Problem 3.2.4: Two parallel conducting plates

Each charged surface alone produces the field of Problem 3.2.3, of magnitude ${\rho }_S/(2{\epsilon }_0)$, pointing away from $+{\rho }_S$ and towards $-{\rho }_S$ (Figure 3.2.20a):

• the sheet $+{\rho }_S$ at $x=d$ gives $-\frac{{\rho }_S}{2{\epsilon }_0}{a}_x$ for $x<d$ and $+\frac{{\rho }_S}{2{\epsilon }_0}{a}_x$ for $x>d$, and
• the sheet $-{\rho }_S$ at $x=0$ gives $+\frac{{\rho }_S}{2{\epsilon }_0}{a}_x$ for $x<0$ and $-\frac{{\rho }_S}{2{\epsilon }_0}{a}_x$ for $x>0$.

For $0<x<d$ both contributions point in the $-x$ direction and add. For $x<0$ and $x>d$ they are opposite and cancel:

The same result follows directly from the fact that the plates are conductors. Take a small pillbox of face area $A$ across the lower surface of the upper plate (Figure 3.2.20b), with one face inside the metal, where $E=0$ in electrostatic equilibrium, and the other face in the gap, whose outward normal is $-a_x$. The side contributes nothing (the field in the gap is normal to the plates), and only the gap face counts:

so that $E_x=-{\rho }_S/{\epsilon }_0$, as above. A pillbox with one face in the metal of the lower plate gives the same value, and a pillbox with one face in the metal and the other outside the pair encloses no charge (the outer faces are uncharged), so the field outside vanishes.

Both routes agree, and the field points from the positive plate ($x=d$) to the negative plate ($x=0$), i.e. along $-a_x$, as it should. The second route shows a general rule: just outside any conductor the normal component of $D$ equals the local surface charge density, $D_n={\rho }_S$ (Section 3.2.5). The field between the plates is uniform as long as $d$ is small compared with the plate dimensions (fringing at the edges neglected). This is the parallel-plate capacitor of Section 3.4.