Capacitors connected in series: Why is the voltage the same?

[Delta2]

In a despaired attempt to use as little math as possible:
The voltage source establishes an electrostatic field inside the resistor. This drives initially the current and charges the capacitor, but as the charge in the capacitor builds up, the capacitor establishes its own electrostatic field inside the resistor. The two electrostatic fields battle each other, with that of the voltage source prevailing and keep driving a decreasing current, but when the capacitor is fully charged, the two electrostatic fields eliminate each other and hence there is zero current inside the resistor.

 

[Leo Liu]

That causes a larger voltage drop across the plates, which means you must have a lower voltage drop across the resistor.

Thanks again for showing me how this works. But shouldn't the voltage difference between the plates increase, which causes the current in the circuit to drop?

You can think of Kirchoff's voltage law for a circuit like this in terms of conservation of energy, i.e. carry a unit charge around the circuit and when you get back to where you started it must have the same potential energy it started with. Then you can deduce the current must go to zero, i.e. the previous paragraph.

I understand this, yet it appears quite unintuitive to me. I use the same method to solve problems on the exams; nevertheless, I really want to know the movement an individual electron in the circuit because I think it is fun to think and makes grasping related concepts much easier.

 

[Leo Liu]

The two electrostatic fields battle each other, with that of the voltage source prevailing and keep driving a decreasing current, but when the capacitor is fully charged, the two electrostatic fields eliminate each other and hence there is zero current inside the resistor.

Thank you for attempting at answering my question. I just want to know through what medium the two electric fields fight with each other.

 

[Delta2]

Thanks again for showing me how this works. But shouldn't the voltage difference between the plates increase, which causes the current in the circuit to drop?

The voltage difference between the plates indeed increases as the capacitor charge build up and that is why (if we apply KVL) the voltage difference across the resistance decreases which is what @etotheipi says.

 

[Delta2]

Thank you for attempting at answering my question. I just want to know through what medium the two electric fields fight with each other.

Well the medium is the interior of the resistor which comprises of positive ions and free electrons.

 

[etotheipi]

I understand this, yet it appears quite unintuitive to me. I use the same method to solve problems on the exams; nevertheless, I really want to know the movement an individual electron in the circuit because I think it is fun to think and makes grasping related concepts much easier.

If you want to know more about the how charges move then I think you must look into the continuity equation and details to do with it. If $\rho$ is the charge density and $\vec{v}$ the drift velocity, then the current density $\vec{J} = \rho \vec{v}$. We know that the current through a surface is the surface integral of the current density. You can imagine, in the ideal case, a homogenous volume of charge drifting around the circuit at a drift speed that is the same everywhere in the series circuit at any given time.

Thank you for attempting at answering my question. I just want to know through what medium the two electric fields fight with each other.

The electric field in the battery does negative work on the charge carriers whilst the electric field in the capacitor would do positive work on the charge carriers (well, charge carriers don't move through the capacitor actually but we can ignore that, the voltage drop definitely still exists!). And when the voltage of the capacitor approaches the voltage across the cell, the work done by the electric field in the resistor must also decrease in order to satisfy $\oint \vec{E} \cdot d\vec{x} = 0$, and microscopically in the resistor $\vec{J} = \sigma \vec{E}$ so the current density needs to decrease in magnitude.

At any given point in the circuit, you only have one electric field. In the ideal (resistanceless) battery, the electric field balances the chemical field. In the ideal wire, there is no electric field. In the capacitor, there is an electric field of magnitude $\frac{\sigma}{\epsilon_0}$. And in the resistor, there is an electric field of magnitude $\frac{1}{\sigma}J$. Since there are no time-varying magnetic fields, the closed loop line integral of these will be zero.

