= Solution
The microscopic pair potential falls as $-r^{-6}$, whereas integrating over two half-spaces gives an interaction \b[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 is
$$
\boxed{\frac FA=-\frac{dW}{dd}=-\frac{A_H}{6\pi d^3}.}
$$
At $d=100\,\mathrm{nm}$ and the supplied effective $A_H$,
$$
\boxed{F/A=-\frac{2.5\times10^{-19}}{6\pi(10^{-7})^3}
=-13.3\,\mathrm{Pa}.}
$$
The minus sign denotes decreasing separation. \b[A total force cannot be calculated without the interacting area]: for area $A$ in square metres, $F\simeq-(13.3\,\mathrm{N\,m^{-2}})A$. This uses the supplied nonretarded half-space model; finite film thickness, geometry and retardation at larger distances can change it.
\Image[/past-exam-of-the-mphil-in-micro-and-nanotechnology-enterprise-course-of-the-university-of-cambridge/2022/paper-2-hamaker-dlvo.png]
{title=Normalized Hamaker distance laws and illustrative salt-dependent DLVO barriers}
{description=Original analytic models. Microscopic energy and plate energy per area are normalized separately. DLVO examples hold an explicitly chosen repulsive amplitude fixed while changing screening length; they are not fits to a specified surface chemistry.}
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