Four point charges $-\mathrm{q},\mathrm{ }+\mathrm{q},\mathrm{ }+\mathrm{q}$ and $-\mathrm{q}$ are placed on y-axis at $\mathrm{y}=-2\mathrm{d},\mathrm{ }\mathrm{y}=-\mathrm{d},$ and $\mathrm{y}=+2\mathrm{d},$ respectively. The magnitude of the electric field $\mathrm{E}$ at a point on the x-axis at $\mathrm{x}=\mathrm{D},$ with $\mathrm{D}\gg \mathrm{d},$ will behave as:

• Correct answer
  A

  $\mathrm{E}\propto \frac{1}{{\mathrm{D}}^4}$
• B

  $\mathrm{E}\propto \frac{1}{\mathrm{D}}$
• C

  $\mathrm{E}\propto \frac{1}{{\mathrm{D}}^3}$
• D

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

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

Hence, the correct answer is option: 'A'.