 Solution

The microscopic pair potential falls as $-r^{-6}$, whereas integrating over two half-spaces gives an interaction energy per area $W(d)=-A_H/(12\pi d^2)$. The plate attraction therefore decays much more slowly with distance because many pairs contribute. The quantities have different dimensions, so the shared sketch uses separately normalized energies at a reference separation rather than pretending that joules and joules per square metre are directly comparable.

The attractive pressure isAt $d=100\mathrm{nm}$ and the supplied effective $A_H$,The minus sign denotes decreasing separation. A total force cannot be calculated without the interacting area: for area $A$ in square metres, $F\simeq -(13.3\mathrm{N}{\mathrm{m}}^{-2})A$. This uses the supplied nonretarded half-space model; finite film thickness, geometry and retardation at larger distances can change it.