# The electric field by an uniformly charged infinite  plane is independent of distance but practically if we move away from the charged plane electric filed should decrease. Why is it so?

Arun Kumar IIT Delhi
8 years ago
Hi
Consider a finite plate as a collection of finite line charge.
Now for a finite line charge and for easier calculation and demonstration assume you finding E at middle.
In practical situation you can't practically assume it to be an infinite line charge without errors.
$\\E_x=\frac{\lambda(sin{\theta_1}+sin\theta_2)}{4\epsilon_0 \pi d}=\frac{\lambda sin\theta}{2\epsilon \pi d} \\E_y=\frac{\lambda(cos{\theta_1}-cos\theta_2)}{4\epsilon_0}=0 \\where all \theta are the angle subtended by perpendicular line joining point to rod and line joining point to one of the ends and incase of point is at middle it its equal \\d is the perpendicular distance from point to line \\so \frac{sin\theta}{d}=\frac{l}{2*d*\sqrt{d^2+\frac{l}{2}^2}} \\so now we can see if we increase the distance field decreases$

Thanks & Regards, Arun Kumar, Btech,IIT Delhi, Askiitians Faculty