Why change in Capacitance? I do know it that the overall capacitance decreases if two or more capacitors are connected in series,but do not understand why?
I do understand it mathematically but don't get the physics behind it.
 A: Well, you have think about the definition of capacitance, as dmckee pointed out in his comment. For two conductors both charged with charge Q and at a potential difference V, capacitance is
$$
C = \frac{Q}{V}
$$
So capacitance is a proportionality constant between charge on two conductor and the potential difference.
Now, if you consider two parallel plate capacitors connected in series instead of a single one, the capacitance goes down because you need more voltage to put the same charge on the plates.
This happens because the same charge on a plate generates an electric field (which is always the same and depends on charge and geometry) but voltage difference also depends on how far the plates are. Indeed voltage can be regarded as energy per unit charge and energy is higher if the plates are farther away, because you have to do work to carry the charges from one plate to the other against the electric field. If you take them far you spend more energy then you would if you take them close by.
So connecting two capacitors is like taking the plates farther and so you need more energy for the same charge or you bring less charge with the same energy. That's the physics.
Thus, capacitance is actually a measure of how much energy I need to charge two conductors per unit charge or voltage. That is clear if you consider the formulas:
$$
U = \frac{Q^2}{2C} \qquad U = \frac{1}{2}CV^2
$$
A: Capacators are basically just parallel plates, so if you have them in parallel, then they act like two large plates (positive and negative terminals) and so can store more charge between them. If they are in series and the voltage drop between the first and last capacitor doesn't change, then the amount of charge you can effecively store in each capacitor will drop so that electrons still "flow downhill" so to speak. What ends up happening is that the smallest capacitor becomes a limiting factor in how much charge can be transferred to the plates. If you have a bunch of enormous capacitors connected in series, but one very tiny capacitor, then the only way you can use all the capaciance of the large capacitors is to have a large charge on the plates...but since they are in series, that requries the small capacitor to also have this charge, which is not possible.
A: Capacitance is defined as $C\equiv Q/V$. An interpretation of this is the following:

Capacitance is the amount of additional charge stored on each plate for every unit of voltage increase across the capacitor.

Capacitance gives you a sense of how much charge you get when you apply some set voltage across the terminals. A high capacitance means you get to have more charge on your capacitor for a given voltage.
Let's switch things around and discuss the inverse of capacitance. Let's call it elastance (this is what it's actually called) and give it the symbol $\Upsilon$ (no idea what symbol is actually used, but various forms of $E$ are confusing). Elastance, then is $\Upsilon\equiv V/Q$. Here's an interpretation:

Elastance is the amount of additional voltage for every unit of charge increase.

Elastance gives you a sense of the voltage that's set up across your terminals when you add a certain amount of charge to one of the plates. High elastance means for even a small amount of charge added, you'll get a big increase in voltage across your capacitor.
Now let's imagine adding charge one bit at a time to a single capacitor (capacitor 1) and simultaneously adding one bit of charge at a time to the combination of two capacitors in series. For capacitor 1, if you add charge $Q$, you get a voltage drop $V=\Upsilon Q.$ No surprise here. What if you add the same amount of charge $Q$ to the combination of series capacitors? Well, if you do manage to add charge $+Q$ to one of the plates, all the other plates of your capacitors will obtain the same charge $\pm Q$. (This has to do with charge conservation.) That means capacitor 2 will have a voltage drop of $\Upsilon Q$ and capacitor 3 will also have the same voltage drop $\Upsilon Q$.
Now add up the voltage drops for your series capacitors. You get a bigger voltage drop. A bigger voltage for the same amount of charge. Sounds like a higher elastance. Higher elastance means a lower capacitance.
In this way, you can view the decrease of capacitance as identical to the increase of elastance. For every bit of charge you add to one of the plates of a capacitor, you end up getting more total voltage with series capacitors, simply because you have more of them contributing.
By the way, using elastance hides the ugliness of the equivalent series capacitance formula:
$$ C_\mathrm{equiv} = \left( \frac{1}{C_1}+\frac{1}{C_2} \right)^{-1}
\rightarrow
\Upsilon_\mathrm{equiv}=\Upsilon_1+\Upsilon_2+\cdots,$$
but it doesn't really save time for calculations.
A: Capacitance is a measure of how much charge is required to make a change in voltage:
$$ C = \frac{Q}{V} $$
As the plates of a capacitor are brought closer together, capacitance increases. This is because the opposite charges on each plate of the capacitor can get closer to each other, and thus cancel each other more completely, and thus the voltage per charge is less.
With two capacitors in parallel, what happens to the plate from each capacitor on the "inside" of the capacitor sandwich?

What you effectively have is one capacitor, made of one plate of C1 and one plate of C2, separated by more distance. The stuff in the middle is just a floating piece of conductor that is stuck inside this bigger capacitor. Since the plates are now separated by a greater distance, you have a smaller capacitance, because for each unit of charge you put on the plates, you will have separated that charge by a greater distance, and thus, a greater voltage.
