Why electric field is maximum at R/√2 i mean generally at first from a look of ring and its axis field value should decrease why there is a maximum in between a point. I know mathematically we can see the graph but how do one perceive that it may have a maximum in between 0 to infinity just by looking?
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$\begingroup$ how do one perceive that it may have a maximum in between 0 to infinity Have you thought about what the field is at the center, and what it is infinitely far away? $\endgroup$– GhosterCommented Oct 1, 2022 at 16:04
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$\begingroup$ Yes it's zero at the center and at an infinitely far away point but it may seem at first the more the distance from center less will be its effect or field value should decrease. I know first x increases E increases after a maximum it again decreases i saw the graph and the formula. $\endgroup$– Gaurav MishraCommented Oct 1, 2022 at 16:23
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$\begingroup$ field value should decrease The magnitude can’t decrease from zero. It can stay 0 or increase. It should be clear that it doesn’t stay 0. $\endgroup$– GhosterCommented Oct 1, 2022 at 23:38
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Say, you're moving on the axis and your position is denoted by $P$. Now suppose the angle formed by the axis(at $P$) and the periphery of the ring is $\alpha$. Now the electric field at $P$ is only due to the cosine components of $\alpha$ as the sine ones cancel out. Keeping this in mind, Let's start from the middle of the ring(centre of the ring in the plane containing the ring) and move axially outwards. When you're at the centre, electric field is zero as $\alpha$ is 90 degrees so no cosine contribution. As you move axially outwards, $\alpha$ starts decreasing from 90 degrees, i.e, $cos\alpha$ starts increasing from zero $\Rightarrow$ cosine component of electric field increases, but at the same time axial distance of $P$ from the centre(say, $x$) is increasing, and electric field falls as $\frac{1}{x^2}$. So, the maxima of electric field at the axis is going to be due to the interplay of increasing $cos\alpha$ which increases the field and increasing $x$ which decreases the field. Till $x=\frac{R}{\sqrt{2}}$ the increase in field due to increasing $cos\alpha$ dominates the decrease in field due to $1/x^2$, so field increases. Keep moving axially outwards and you'll see that after $x=\frac{R}{\sqrt{2}}$ the decrease in field due to $1/x^2$ dominates the increase due to $cos\alpha$, so field starts decreasing.