# Voltage drop across capacitors in series, why?

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.

• Can you elaborate a bit? – user36790 Mar 27 '16 at 13:11
• @Mohammed In between the capacitor plates electric field is generated , and in the direction of this field potential decreases. – JM97 Mar 27 '16 at 13:44
• I mean to say that the net voltage when capacitors are connected in series is V=V1+V2+.... which indicates that voltage drop does occur when capacitors are connected in series, but why and how does it occur? I don't need an answer based on formula i want to know what interactions between charges result in voltage drops across capacitors? – Shahbaaz1104 Mar 27 '16 at 13:45
• @MohammedShahbaazAli if you see the diagram , then you'd see the plates are charged oppositely and electric field exists between the plates so potential decreases. – JM97 Mar 27 '16 at 13:49
• V=Q/C so if you understand why the charge associated with the capacitors is the same, then you see the voltages must be proportional to 1/C. – The Photon Mar 27 '16 at 16:35

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 $\frac 5 Q = \frac 2 Q + \frac 2 Q \Rightarrow \frac 1 C = \frac 1 C_1 + \frac 1 C_2$.

• I got it sir, thanks so when we connect capacitors in parallel its similar to increasing the net area of capacitors hence the capacitance also increases, right ? – Shahbaaz1104 Mar 28 '16 at 18:10
• In this case you can think of the series capacitors having the same area but the separation has been increased. – Farcher Mar 28 '16 at 18:57

When you charge a capacitor negative charge builds up on one plate and (an equal magnitude) of positive charge builds up on the other plate. This means (as has been pointed out) that there is an electric field between the plates, and you can only have fields between places of different electric potential. You can think of the electric field as a force and when you use a force to push a charge over distance this is energy. In electricity, energy divided by charge is called potential.

Let us see in detail the charging of two capacitors in Series:

Let us consider two capacitors c1 and c2 connected in series: i.e., such that the positive plate of one is attached to the negative plate of the other.

Let us suppose that the positive plate of capacitor c1 is connected to the input wire, the negative plate of capacitor c1 is connected to the positive plate of capacitor c2, and the negative plate of capacitor c2 is connected to the out put wire.

It is vital to realize that the charge Q stored in the two capacitors is the same. This can be seen by considering the internal plates:

i.e., the negative plate of capacitor c1, and the positive plate of capacitor c2.

These plates are physically disconnected from the rest of the circuit, so the total charge on them must remain constant.

Assuming, that these plates carry zero charge when zero potential difference is applied across the two capacitors. Therefore in the presence of a non-zero potential difference the charge +Q on the positive plate of capacitor c2 must be balanced by an equal and opposite charge -Q on the negative plate of capacitor c1.

Since the negative plate of capacitor c1 carries a charge -Q, the positive plate must carry a charge +Q. Likewise, since the positive plate of capacitor c2 carries a charge +Q , the negative plate of capacitor c2 must carry a charge -Q. The net result is that both capacitors possess the same stored charge Q .

The potential drop across two capacitors must be different as their charges are same but their capacitance have different values but the total potential drop applied across the input and output wires must be V= V1 + V2