# Electric field falls off faster than $\frac{1}{r^2}$ for large distances

An excerpt from a book;

The electric field due to a charge configuration with total charge zero, is not zero; but for distances large compared to the size of the configuration, its field falls off faster than $\frac{1}{r^2}$ , typical of field due to a single charge. An electric dipole is the simplest example of this fact.

1. Why is the field not zero even if the net charge is zero? Wouldn't the field cancel out?

2. What does it mean by - the field falls off faster than $\frac{1}{r^2}$ for large distances.

Does it mean that the field intensity decreases at a faster rate at large distances? If yes, then why does that happen?

1. Why is this typical of a field due to a single charge?

2. How is the dipole an example of this fact?

I would appreciate if the answer is aimed for a highschool student with only basic knowledge of electrostatics and doesn't involve complicated equations.

Imagine two charges, one $+q$ and another $-q$, separated by $1$m. As a system the net charge is $0$, but clearly the field will not be $0$ a distance of $10$m away: the fields of each charge goes off like $1/r^2$ but only partially cancel out when you add them because of the distance between the charges; the partial cancellation will be quite direction dependent.

Imagine now you’re not $10$m but $10^9$m away. For large distances one can show the field of the dipole falls off like $1/r^3$ because the fields of the two charges almost cancel out but not completely. There is also: a directional effect, and the field strength depends on the ratio $d/r$ of the distance $d$ between the two charges to the distance $r$ between the center of the charges and the point where you evaluate the field.

It depends basically if you consider the constituents into a single net point source or if you still consider them separately.

Just by looking at the field pattern you can see that there is no place where there is going to be complete cancellation?

In simple terms the reason for the drop off of the resultant field being greater than $\frac{1}{r^2}$ goes something like this.

Imagine a positive and a negative charge producing a magnetic field at position $C$ as shown in the diagram below.

The field from each of the charges $\vec E_+$ and $\vec E_-$ can be resolved into components $E$ and $e$.

The components $E$ cancel out and you are just left with a field of magnitude $2e$.

As the distance $x$ increases the magnitude of component $e$ gets smaller and smaller relative to $E_+$ which is shown by using similar triangles

$\dfrac{e}{E_+} = \dfrac d x \Rightarrow e = \dfrac d x E_+$

However the magnitude of $\vec E_+$ is proportional to $\dfrac {1}{x^2}$ so the resultant field $2E \propto \dfrac {1}{x^3}$.

1. If we consider an electric dipole - a positive and a negative charge, separated by distance r - at small distances ( $<r$ ) the field will be dominated by whichever charge you are closer to.

2. On the other hand, at very large distances then, as you suggested, the fields cancel out and so the overall attraction falls off faster than than $1/r^2$.

3. I think it's saying that the $1/r^2$ relationship is typical of a single charge, in contrast to the dipole. However, the answer is written ambiguously.

It's called a multipole expansion. The first term is the usual $1/r^2$ term and it scales with the total charge. It follows an infinite series of higher order terms with power $1/r^n, n>2$ that have different factors, called "multipole moments". The next one is the dipole moment, then there's the quadrupole moment, and so on. These moments aren't zero even though the total charge is zero because the charges don't have to sit on top of each other. Roughly speaking the higher moments tell you how the charges are distributed.

1. Let$\vec{E}_{+(-)}$ be the electric field created by a positive (negative) charge. Often, one of the charges will be a little bit closer to you than the other. Since $\vec{E}=\vec{E}_+ + \vec{E}_-$, the only scenario when $\vec{E}=\vec{0}$ will occur when the two point charges coincide (where we would have $\vec{E}_+ + \vec{E}_-=\vec{0}$). Any other configuration would have a non-zero $\vec{E}$.

2. Think about the functions $f(x)=1/x^2$, $g(x)=1/x^3$, and $h(x)=1/x^4$. All approach zero as $x$ becomes large, but for any given x (greater than 1), we always have $h(x)<g(x)<f(x)$.

3. I'm not sure I understand what you're asking in # 3.

4. The magnitude of $\vec{E}$ created by a dipole (point charges of opposite sign that are close together) will decrease faster than it would for a single point charge. I like to visualise this as a result of the fact that the two charges are close together (compared to the distance from us to the charges). Because they almost coincide, the fields from each charge ($\vec{E}_+$ and $\vec{E}_-$) almost cancel each other out, with a small leftover electric field $\vec{E}$, (as mentioned in # 1 above). It turns out for a dipole, this leftover electric field $\vec{E}$ is $\propto 1/r^3$ (this is a typical 2nd semester undergrad calculation). To arrive at this result, do a Taylor Expansion assuming $r>>d$ (where d is the separation between the charges and r is the distance from you to the point halfway between the charges).