Figure shows two capacitors connected in series and joined to a battery. The graph shows the variation in potential as one moves from left to right on the branch containing the capacitors.
We know that,
Voltage dropor potential difference) across capacitor is given by
Where,
Q=charge on the capacitor
C=capacitance of the capacitor
By looking at the graph,
We can see that first increment in voltage is greater than the second increment. Therefore,
we can conclude that voltage drop across capacitor C1 is greater than the voltage drop across capacitor C2
on moving left to right C1 comes first)
Since charges on the capacitors in series are same ,
∴ Q1=Q2
The potential drop across the capacitor C1 is more than Capacitor C2.
V1>V2
Putting the values of V, we get
On solving, we get
Or
C2>C1
Thus, the capacitance of the capacitor C1 is less than C2