There're some easy concepts which I don't get exactly in my mind. I'm afraid I've been studying these things for two years of my engineering but they still bother me. Capacitor is one of them. Can someone explain?

  • What does a capacitor do? Does it store charges? If so, then how does it do?

I have searched it on Google and Yahoo but didn't find any helpful thing there (for me). So I will be glad if I got my problem solved here.

P.S. I hope that the question would not again be an off-topic, as it always does and moreover people don't suggest then where to go. It's a real sad thing.

  • \$\begingroup\$ Only ask one question. They are too broad, also. \$\endgroup\$ Commented Jul 7, 2013 at 18:10
  • \$\begingroup\$ I thought these were small topics to discuss, at least for me as I need just the basics for them. \$\endgroup\$
    – Syed Sahl
    Commented Jul 7, 2013 at 18:12
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    \$\begingroup\$ When a question is off-topic here you can't require me to know where to put the question - I don't know more about other fora than you do! \$\endgroup\$ Commented Jul 7, 2013 at 18:28
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    \$\begingroup\$ Have you read the top two google hits howstuffworks.com/capacitor.htm and en.wikipedia.org/wiki/Capacitor ? If there are terms there you don't understand, have you researched those? \$\endgroup\$
    – pjc50
    Commented Jul 7, 2013 at 18:29
  • \$\begingroup\$ Is this really still too broad a question? It is rather basic, but other similar question have been very well received (5+ upvotes) here (cf electronics.stackexchange.com/questions/8745 and electronics.stackexchange.com/questions/4788 ). Just because coil guns have more geek chic than capacitors doesn't mean that "How does a coil gun work?" is an okay question while "How does a capacitor work?" is not. \$\endgroup\$
    – us2012
    Commented Jul 7, 2013 at 18:34

2 Answers 2


If by charges you mean electric charges, then no, a capacitor does not store charges. This is a common misconception, maybe due to the multiple meanings of the word charge. When some charge goes in one terminal of a capacitor, an equal amount of charge leaves the other. So, the total charge in the capacitor is constant.

What capacitors store is energy. Specifically, they store it in an electric field. All the electrons are attracted to all the protons. At equilibrium, there are equal numbers of protons and electrons on each plate of the capacitor, and there is no stored energy, and no voltage across the capacitor.

But, if you connect the capacitor to something like a battery, then some of the electrons will be pulled away from one plate, and an equal number of electrons will be pushed on to the other plate. Now one plate has a net negative charge, and the other has a net positive charge. This results in a difference in electrical potential between the plates, and an increasingly strong electric field as more charges are separated.

The electric field exerts a force on the charges which attempts to return the capacitor back to equilibrium, with balanced charges on each plate. As long as the capacitor remains connected to the battery, this force is balanced by force of the battery, and the imbalance remains.

If the battery is removed, and we leave the circuit open, the charges can't move, so the charge imbalance remains. The field is still applying a force to the charges, but they can't move, like a ball at the top of a hill, or a spring held under tension. The energy stored in the capacitor remains.

If the capacitor terminals are connected with a resistor, then the charges can move, so there is a current. The energy that was stored in the capacitor is converted to heat in the resistor, the voltage decreases, the charges become less imbalanced, and the field weakens.

Further reading: CAPACITOR COMPLAINTS (1996 William J. Beaty)

  • \$\begingroup\$ And if I attach the capacitor with a load(any appliance) it will provide the appliance a DC ? \$\endgroup\$
    – Syed Sahl
    Commented Jul 8, 2013 at 8:08
  • \$\begingroup\$ @SyedSahl a resistor is a simple load. It doesn't matter what the load is; if there is stored energy in the capacitor then there is a voltage, and if charges can move, there will be a current. \$\endgroup\$
    – Phil Frost
    Commented Jul 8, 2013 at 11:07
  • \$\begingroup\$ I mean what type of current will flow? AC or DC? \$\endgroup\$
    – Syed Sahl
    Commented Jul 8, 2013 at 11:22
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    \$\begingroup\$ @SyedSahl it depends on the load, and how you define AC and DC. If the load is purely resistive, then you get an exponentially decaying voltage and current. If the load is a current sink, then you git a linearly decaying voltage and constant current. If the load is an inductor, then the stored energy bounces back and forth between the capacitor and inductor and you get AC until something (wire resistance, EM radiation...) absorbs the energy (LC tank circuit). A boost converter load will draw current in pulses which could be considered AC. It depends on the load. \$\endgroup\$
    – Phil Frost
    Commented Jul 8, 2013 at 11:35
  • \$\begingroup\$ Mmmmh...perhaps it is a kind of "hairsplitting", on the other hand: Is it really wrong to say "Yes, the capacitor can store charges - positive charges on one plate and negative charges on the other one. Hence, for answering the question, is it really necessary to add up - automatically - both sides? \$\endgroup\$
    – LvW
    Commented Apr 5, 2016 at 13:47

