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In the problem as shown below enter image description here

The electric field comes out to be

$$ E_y = \frac{1}{4\pi\epsilon_0} \frac{2\lambda\sin\theta}{y} = \frac{1}{4\pi\epsilon_0}\frac{2\lambda}{y}\frac{\ell/2}{\sqrt{y^2 + (\ell/2)^2}} $$

In the limit where $y\gg\ell$, the above expression reduces to the "point-charge" limit,

$$ E_y \approx \frac{1}{4\pi\epsilon_0}\frac{2\lambda}{y}\frac{\ell/2}{y} = \frac{1}{4\pi\epsilon_0}\frac{\lambda\ell}{y^2} = \frac{1}{4\pi\epsilon_0}\frac{Q}{y^2} $$

On the other hand, when $\ell \gg y$, we have

$$ E_y = \frac{1}{4\pi\epsilon_0}\frac{2\lambda}y $$

Then a graph is shown

enter image description here The graph is of $E_y/E_0 $ as a function of $y/l$. What does it signify? Why not the graph were of $E$ versus $y$ or $l$?

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  • $\begingroup$ Vote to close? In no way is this a question asking for help on homework. The problem as presented is completely worked out! $\endgroup$ – garyp Mar 26 '18 at 2:36
  • $\begingroup$ @garyp : how is this not a homework question showing little to no effort? Not commenting on the potential relevance or interest of the question, just commenting on the ask format. $\endgroup$ – ZeroTheHero Mar 26 '18 at 3:48
  • $\begingroup$ @ZeroTheHero He or she is not asking for help solving the question. The question is about a style of presentation. $\endgroup$ – garyp Mar 26 '18 at 11:25
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This type of graph is very useful, and I applaud the text for doing this.

By dividing the field strength by a strength characteristic of the system, and a dividing the length by a characteristic length, the graph becomes universal. The graph can be applied to a rod of any length and any charge density.

Among other things, the numerical values on the $x$ and $y$ axes are close to 1. Without this normalization the axes values might be around $10^{13}$ or $10^{-34}$ or anything at all ... depending on the length of the rod and the charge density. On a normalized universal plot the values are always around 1.

Most plots in professional publications are presented in this fashion.

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  • $\begingroup$ Thanks. But what is meant by "strength characteristic" of the system? That is $ E_0 $, if I am right $\endgroup$ – Ravi Prakash Mar 26 '18 at 12:35
  • $\begingroup$ $E_0=Q/4\pi\epsilon_o \ell^2$ $\ldots$ the only electric field that can be constructed from the parameters of the system, $Q$ and $\ell$ (or equivalently, $\lambda$ and $\ell$). It's thus called the characteristic field strength. With $y\approx\ell\;$ ($y/\ell \approx 1$), we have $E \approx E_0$ ($E/E_0 \approx 1$). $\endgroup$ – garyp Mar 26 '18 at 13:06
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The interest of this question is that it shows that most quantities have scales associated with them, and that there is some physics tied to the choice of this scale.

The natural scale of the problem has been chosen as the magnitude of $E_y$ at a distance $\ell$ from a point charge. You therefore have $$ E_0=\frac{Q}{4\pi \epsilon_0 \ell^2}\, ,\qquad E_y= \frac{Q}{4\pi \epsilon_0 y \sqrt{y^2+(\ell/2)^2}} $$ so the ratio $$ \frac{E_y}{E_0}=\frac{\ell^2}{y\sqrt{y^2+(\ell/2)^2}} $$ can be expressed in terms of the dimensionless quantity $\xi=y/\ell$ as $$ E_y=\frac{2}{\xi\sqrt{1+\xi^2}}\,E_0 $$ with $\xi=y/\ell$.

This gives you a way of comparing the field for the distributed charges to the field of a point charge since the quantity on the right is a multiple of the reference field $E_0$. For small $y/\ell$, the field is not like the field at a distance $\ell$ from point charge, and for large $y/\ell$ it isn't like this field either.

I don't see any obvious advantages to choosing this scale, but that's one that has been chosen. As illustrated by the graph, in the limit where $y/\ell\to\infty$, the field $E_y\to 0$ and in the limit where $y/\ell \to 0$ the field $E_y\to \infty$, so neither limit relates particularly well with the reference field $E_0$.

Another choice of scale would be $$ \bar{E}_0=\frac{\lambda}{4\pi\epsilon_0 y}\, , $$ which is the field of an infinitely long line with constant charge density $\lambda$.
\begin{align} \frac{E_y}{\bar{E}_0}&=\frac{1}{4\pi\epsilon_0}\frac{2\lambda}{y} \frac{\ell/2}{\sqrt{y^2+(\ell/2)^2}}\times \frac{4\pi\epsilon_0 y}{2\lambda}\\ &=\frac{\ell/4}{\sqrt{y^2+(\ell/2)^2}}\, \\ &= \frac{1}{\sqrt{1+\xi^2}}\, , \\ E_y&= \frac{1}{\sqrt{1+\xi^2}}\bar{E}_0 \end{align} with again $\xi=y/\ell$. With this choice of scale, the field for $y/\ell\to 0$, i.e. the magnitude $E_y$ when the length $\ell$ of the stick is much greater than $y$, is basically $E_0$, the field an infinite rod. This would also be the regime where $y$ is very "close" to the stick, in the sense that $y/\ell <<1$, with $\ell$ the natural length scale for this problem.

The limit $y/\ell\to\infty$ correspond to being very far from the stick. The field at such a large distance is expected to be $0$, hence it would be much smaller than the field very far away from an infinitely long stick, which itself is $0$.

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