First of: the energy W does not increase exponentially it increases quadratic!
Second:
What do you want?
Energy density!
So to get good high energy density you need high C because you can (as you already stated) only go to a certain amount of voltage until you get a breakdown. The problem is, C does not change easily. If you take a simple Plate - Plate capacitor
$C \propto \frac{A}{d}$ where A is the Area of the plates and d the distance between them. The smaller the distance the higher the electric field gets between the plates gets relative to Voltage ($|\vec E| \propto \frac{U}{d}$) the faster you get a spark. So the only thing you can actually increase safely is the area. That of course increases the wight/size of you capacitor, reducing the Energy density. This generally applies to all capacitors
Wikipedia does have an excellent article about capacitors:
http://en.wikipedia.org/wiki/Capacitor
Third:
About the capacitance:
A spring is a very good analog to your problem:
A newtonian spring has a certain flexibility, describing the force you have to pull it relative to the expansion of the spring:
$F=k \times x$
Where F is the Force k is the flexibility and x the way. So the capacitance is actually k in this analog. But F is not the Energy W.
$E = \int_0 ^x F(x) dx$
So the more you pull the harder it gets to pull because the force from the spring increases. So to get the energy you can not just calculate $E = F \times x$ but you rather have to integrate the way.
That of course gets you to
$E = \frac{1}{2} k x^2$
Back to the Capacitor:
The more electrons you put in it, the stronger becomes the electric field between the plates of the capacitor. So you have to increase the voltage. The capacitance just tells you how high your voltage has to be (how hard you have to push) to put more electrons on the plate.
The circuit is drawing infinite amount of current from the battery. Have I correctly understood this ?
Yes.
There will be a potential difference across the resistor in parallel to capacitor and that potential difference will be resposnsible for charging it
The potential across the capacitor can't change instantaneously.
Therefore in the time immediately after the switch closes, the voltage across the resistor (the one in parallel with the capacitor) is zero. From Ohm's law, then, there is no current through this resistor in that instant.
To find the current that is charging the capacitor (in the instant immediately after closing the switch), you can use KCL at the node where the capacitor and the two resistors are all connected.
Alternately, you can replace the voltage source and the two resistors with a Thevenin equivalent circuit, and again find the charging current as time progresses after the switch is closed.
Best Answer
To understand the answer, you need to be aware of the concept of electric potential. Electric potential is a scalar quantity. In any circuit, there is a potential at any given point on the wire. The difference in potential between any two points in this circuit is sometimes called potential difference or voltage. You can understand the difference between potential and potential difference better here.
Imagine you had a circuit as shown below:
Assume that the potentials at $A,B,C$ are $V_A,V_B, V_C$. The potential difference between A and B is $V_A - V_B$. The potential difference between B and C is $V_B - V_C$. Adding the two, we get the expression $V_A - V_C$ which is the correct term for potential difference across AC.
I hope that this makes it clear that potential differences between two sets of points can be added to obtain the potential difference between the first and last point.