Four point charges - q ,   + q ,   + q and - q are placed on y-axis at y = - 2 d ,   y = - d …

1. $\mathrm{E}\propto \frac{1}{{\mathrm{D}}^4}$
2. $\mathrm{E}\propto \frac{1}{\mathrm{D}}$
3. $\mathrm{E}\propto \frac{1}{{\mathrm{D}}^3}$
4. $\mathrm{E}\propto \frac{1}{{\mathrm{D}}^2}$

Solution

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 ${\mathrm{\theta }}_2\approx 0$ .
$\therefore {\mathrm{E}}_{\mathrm{n}\mathrm{e}\mathrm{t}}={\mathrm{E}}_{+\mathrm{q}}-{\mathrm{E}}_{-\mathrm{q}}\approx \frac{\mathrm{K}\mathrm{q}}{{\mathrm{D}}^2+{\mathrm{d}}^2}-\frac{\mathrm{K}\mathrm{q}}{{\mathrm{D}}^2+4{\mathrm{d}}^2}$
$=\frac{\mathrm{K}\mathrm{q} 3{\mathrm{d}}^2}{\left({\mathrm{D}}^2+{\mathrm{d}}^2\right)\left({\mathrm{D}}^2+4{\mathrm{d}}^2\right)}\approx \frac{3\mathrm{K}\mathrm{q}{\mathrm{d}}^2}{{\mathrm{D}}^4}$
$\mathrm{E}\propto \frac{1}{{\mathrm{D}}^4}$

Asked in: JEE Main 2019 (09 Apr Shift 2)

Practice more Electrostatics questions on Aicharya