# Why does the electric field intensity increase (for some distance) as we go further from the center of a uniformly charged ring?

Well I am in class 12th and currently reading about the electric field intensity of various systems with continuous charge distribution.

So I read about the electric field intensity of a uniformly charged ring (say of total charge $$Q$$ and radius $$R$$) at a distance $$x$$ from its center on its axial line which is given by this formula

$$\vec E= \frac{Qx}{(4\pi \epsilon_o)(x^2+R^2)^{\frac{3}{2}}}$$

This relation indicates that the field intensity increases till $$x=\frac{R}{\sqrt 2}$$ which is something I didn't expect to get.

Mathematically this is very satisfying but physically it makes no sense to me.

We are going away from the charged body, so how come the electric field intensity increase even for some given finite distance ?

• because of symmetry $|E(x=0)|=0$ and $|E(x\ne 0)\ >0$ then as you move away from the center $|E|$ must also increase for a while... Jul 5, 2021 at 15:31
• Jul 5, 2021 at 16:27

This is because the electric field is a vector. In the center, the field cancels out because the vector sum of the elementary fields is zero: two opposite elements cancel each other. As one rises along the axis, the components along the axis add up and the field, which was zero, increases although the distance increases. But ultimately, it's the distance that wins and the field decreases.

To add some plots to @VincentFraticelli's answer, the electric field for such a ring can be written as:

$$\mathbf{E} \sim \frac{1}{(x^2 + R^2)} \cos\theta, \tag{1}\label{1}$$

where $$\cos\theta$$ is the angle that the edge of the ring subtends at the point $$P$$ where the field is being calculated. The vertical components all cancel out by symmetry.

Let's imagine the ring to be a bunch of infinitesimal "point" charges $$\text{d}q$$ close to each other. Let's look at one of these little charges. As you move further away from the ring, two things are happening:

1. First, the fraction of this point charge's field which points along $$x$$ and $$y$$ changes. The $$y$$ component starts off (at $$x=0$$) as being the only component, since at the centre of the ring all the field lines point along the $$y-$$axis. However, as $$x$$ increases, each of these infinitesimal point charges $$\text{d}q$$ contribute more along the $$x-$$axis than along the $$y-$$axis, because $$\theta \to 0$$.

2. Second, the overall electric field falls off as $$\sim 1/x^2$$, since the field of each individual point charge falls off as the inverse of the square of the distance.

Of course, by symmetry, the field along the $$y-$$axis always cancels out, no matter how large or small it is.

As @Vincent points out, the first term in Equation (\ref{1}) (which arises from the $$1/x^2$$ behaviour of the electric field) decreases as you move away from the ring. However, the second term (the cosine) actually increases, since as you move further away from the ring, $$\theta \to 0$$, which implies that $$\cos \theta \to 1$$. As I pointed out, this is because as you move away from the ring, a larger fraction of the electric field of each "bit" of charge $$\text{d}q$$ points in the $$x$$ axis. (So at infinity, all of the field of the ring is pointing along the $$x-$$axis, even if its magnitude is zero.)

If you plot the two functions against $$x$$, you see the following curves:

As you should be able to make out, very close to the ring (when $$x\ll R$$), it is basically the cosine term that dominates how the field changes as you change $$x$$. However, far away, it's the $$\sim 1/x^2$$ term that dominates, as you would expect.

• thanks a lot for the graphical representation. Helped a lot.. +1.... Conclusion : this whole thing depends on the rate of change of the factors .. initially the rate of increase of cosine is great and hence it dominates.. but by some distance the rate of decrease of the other factor is greater than the other.. Jul 5, 2021 at 17:58