Basically when you connect more than 1 capacitor in series then the charge on each capacitor is same but there is a voltage drop across each capacitor. I have no intuition as to why the voltage drop occurs. Please help me visualize the situation and understand why is there a voltage drop across capacitors.
[Physics] Voltage drop across capacitors in series, why
capacitanceelectrostaticsvoltage
Related Solutions
Charges on all plates for capacitors in series must be equal. Q=CV for each capacitor. If you look at the charge per plate before you connect them in series, both have the same charge. So, after you connect them, nothing will change (no charge will flow). Since the capacitance of each capacitor doesn't change either, they keep the same voltages.
then why is there no potential difference between the two capacitors
It's not quite clear what you mean here but do understand that charged capacitors are electrically neutral.
When a capacitor is "charged", it is not electrically charged, it is energy charged in the same sense as when we say a battery is charged.
There is nothing mysterious about two series connected circuit elements having different voltage drops. Think of two series connected resistors with different resistor values.
I would have thought that as plate B is positively charged and plate C is negatively charged, there would be a potential difference between the two
You're forgetting something fundamental: The plates B and C along with the wire that connects them are conductors. But, for an ideal conductor, charge distributes itself so that there is no (static) potential difference across the conductor.
The voltage between the bottom plate of C1 and the top plate of C2 is zero precisely because a conductor connects the two plates.
Best Answer
Here is a slightly different way of considering two capacitors in series.
Diagram 1 shows an ideal parallel plate capacitor with a potential difference of 5 V across its plates $AA'$ and $BB'$.
The capacitance of this capacitor is $C = \frac Q 5 $
Also shown in red are some equipotential surfaces one example being labelled $DD'$.
If an uncharged, very thin conducting plane is introduced on an equipotential surface then charges are induced on the surface of the conducting plane as shown in diagram 2.
The charge must be induced to ensure that the electric field within the conducting plane is zero.
The introduction of an uncharged, very thin conducting plane does not change anything else.
Now there are two parallel plate capacitors of capacitance $C_1 = \frac Q 2$ and $C_2 = \frac Q 3$
So there you have the voltage drop and zero net charge on plate $DD'$
Furthermore $\frac 5 Q = \frac 2 Q + \frac 2 Q \Rightarrow \frac 1 C = \frac 1 C_1 + \frac 1 C_2$.