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Step-by-Step Solution
Step 1: Identify the Configuration of Charges
We have four charges placed along the y-axis:
Charge –q at y = –2d
Charge +q at y = –d
Charge +q at y = +d
Charge –q at y = +2d
We want the net electric field at a point on the x-axis at x = D (where D ≫ d).
Step 2: Express the Electric Field Components
Each charge creates an electric field at the point (D, 0). Because all charges lie on the y-axis, their contributions to the electric field along the x-axis involve the cosine of the angle between the line joining the charge to (D,0) and the x-axis.
If $r$ is the distance between a charge and the point (D, 0), then the magnitude of the electric field due to that charge is
$E = \frac{k\,|q|}{r^2}\,,$
and the x-component is
$E \cos \theta = \frac{k\,|q|}{r^2} \cdot \frac{D}{r}\,=\,\frac{k\,|q|\,D}{r^3}\,.$
Step 3: Group the Charges into Two Pairs
Notice there are two pairs:
Pair 1: (–q at y = –2d) and (+q at y = –d)
Pair 2: (+q at y = +d) and (–q at y = +2d)
Within each pair, the charges are relatively close to each other along the y-axis compared to the point of observation at (D, 0). We will sum the respective x-components of the electric fields from these pairs.
Step 4: Write the Net Field Along the x-axis
Using the expression for each charge’s x-component and summing carefully, the net electric field at (D, 0) can be written as:
$
E_x = 2E_1 \cos \theta_1 \;-\; 2E_2 \cos \theta_2
$
After substituting $E_1 = \frac{k\,q}{(d^2 + D^2)}$ (with appropriate geometry factors) and
$E_2 = \frac{k\,q}{((2d)^2 + D^2)}$, the final simplified form along the x-axis becomes:
$
E_x = 2kqD\,\Bigl[\bigl(d^2 + D^2\bigr)^{-\tfrac{3}{2}} \;-\;\bigl(4d^2 + D^2\bigr)^{-\tfrac{3}{2}}\Bigr].
$
Step 5: Apply the Binomial Approximation
Since $d \ll D$, we can write
$
\bigl(1 + \tfrac{d^2}{D^2}\bigr)^{-\tfrac{3}{2}}
\approx 1 - \tfrac{3}{2}\,\tfrac{d^2}{D^2}
$
and
$
\bigl(1 + \tfrac{4d^2}{D^2}\bigr)^{-\tfrac{3}{2}}
\approx 1 - \tfrac{3}{2}\,\tfrac{4d^2}{D^2}.
$
Subtracting these two expressions and simplifying leads to:
$
E_x \;\approx\; \frac{2\,k\,q\,D}{D^3}\,\Bigl[\Bigl(1 - \tfrac{3\,d^2}{2\,D^2}\Bigr)
-\Bigl(1 - \tfrac{3\times 4\,d^2}{2\,D^2}\Bigr)\Bigr].
$
Further simplification yields
$
E_x \;\propto\; \frac{d^2}{D^4}.
$
Step 6: Conclude the Dependence on D
Hence, the net electric field at (D, 0) scales as
$
\frac{1}{D^4}
$
when $D \gg d$. Thus, we arrive at the final result:
$\displaystyle E \,\propto\, \frac{1}{D^4}.$
