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

Answer 1

Answer: (D)

Electric field at $p=2E_1\cos {\theta }_1-2E_1\cos {\theta }_2$
$=\frac{2Kq}{\left(d^2+D^2\right)}\times \frac{D}{{\left(d^2+D^2\right)}^{1/2}}-\frac{2Kq}{\left[{\left(2d\right)}^2+D^2\right]}\times \frac{D}{{\left[{\left(2d\right)}^2+D^2\right]}^{1/2}}$
$=2KqD\left[{\left(d^2+D^2\right)}^{-\frac{3}{2}}-{\left(4d^2+D^2\right)}^{-\frac{3}{2}}\right]$
$=\frac{2KaD}{D^3}\left[{\left(1-\frac{d^2}{D^2}\right)}^{-3/2}-{\left(1+\frac{4d^2}{D^2}\right)}^{-3/2}\right]$
Applying binomial approximation $\because d<<D$
$=\frac{2KqD}{D^3}\left[1-\frac{3}{2}\frac{d^2}{D^2}-\left(1-\frac{3\times 4d^2}{2D^2}\right)\right]$
$=\frac{2KqD}{D^3}\left[\frac{12}{2}\frac{d^3}{D^2}-\frac{3}{2}\frac{d^2}{D^2}\right]$
$=\frac{9kq{d}^2}{D^4}$