Approximating and simplifying an electric field I calculated the electric field exerted by a square lying on the $xy$ plane with side length $a$ on a random point in space $(x,y,z)$, and got this expression:
$$\vec{E}=\frac{\sigma}{4\pi\epsilon_{0}}\int_{-\frac{a}{2}}^{\frac{a}{2}}\int_{-\frac{a}{2}}^{\frac{a}{2}}\frac{(x-\tilde{x})\hat{x}+(y-\tilde{y})\hat{y}+z\hat{z}}{((x-\tilde{x})^{2}+(y-\tilde{y})^{2}+z^{2})^{\frac{3}{2}}}d\tilde{x}d\tilde{y}.$$
Now I wanted to make an approximation to the electric field far away from the square, meaning $r\gg a$ where $r$ is the distance from the origin.  I expect to get an electric field of a point charge in the origin with the charge of the whole square. How would you go about it? I tried defining $\varepsilon=\frac{a}{r}$ and using taylor series to approximate but I get to nothing. Is this the right way to tackle these kind of problems?
I'm not asking for a full solution, and it's not a solve my homework kind of post. I often fall into trouble trying to take different limits to answers. This is just a single example to work with and hopefully be able to apply the techniques in many different kind of problems I face in physics.
 A: Let $\vec{r}=x\hat{x}+y\hat{y}=r\hat{r}$ and $\vec{\tilde{r}}=\tilde{x}\hat{x}+\tilde{y}\hat{y}$. Then the integrand becomes
$$\frac{\vec{r}-\vec{\tilde{r}}+z\hat{z}}{(|\vec{r}-\vec{\tilde{r}}|^2+z^2)^{3/2}}$$
Since the square of charge is far away from the origin, when the point $(x,y,z)$ is very far from the square of charge, it will also be very far from the origin. In that case, then one or more of the following must be true:


*

*$r\gg a$

*$z\gg0$
So there are three cases: 1) $r$ is big and $z$ is small, 2) $z$ is big and $r$ is small, 3) both $r$ and $z$ are big.

If $r\gg a$, then it must be true that $\vec{r}-\vec{\tilde{r}}\approx \vec{r}$. To prove this, we first use the Law of Cosines:
$$|\vec{r}-\vec{\tilde{r}}|^2=r^2+\tilde{r}^2-2r\tilde{r}\cos{\theta}$$
where $\theta$ is the angle between the two vectors. Since we know that $r\gg a$, we also know that $r\gg \tilde{r}$, so $r^2\gg \tilde{r}^2$ and $r^2\gg r\tilde{r}$. This means that $|\vec{r}-\vec{\tilde{r}}|\approx r$.
Likewise, to calculate the direction of the resultant unit vector, we can use the Law of Sines:
$$\frac{\sin\alpha}{\tilde{r}}=\frac{\sin\theta}{|\vec{r}-\vec{\tilde{r}}|}$$
where $\alpha$ is the angle between $\hat{r}$ and the new unit vector. We already know that $|\vec{r}-\vec{\tilde{r}}|\approx r$, so:
$$\sin\alpha\approx\frac{\tilde{r}}{r}\sin\theta$$
Since we also know that $\tilde{r}<a$, so $\frac{\tilde{r}}{r}\ll1$, we therefore have that, to first order,
$$\sin\alpha\approx 0$$
Since $\alpha$ was the angle between $\hat{r}$ and the new resultant vector, we can now say that the resultant's unit vector is approximately $\hat{r}$. So it has a magnitude of $r$ and a direction of $\hat{r}$, meaning that $\vec{r}-\vec{\tilde{r}}\approx\vec{r}$.
If we plug that in to the original expression, we get (when $r$ is big):
$$\frac{\vec{r}-\vec{\tilde{r}}+z\hat{z}}{(|\vec{r}-\vec{\tilde{r}}|^2+z^2)^{3/2}}\approx \frac{\vec{r}+z\hat{z}}{(r^2+z^2)^{3/2}}$$
No matter what size $z$ is, this is now identical to the expression for a point charge in cylindrical coordinates. Since the integrand is now constant, integrating it now reduces to multiplying by $a^2$. So this covers the first and third cases.

For the second case, when $r$ is small but $z$ is big, the following must be true:


*

*$z\gg x$

*$z\gg y$

*$z\gg \tilde{x}$

*$z\gg \tilde{y}$
This also means that $z\gg(x-\tilde{x})$ and $z\gg(y-\tilde{y})$. By a similar procedure as the above, you can prove using the Law of Cosines and Law of Sines that $(\vec{r}-\vec{\tilde{r}})+z\hat{z}\approx z\hat{z}$, which also means that $|\vec{r}-\vec{\tilde{r}}|^2+z^2\approx z^2$. Plugging these in, we have that:
$$\frac{\vec{r}-\vec{\tilde{r}}+z\hat{z}}{(|\vec{r}-\vec{\tilde{r}}|^2+z^2)^{3/2}}\approx \frac{z\hat{z}}{(z^2)^{3/2}}=\frac{\hat{z}}{z^2}$$
which is exactly the expression for a point charge along the $z$ axis. Since the integrand is, again, a constant, integrating amounts to multiplying by the integration area $a^2$.
