Hi Kirubha,

Mathematicaly, you can deduct this from 2 equations:

1) Qc=C x V (charge in capacitor)

if differential operator (d/dt) is applied, then

Ic=C x dv/dt

while in DC circuit we have dv/dt=0, so Ic=0

or

2) Impedance in capacitor is define by:

Xc=1/(2 x pi x f x C)  (f=frequency)

because in DC circuits there is no frequency involve (f=0), then Xc is theorically infinite. This mean a large impedance. Thus, Ic=0 (DC circuit).

Hi Kirubha,

Mathematicaly, you can deduct this from 2 equations:

1) Qc=C x V (charge in capacitor)

if differential operator (d/dt) is applied, then

Ic=C x dv/dt

while in DC circuit we have dv/dt=0, so Ic=0

or

2) Impedance in capacitor is define by:

Xc=1/(2 x pi x f x C)  (f=frequency)

because in DC circuits there is no frequency involve (f=0), then Xc is theorically infinite. This mean a large impedance. Thus, Ic=0 (DC circuit).

First I noted here that, Why Capacitor allowed AC ? After getting this you will come to know about answer.

-------------------| |------------------

Capacitor having two parallel plates. When AC voltage is applied across the plates of parallel plate capacitor, plate 1 will start to get charge till peak voltage attained and plate 2 of capacitor will get negative charge. But after maximum voltage, as the voltage of source is less than the voltage across the capacitor plates, the capacitor will start discharging till source voltage becomes zero. After that as the source voltage is going negative, the Plate 1 will now become negatively charged while plate 2 will be positively charged till negative peak of source voltage but once negative peak of applied voltage crosses, the capacitor will again start to discharge as the potential difference across the plates of capacitor is more than the source voltage. In this way Capacitor continuously charge and discharge for applied AC and hence we say Capacitor is allowing AC to flow.

I completely agree with Franco Chairella....!

Because of zero frequency in DC system, it behaves like an open circuit, as you can see in the formula, Xc=1/(2 * Pi * f *c)

A capacitor is formed by charges being accumulated (on the plates/surface / any surface with an insulator in between) ,  DC has direction (unlike AC they change or  characterized as frequency ) , so with DC charges settle (forming polarity or a capacitor ) 

Perhaps Kirubha is looking for an explanation of how capacitors block DC, rather than math models that quantify the phenomenon?

A capacitor is an open circuit. Just like an open switch. It blocks DC, and it can also block at least some AC frequencies, because it's an open circuit!

A better question might be, why don't capacitors always block all frequencies? That's because capacitors are designed to allow for accumulation of charge. Very much like a water dam allows water to accumulate upstream of the dam. If you build a dam and close it tight, water from upstream will not instantly stop flowing, right? It will instead build up in a pool, much like charge builds up in the capacitor when DC current flows.

When the accumulated charge in the capacitor reaches the same voltage as the voltage source, that's when current flow stops. Switches too are capacitors in a sense, except the charge they can store on their contacts is minimal. So you will notice a much more rapid stopping of current flow than you would through a capacitor of any typical value.

AC current flow would be like a river changing direction periodically. Imagine a dam in such a river, with a pool on each side of the dam. If you keep changing the direction of water flow, although no water ever actually goes through the dam, the flowing water will periodically fill and empty the pools on each side of the dam, making it appear as if water were going through the dam.

Similarly, capacitors don't block AC, even though the charges are not actually going through the capacitor. 

A capacitor blocks DC as once it gets charged up to the input voltage with the same polarity then no further transfer of electrons can happen accept to replenish the slow discharge due to leakage if any. hence the flow of electrons which represents electric current is stopped.

Capacitors will be used for different purpose in a electrical circuit. I would like to answer this question in an electrical way, impedance.Z=U/I, Z is the impedance.

Impedance for R, L, C  can be given in R, jwL, 1/(jwC), where w=2*PI*f, f  is the signal frequency in electrical circuit. If we only care about the amplitude, then we can ignore the complex index j. So for capacitor, it has an impedance of 1/(wC), for DC signal, w=0, then the impedance is very large, which means open circuit.

Do not get confused, that you can charge a capacitor with a constant current, in this case, the voltage of the capacitor has been changed, like explained by Franco.

But physical understanding is very important, which has been explained very well above, like Albert and others.  

I think the answer sought may be simpler than those given, a capacitor blocks DC because it is constructed with an insulator separating the two conductive plates.

Yes, a capacitor can block DC as capacitor plates are separated by insulator still initial charging current flows till voltage on plates equals applied DC voltage. After charging DC current does not flow as DC frequency is zero so capacitor offers theoretically infinite impedance. Practically there is some leakage current that is limited by actual resistance provided by insulating material between the plates.

It is a good q. I suppose what is sought is a physical explanation Consider the following:

1. Normal intuition of current is flow of charge. This is usually modeled in conductors as a flow of electrons. In  a finer view it is transfer of electrons in the conduction band for usual conductors such as copper wires. i imagine this can be experimentally shown. So flow of charge is an experimental fact  that can be demonstrated via its effect ie moving charges produce magnetic effects.

2. A separate phenomena which was based on mathematical considerations due to Maxwell was that change in field lines wrt to time is same as a current .. In other words  if field lines increase then there must be an increase in the charge covered by a surface which is normal to the field. This was the beginning of Em wave theory.It is a fact that em waves can be transmitted without a material medium.

3. From 2, between capacitor plates, although actual electrons . cannot flow(assuming perfect insulator w) electric field can exist. So any variation of accumulation of charges on the plates, effect a corresponding change in the electric field( usual D= electric flux). . Thus we can model the change in the electric field inside the capacitor as a current = d/dt( D) , famous displacement current..the one single addition made by Maxwell which brought in the era of waves viz electromagnetic waves.and changed mankind.

