Physics - Electrostatics Question with Solution | TestHub

The electric field at$\mathrm{P}$is superposition of the fields due to all four charges. The electric field due to the two$+\mathrm{q}$charges is

${\mathrm{E}}_{+\mathrm{q}}=\frac{\mathrm{K}\mathrm{q}}{{\mathrm{D}}^2+{\mathrm{d}}^2}{\cos \mathrm{\theta }}_1\times 2$
${\mathrm{E}}_{-\mathrm{q}}=\frac{\mathrm{K}\mathrm{q}}{{\mathrm{D}}^2+4{\mathrm{d}}^2}{\cos \mathrm{\theta }}_2\times 2$
The field due to$+\mathrm{q}$is towards right and the field due to$-\mathrm{q}$is towards left. Also${\mathrm{\theta }}_1\approx 0$and${\mathrm{\theta }}_2\approx 0$.
$\therefore {\mathrm{E}}_{\mathrm{n}\mathrm{e}\mathrm{t}}={\mathrm{E}}_{+\mathrm{q}}-{\mathrm{E}}_{-\mathrm{q}}\approx \frac{\mathrm{K}\mathrm{q}}{{\mathrm{D}}^2+{\mathrm{d}}^2}-\frac{\mathrm{K}\mathrm{q}}{{\mathrm{D}}^2+4{\mathrm{d}}^2}$
$=\frac{\mathrm{K}\mathrm{q} 3{\mathrm{d}}^2}{\left({\mathrm{D}}^2+{\mathrm{d}}^2\right)\left({\mathrm{D}}^2+4{\mathrm{d}}^2\right)}\approx \frac{3\mathrm{K}\mathrm{q}{\mathrm{d}}^2}{{\mathrm{D}}^4}$
$\mathrm{E}\propto \frac{1}{{\mathrm{D}}^4}$