How can a conductor with high capacitance store more charge at less potential? In my textbook a statement is given that goes like "for large C , V is small for a given Q". I am not getting it if charge on a capacitor is increasing wouldn't it increase potential between them. why does it even depend on C?. can someone explain it intuitively.
 A: 
for large C , V is small for a given Q

One has to be clear about what is increasing and what is kept constant here.
The basic relation is $$Q=CV.$$ Thus, if we have two capacitors with capacitances $C_1, C_2$, such that $C_1<C_2$, and we charged them to hold the same charge ("given $Q$"), then the capacitor with larger capacitance need smaller voltage for this:
$$
V_1=\frac{Q}{C_1}>\frac{Q}{C_2}=V_2
$$

if charge on a capacitor is increasing wouldn't it increase potential between them

On the other hand, if we consider a single capacitor ("given $C$") and  increase the charge on it, then it will have higher potential difference:
$$
Q_2>Q_1 \Rightarrow V_2=\frac{Q_2}{C} > \frac{Q_1}{C}=V_1.
$$
A: 
Diagram 1
$q = c\,V$
Diagram 2
The area of the plates is doubled and the separation is kept constant so $C= 2\,c$ and the charge stored on the capacitor is still the same, $q$.
The surface charge density has decreased by a factor two so the electric field strength between the plates has decreased by a factor two. (Think of electric field lines starting on a positive charge and ending on a negative charge with the electric field strength being a measure of the density of electric field lines.)
The electric field strength is proportional to the potential gradient between the plates and as the separation of the plates has stayed the same the potential voltage across the plates has decreased by a factor of two, $v=V/2$.
$q\text{(same)} = C\text{(doubled)}\,v\text{(halved)}$
which is what is stated in your question  "for large C , V is small for a given Q".
Diagram 3
Double the charge stored, $Q=2q$, and the electric field becomes the same as in diagram 1.
Thus, the voltage across the plates is now $V$ (double that in Diagram 2).
$Q\text{(doubled)}=C\text{(doubled)}\,V\text{(same)}$.