 

[hutchphd]

In the ideal (resistanceless) battery, the electric field balances the chemical field

I very much dislike this terminology. WTF is a "chemical field"?
How about:
The ionic processes in the battery endeavor to maintain a fixed potential difference at its terminals. This open circuit potential is commonly called the battery "voltage".
Please no chemical fields. No morphogenic fields. No "fields of psychic energy"

 

[etotheipi]

I very much dislike this terminology. WTF is a "chemical field"?
How about:
The ionic processes in the battery endeavor to maintain a fixed potential difference at its terminals. This open circuit potential is commonly called the battery "voltage".
Please no chemical fields. No morphogenic fields. No "fields of psychic energy"

Okay yes that was sloppy. Let me define a vector field $\vec{f}_{chem}(\vec{r})$ s.t. $q\vec{f}_{chem}(\vec{r}) = \vec{F}_{chem}(\vec{r})$ is the chemical force on a charge carrier in the battery. That is my chemical field .

I guess this is fine for Physics purposes, and for calculating EMF, etc. But in reality like you say, the chemistry and processes at the electrodes is much more complicated than just a single tenuous field.

 

[vanhees71]

Instead of many words, simply do the math. As explained several times above a series of a battery (EMF $U$), resistor, $R$, and a capacitor, $C$, using Faraday's Law of induction in the quasistatic approximation, i.e., neglecting the self-inductance of the circuit, leads to
$$U_C+R i -U=0.$$
Now
$$U_C=Q/C, \quad i=\dot{Q},$$
where $U$ is the charge on one of the plates of the capacitor and thus
$$\frac{1}{C} Q + R \dot{Q}=U.$$
Initial condition: $Q(0)=0$. The general solution of this ODE is the superposition of a particular solution of the inhomogeneous equation, which obviously is given by
$$Q_{\text{inh}}(t)=C U=\text{const}.$$
and the general solution of the homogeneous equation, which is
$$Q_{\text{hom}}(t)=A \exp\left (-\frac{t}{R C} \right),$$
i.e.,
$$Q(t)=A \exp \exp\left (-\frac{t}{R C} \right) + CU .$$
The initial condition determines the integration constant $A=-CU$:
$$Q(t)=CU \left [1-\exp \exp\left (-\frac{t}{R C} \right) \right].$$
So the charge and thus $U_C=Q/C$ reaches exponentially its stationary limit (relaxation time $\tau=R C$). The current through the circuit is
$$i(t)=\dot{Q}(t)=\frac{U}{R} \exp \left (-\frac{t}{R C} \right).$$
So it starts at $t=0$ as if there's no capacitor in the circuit. In other words, in the first moment an uncharged capacitor acts like a short-circuit, but the current drops exponenitialle again with the said relaxation time $\tau$.

 

[sophiecentaur]

I just want to know through what medium the two electric fields fight with each other.

That's a bit too anthropomorphic for my liking!

Using the Fields approach is needlessly difficult, compared with looking at the Potential Energy situation. Nothing is fighting anything. The increasing charge in the capacitor is increasing the potential across the terminals so the PD across the resistor (its share of the available battery PD) is decreasing - sooooo the current through it also decreases.
There is absolutely no difference in the validity of a Field or Potential treatment so it's perfectly OK to choose the one which makes most sense. Many people seems to find it important to justify Electrical calculations by using Fields - why? It's just got to be harder when you think that a Vector has two values associated with it but a Potential has only one.
In the analagous situation of water flowing between two tanks, with different water levels, through a constriction, the flow is proportional to the Difference in Heights between the tanks. Nothing fights anything.

 

[iVenky]

I don't think the voltage is the same for all. The charge is same for all, as the internal nets are floating and charge can't come from anywhere else (-Q+Q=0) i.e. the charge conservation holds true.

Q=CV=> implies if C is lower then V is higher

V1+V2+V3=V due to KVL.

An equivalent analogy is flow of fluid through a pipes P1,P2,P3 of varying diameters as shown.

The mass flow (mass of liquid / second across a cross section area) should be the same for all points (similar to charge)

What changes is the velocity and pressure of pipes P1, P2 and P3 (like voltage described above)

 

[Dale]

Q=CV=> implies if C is lower then V is higher

V1+V2+V3=V due to KVL.

You are talking about the voltages across the capacitors. The OP is asking about the voltage across the wire connecting two capacitors. That voltage is zero so the connected plates are at the same voltage.