# Electric Field due to line charge

There is a particular paragraph in Electricity and Magnetism by Purcell that I'm not able to understand. It's the last para in section 1.13, pg-30 which goes like this

The field of an infinitely long line charge, we found, varies inversely as the distance from the line, while the field of an infinite sheet has the same strength at all distances. These are simple consequences of the fact that the field of a point charge varies as the inverse square of the distance.

If that doesn’t yet seem compellingly obvious, look at it this way: roughly speaking, the part of the line charge that is mainly responsible for the field at P in Fig. 1.24 is the near part – the charge within a distance of the order of magnitude r. If we lump all this together and forget the rest, we have a concentrated charge of magnitude $$q \approx \lambda r$$, which ought to produce a field proportional to $$\frac{q}{r^2}$$,or $$\frac{ \lambda}{r}$$. In the case of the sheet, the amount of charge that is “effective,” in this sense, increases proportionally to $$r^2$$ as we go out from the sheet, which just offsets the $$\frac{1}{r^2}$$ decrease in the field from any given element of the charge

I don't get anything. First that near part approximation and then that lumping stuff.

Edit: The electric field due to the element $$\lambda dx$$ given by

$$dE_y\propto \frac{\lambda dx}{(r^2+x^2)}\cos\theta=\frac{\lambda dx\cdot r}{(r^2+x^2)^{3/2}}$$ $$\frac{dE_y}{dx}\propto \frac{ r}{(r^2+x^2)^{3/2}}$$ If you plot the function on the right, you get a plot that has a peak around $$x=0$$, So That's clear that the contribution is coming around this part. $$\frac{dE_y}{dx}\propto \frac{ r}{(r^2+x^2)^{3/2}}\approx \frac{1}{r^2}\left(1-\frac{3x^2}{2r^2}\right)$$

But still, I don't get the fact why we should take the magnitude of order $$r$$. If I take it for a grant then lumping can be understood. $$q = \int dq\approx \int_0^r\lambda dx=\lambda r$$

• To be clear, could you provide a bit more context as to what is going on here? Are you trying to calculate the electric field due to an infinite line charge? May 9, 2021 at 13:26
• I think you should add your own thoughts so that the question isn't closed.
– Babu
May 9, 2021 at 13:59
• @Buraian I have added a little explanation. May 9, 2021 at 14:47
• Hmm did my answer help? Here is one way to think about it, what charge should you replace the length segment with such that you can simulate the same field as the length segment. What do you think of this?
– Babu
May 9, 2021 at 16:29
• Well, What I don't get is that order stuff. How do we know that we need to take up to order of $r$? If this gets fixed, then I don't find any problem with lumping the whole charge. What I think about is the same, that is to replace the line charge with two charges on opposite side. May 9, 2021 at 17:20

Yeah, this is a common doubt. Consider the situation as shown in the figure posted by you.
We have to calculate electric field at a distance $$r$$ from the line charge.
We can see that as the line charge is infinite. So the charge elements which are very far from P, contributes negligibly to the electric field at P (as $$F\alpha\frac{1}{r^2}$$). Only those charge elements will contribute more which is close to P (upto $$r$$ or $$2r$$ length of the line charge). Then the charge in this length is $$\lambda r$$.
As we know electric field is proportional to $$\frac{q}{r^2}$$, so in this case electric field is proportional to $$\frac{\lambda r}{r^2}=\frac{\lambda}{r}$$.

Similar is the case for infinite sheet of charge.

Hope that helps!

I have to agree that this is indeed a confusing paragraph, but here is how I understood it. Imagine instead of a continuous density, we could replace that with a single discrete entity which causes the same effect. The philosophy is remniscent of how we lump circuit elements. This is well discussed in the Feynnman lectures.

The way we have described the ideal inductance illustrates the general approach to other ideal circuit elements—usually called “lumped” elements. The properties of the element are described completely in terms of currents and voltages that appear at the terminals. By making suitable approximations, it is possible to ignore the great complexities of the fields that appear inside the object. A separation is made between what happens inside and what happens outside. Source

Once, we got that replacing idea, we can more or less under stand the reasons why Purcell had taken the values that he did for the charge and all.

I think what he's trying to say is that, if a point charge gives off an electric field like $$E \sim q/r^2,$$ then we can look at a distribution of such point charges and ask:

"how many of these charges do I 'see' in my field of vision? I'm gonna say that $$r$$ is the distance between me and the nearest charge in the distribution, and treat the field from each nearby charge I 'see' as $$\sim q/r^2$$, neglecting those that are farther away. If this is the case, then how does my field of vision as a function of $$r$$ affect the field I end up getting?"

For a 1-dimensional line distribution, let's say my line of sight is given by a distance $$\ell$$. Well, as I step farther and farther away from the line ($$r$$ increasing), I'm going to end up "seeing" more of the line ($$\ell$$ increases). As a rough approximation, I can take my field of vision to scale like $$\ell \sim r$$ so that $$q \sim \lambda \ell \sim \lambda r \qquad \longrightarrow \qquad E \sim q/r^2 \sim \lambda r/r^2 \sim \lambda/r$$

(obviously an oversimplification; it's more like the closer we are to the wire, the more the stuff directly in front of us will dominate things, while the stuff laterally farther away and "outside our line of sight" contribute less by comparison. As we move back away from the wire, this lateral distance becomes less important, and things laterally farther away enter our line of sight, contributing almost as much as the stuff right in front of us)

For a 2-dimensional sheet of charges, my areal field of vision scales more like $$A \sim r^2$$ so that $$q\sim \sigma A \sim \sigma r^2 \qquad \longrightarrow \qquad E \sim \sigma A/r^2 \sim \sigma r^2/r^2 \sim \text{constant}$$

The "near part" is basically the fact that I only include those charges lying in my field of vision, a field which is determined roughly by the nearest distance $$r$$. "Lumping together" means that I'm treating all of them on the same footing, with the same $$E \sim q/r^2$$ contribution.

Sorry if this is more confusing than helpful; I'm just trying to stick to a rough and general physical explanation like Purcell's doing

Electric field lines help. The separation of the field lines shows the strength of the electric field.

For an infinite line charge, the field lines must point directly away from it. Let's say there are 36 field lines leaving a given point on the line charge, with a $$10^\circ$$ spacing. The separation of the field lines increases linearly with distance from the line charge - and so the electric field strength decreases linearly with distance.

For a sheet charge the field lines must again point directly away from the sheet (due to symmetry, there is no reason for them to have any other component of direction). The separation of any two field lines thus remains constant, so the electric field strength is constant with distance from the sheet.