Question

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

• A. $E\propto \frac{1}{D^2}$
• B. $E\propto \frac{1}{D^3}$
• C. $E\propto \frac{1}{D}$
• D. $E\propto \frac{1}{D^4}$

Answer 1

The correct option is D $E\propto \frac{1}{D^4}$


The net electric field at point $\text{P}$ is,

$E=2E_1\cos {\theta }_1-2E_2\cos {\theta }_2$

$=\frac{2kq}{(D^2+d^2)}\times \frac{D}{\sqrt{D^2+d^2}}-\frac{2kq}{D^2+(2d{)}^2}\times \frac{D}{\sqrt{D^2+(2d{)}^2}}$

$=\frac{2kqD}{(D^2+d^2{)}^{3/2}}-\frac{2kqD}{[D^2+(2d{)}^2{]}^{3/2}}$

$E=2kqD\left[\right(d^2+D^2{)}^{-3/2}-(4d^2+D^2{)}^{-3/2}]$

Since, $d<<D$, so applying binomial expansion we get,

$E=\frac{2kqD}{D^3}[(1-\frac{3}{2}\frac{d^2}{D^2})-(1-\frac{3}{2}\frac{4d^2}{D^2})]$

$E=\frac{9kq{d}^2}{D^4}$

$\Rightarrow E\propto \frac{1}{D^4}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.