Two parallel-plate capacitors are arranged perpendicular to the common axis. The separation ‘d’ between the capacitors is much larger than the separation '$l$' between their plates and than their size. The capacitors are charged to $q_{1}and{q}_2$  respectively. The magnitude of force F of interaction between the capacitors is $\frac{\left(x\right)q_1{q}_2{l}^y}{2\pi {\epsilon }_0{d}^z}$ . Find (x + y + z)

Let us consider the case when the capacitors are oriented so that the plates with like charges face each other. The field produced by the first capacitor on the axis at a distance x, from the positive plate is 

$E_x=\frac{q_1}{4\pi {\epsilon }_0}\left[\frac{1}{x^2}-\frac{1}{{(x+l)}^2}\right]\approx \frac{2q_{1}l}{4\pi {\epsilon }_0{x}^3}(x>>l)$

The force acting on the second capacitor situated at a distance d from the first one is

$F=q_2\left[E_d-E_{\left(d+l\right)}\right]=\frac{2q_1{q}_{2}l}{4\pi {\epsilon }_0}\left[\frac{1}{d^3}-\frac{1}{{(d+l)}^3}\right]$

$\approx \frac{3q_1{q}_2{l}^2}{2\pi {\epsilon }_0{d}^4}$

Alternatively:
We can consider the two capacitors as two short dipoles separated by a relatively large distance.

$F=\frac{-dU}{dr}$
$U=-\overrightarrow{p_2}·\overrightarrow{E_1}$ , where $\overrightarrow{E_1}$  is the electric field due to dipole of dipole moment $\overrightarrow{p_1}$  at the location of dipole of dipole moment  $\overrightarrow{p_2}$

$U\mathit{=}\mathit{-}\left(-p_2\widehat{i}\right)\mathit{·}\left(\frac{2K{p}_1}{r^3}\widehat{i}\right)\mathit{=}\frac{\mathit{2}\mathit{K}{\mathit{p}}_{\mathit{1}}{\mathit{p}}_{\mathit{2}}}{{\mathit{r}}^{\mathit{3}}}$

$F=\frac{-dU}{dr}=\frac{3K{p}_1{p}_2}{2r^4}$  (positive sign indicates repulsive force) 

$F=\frac{3q_1{q}_2{l}^2}{2\times 4\pi {\epsilon }_0{d}^4} \left(x=3,y=2,z=4\right)$