the electric field asymptotes to En, + and En,-, respectively, along the positive z or negative z direction, respectively. Show that the total electric force per unit area along the normal direction acting on this charge distribution is fn = (En, + + En,-) 0, (4)where 0' = f dz p(z) is the surface charge density integrated across the surface layer. [Hint Consider applying Gauss's 1aw.]