Approximations of the kind $x \ll y$ I have an expression for a force due to charged particle given as
$$F=\frac{kQq}{2L}\left(\frac{1}{\sqrt{R^2+(H+L)^2}}-\frac{1}{\sqrt{R^2+(H-L)^2}}\right) \tag{1}$$ where $R$, $L$ and $H$ are distance quantities.
Now I want to check what happens when:


*

*$H\gg R,L$

*$R,H\ll L$


How can I work out the approximation of this force?
Do I have to write it slightly different into form (2) to get it right?
$$ ~F=\frac{kqQ}{2LR}\left(\frac{-1}{\sqrt{1+\left(\dfrac{H+L}{R}\right)^2}}+\frac{1}{\sqrt{1+\left(\dfrac{H-L}{R}\right)^2}}\right) \tag{2}$$ (which is the same expression just written out differently). Any  explain about this subject would be very helpful.
 A: For $H\gg R,L$ And for $L\gg R,H$ you get pretty much the same thing.
First off, $(H+L)^2\sim H^2$ and the same goes for $(H-L)^2$. That means that $(H+L)^2+R^2\approx (H+L)^2$. However, $(H+L)\not\approx H$, which means that $\sqrt{R^2+(H+L)^2}\approx H+L$. This makes the first approximation have $\frac{1}{H+L}-\frac{1}{H-L}$ in it.
The second approximation proceeds much the same way. $R$ is approximated away and you are left with the root of $(H\pm L)^2$. The resulting approximations should be identical
A: Let's focus just on the interesting bit of the equation:
$$\frac{1}{\sqrt{R^2+(H+L)^2}} - \frac{1}{\sqrt{R^2+(H-L)^2}} =\\
\frac{1}{\sqrt{R^2+H^2+2HL+L^2}} - \frac{1}{\sqrt{R^2+H^2-2HL+L^2}}$$
Now if $H>>L,R$, we are left just with the terms with $H$:
$$\approx \frac{1}{\sqrt{H^2+2HL}} - \frac{1}{\sqrt{H^2-2HL}}\\
=\frac{1}{H}\left(\frac{1}{\sqrt{1+2\frac{L}{H}}}-\frac{1}{\sqrt{1-2\frac{L}{H}}}\right)\\
\approx \frac{1}{H}\left(1-\frac12 \cdot 2\frac{L}{H}-\left(1+\frac12\cdot 2 \frac{L}{H}\right)\right)\\
=-\frac{2L}{H^2}$$
Almost exactly the same approach works for $L>>R, H$ (I will leave the details up to you).
A: Your final expression is off by a minus sign, and is not what you want.  For #1, $H$ is large, so I'd factor $H$ out.  For #2, $L$ is large, so I'd factor $L$ out.
When you factor out a large thing, the remaining things are either numbers (which are what they are) or small things.  And when something is small you can approximate it by comparing it to other things.
A: When considering these things, at least as a warm up to a more rigorous answer, it is worth thinking about what it means to say that $H>>R$. I take this to mean that, roughly, if I add $H$ to $R$ I'm going to get something close to $H$, as it is much larger. For example $1000000>>1$ so $1000000+1\simeq1000000$. When the variables are squared, as you have, this difference is even more marked. 
So, if $H>>R,L$, $(H+L)^2+R^2 \simeq (H+L)^2$, so the first term in your brackets becomes ${1 \over H+L}$ (or ${1 \over |H+L|}$). Similarly the second term, and then add the two to get something non-zero.
You can use a similar process for the second question. 
A more rigorous way to do it would be to do something similar to what you and other answerers have done, which is to take out a factor from the front of your square roots which corresponds to the large variable in each case, so that you're left with dimensionless terms inside like $1, L/H$, and $R/H$. These last two can be neglected compared to 1 and you should get the same results.