4.   3above  would explain the transient current associated with initial charging  of the capacitor .  .If a steady dc is maintained then after enough accumulation of charges the applied voltage is supported by the voltage due to the charge on the plate. There is no further change in charge on the plate , no change in D so current =d/dt D=0.. The connection between D and and the electric field is given by Permittivity.eps0in vacuum.

Cheers

 Well, he does not actually go through the capacitor the alternating current (electrons through the dielectric) - it's just "induction" :-)

The capacitor blocks the DC current because it is constructed from two metal plates separated by a thick insulator. The ideal insulator has an infinite resistivity and therefore the capacitor has infinite resistance to the passage of DC currents. Practical capacitors with real insulators have limited but very high resistance and therefore there is a very small leakage DC current in the capacitors which is extremely small.

If the insulator is made very thin in the order of nanometers, electrons can tunnel through this type of capacitors and the conduction of the DC currents becomes appreciable. This occurs in the the MOS capacitors of the MOS transistors when its insulator thickness becomes in the range of nanometers.

It also can conduct DC currents when the applied DC voltage on the capacitor approaches its breakdown voltage. Under this condition avalanche breakdown takes place which results in an exponential growth of the free charge carriers in the insulator of the capacitor.

Best wishes

This is true in the steady state (when time is relatively big), but in the transioire regime (near t = 0) the capacitor does not switch on the DC current.

See also the question below.

https://www.researchgate.net/post/Does_the_current_flow_through_a_capacitor_and_if_so_why

My intuitive guess is this: When you apply a DC voltage to a capacitor, there could be a minimum time when you could detect a current until the plate "gets full of electrons". After that no more of them could go there, so no more induction variations will happen. When an AC is applied, there will be "electron´s movement", so there will be current.

Mathmatically, Franco Chiarella explained great.

Wikipedia has an excellent analogy for understanding this. In the analogy, the copper wire is a pipe, electrons are water. A capacitor is an elastic band that expands when you push water from one direction to the other. Because he elastic band is blocking the direct pathway for water from one direction to the other, water can never pass to the other side. So, there is no "direct current (DC)" . Applying higher pressure only increases the tension on the elastic band. However, if you keep pushing, pulling, pushing, pulling water from the first side, the elastic band pushes, pulls, pushes, pulls the water from the other side, therefore creating an AC water flow. Although both sides are DC BLOCKED, one can easily create an AC CURRENT.

Capacitors are essentially two plates that are mounted next to each other, with a gap between them so that the plates don't touch. That's why it's drawn as --| |-- on a diagram.

Direct current can't jump the gap between plates, because it would take a massive amount of voltage to force the electron to jump the gap between plates. The electrons hit the plate and stop.

Alternating current, on the other hand, is moving the electrons back and forth in place -- so the plate on one side of the capacitor is constantly having electrons pushed in and then pulled back out. This motion creates a small electric field which induces the same alternating current in the other plate, because electric fields can jump the gap between plates.

Hope that helps with your general understanding. Other people have posted lots of great math, but I didn't see much in the way of conceptual understanding of the physics at play.

Hi Tolga, the picture (of William Beaty) is for you...

Here are some related questions, too...

https://www.researchgate.net/post/What_do_coupling_capacitors_really_do_in_AC_amplifiers_Are_they_rechargable_batteries_conveying_voltage_variations_or_diverting_biasing_currents

https://www.researchgate.net/post/How_do_inductor_and_capacitor_oppose_input_voltage_sources_Can_we_consider_them_as_time-variable_voltage_sources

My answer is going to match with quite a few answers. I would like to explain in my own way.

Xc = (1/ 2* (pi)*f*C), where f= frequency of the impinging signal and C= capacitance value.

When f=0 for DC, Xc tends to infinity. So, C acts as an open circuit for DC.

The answer depends on what is meant by DC. It's true that the impedance of a capacitor goes to infinity as the frequency goes to zero. But the concept of impedance assumes sinusoidal steady state and so isn't quite applicable to a genuine DC circuit.

It's true that, for a circuit in DC steady state (all voltages and currents are constant), any capacitor in the circuit has zero current through. But does this mean the capacitor blocks DC current?

Consider a simple circuit consisting of an (ideal) constant current source I and (ideal) capacitor C. Does the capacitor block this DC current? For concreteness, stipulate that the initial voltage across is zero. It follows that the voltage across the capacitor is the linear function of time V = (I/C) t and the current through is I, i.e., the (ideal) capacitor does not block (present an open circuit) to a DC (constant) current. Since this circuit has no DC steady state - the voltage across is never constant - this is not a contradiction.

Now, in the non-ideal world, capacitors are often used to couple a voltage signal with non-zero mean to another circuit or load that requires a zero mean voltage signal. In such a a case, the capacitor charges to the mean of the voltage signal and this mean voltage is dropped across the capacitor leaving only the AC component for the following circuit. These are often called "DC blocking capacitors". I would prefer "DC dropping capacitors" instead.

So, why in cyclic voltammetry of a capacitive material, the E=f(I) curve is rectangular, meaning that the current is the same on an interval of potential. How can you explain that please ?

To do a measurement it is usually necessary to change something. If the current or voltage changes it is not DC. I am not sure what cyclic voltammetry of a capacitive material is, but when you go from one voltage direction to the other, charges in a dielectric move slightly, from one side of each molecule or atom to the other, so there is a momentary current, in the same way as the diaphragm in the example above bulges in the opposite direction, so a bit or water moves.