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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?
5 years ago

Arun Kumar
IIT Delhi
256 Points

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

5 years ago
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Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions