Physics - Elecrostatics Question with Solution | TestHub

$\mathrm{U}=\frac{\mathrm{k}\left(-\mathrm{q}\right)\left(\mathrm{q}\right)}{\mathrm{d}}+\frac{\mathrm{k}\left(-\mathrm{q}\right)\left(\mathrm{Q}\right)}{\left(\frac{\mathrm{d}}{2}-\mathrm{D}\right)}+\frac{\mathrm{k}\left(\mathrm{q}\right)\left(\mathrm{Q}\right)}{\left(\mathrm{D}+\frac{\mathrm{d}}{2}\right)}$
$=\frac{-\mathrm{k}{\mathrm{q}}^2}{\mathrm{d}}-\frac{\mathrm{k}\mathrm{Q}\mathrm{q}}{\left(\mathrm{D}-\frac{\mathrm{d}}{2}\right)}+\frac{\mathrm{K}\mathrm{Q}\mathrm{q}}{\left(\mathrm{D}+\frac{\mathrm{d}}{2}\right)}$
$=\frac{-\mathrm{k}{\mathrm{q}}^2}{\mathrm{d}}+\mathrm{k}\mathrm{Q}\mathrm{q}\left(\frac{-\mathrm{D}-\frac{\mathrm{d}}{2}+\mathrm{ }\mathrm{D}-\frac{\mathrm{d}}{2}}{{\mathrm{D}}^2-\frac{{\mathrm{d}}^2}{4}}\right)$
$=\frac{-\mathrm{k}{\mathrm{q}}^2}{\mathrm{d}}-\frac{\mathrm{k}\mathrm{Q}\mathrm{q}\mathrm{d}}{{\mathrm{D}}^2-\frac{{\mathrm{d}}^2}{4}}$
Since, D>>d
$\mathrm{U}=\frac{-\mathrm{k}{\mathrm{q}}^2}{\mathrm{d}}-\frac{\mathrm{k}\mathrm{Q}\mathrm{q}\mathrm{d}}{{\mathrm{D}}^2}=\frac{1}{4\mathrm{\pi }{\mathrm{ϵ}}_0}\left[-\frac{{\mathrm{q}}^2}{\mathrm{d}}-\frac{\mathrm{q}\mathrm{ }\mathrm{Q}\mathrm{ }\mathrm{d}}{{\mathrm{D}}^2}\right]$