In short, capacitors are two conductive objects, often small plates, separated by something that insulates, known as a dielectric. Much like the static buildup that occurs if you rub a balloon on your arm and put it to your hair, opposite charges build up on either plate, allowing it to store energy in the form of charge. There are 2 other key factors that influence capacitor behaviour and make them so useful. -They charge exponentially, not linearly. Say I charge a capacitor with a constant voltage, and I measure the voltage across the capacitor (which corresponds to the charge held within it) every say, 5 seconds. It would not go up in say, 0.1 every 5 seconds. Instead it increases by a set percentage of the total capacity per unit time. This is effectively the same principle (except in reverse) to radioactive decay - the "half-life" is an intuitive concept, corresponding to the time taken for the amount to reduce to 50% of what it was - but not for it to lose a set amount (i.e. it's not 50 molecules per second, it's 50% per second). It looks something like this: test

As you can see it charges fast at the start, but then slows down as charge accumulates.

-The second is the consequences of this charge accumulation. As the voltage increases, the current "through" the capacitor drops - apparently increasing the electrical resistance of the capacitor. However, if we were to reverse the polarity of the input power supply, switching them around, it has the effect of "decreasing" the resistance - the charge, rather than being squeezed into the capacitor, can easily flow out and in fact effectively boosts the effective voltage. The main consequence of this is that the capacitor resists DC, but allows AC. More concretely, the higher the frequency of the voltage polarity switching (i.e. AC), the less the capacitor will impede the flow of current in circuit. The capacitor can be thought of as an electric spring. You push down on it, symbolising current flowing into it. At first it offers little resistance. However, as you keep pushing, the spring pushes back harder, until you can effectively push no more. This is equivalent to the voltage across the capacitor (again equivalent to charge stored within it) getting close to the input voltage - like the upward force of the spring balancing out against your weight. Now what happens if you push in the opposite direction? The spring works with you instead of against you, increasing the output force past what you could hope to achieve with your muscles and weight alone.

So how can we exploit this? There are two main types of capacitor usage depending on how they are arranged in a circuit - "coupling", where the capacitor is in series, and "decoupling", capacitor in parallel. Both make use of these aforementioned principles.

Coupling are used in blocking DC - this is most often found in signal processing and radios. The smaller the capacitor, the higher the frequency it impedes (as it charges faster), so by adjusting capacitance, we can adjust frequencies blocked. When used with an inductor (the diametric opposite of a capacitor) - the most relevant property of which is the blocking of HIGH frequencies, we can restrict signals into a particular "band" of frequencies - a "band pass" circuit. This is critical in radios to transmit or receive at the desired frequency.

Coupling capacitors are also used in timing circuits - since transistors (electronic switches) turn on at a known voltage, and capacitors charge at a known rate, they can be used to only turn the transistor on at a certain time (or frequency).

Decoupling capacitors are used either for energy storage or for electrical "damping". Again, it helps to think about it in terms of a spring.

A spring in a pellet gun shows the energy storage perfectly. The spring is pulled back, analogous to the capacitor being charged, then released, allowing it to discharge its energy into a "load" - mechanically speaking, the pellet (or other ammunition), electrically, a component, say, a light. Capacitors are ideal for situations in which lots of energy is needed in a short period of time, since they discharge extremely quickly - for instance, a defibrillator. The battery alone could not possibly discharge all the required energy so quickly, so the internal capacitor instead stores it and releases as required.

For the damping, it's best to think of the capacitor/spring analogy as the spring in car suspension. Car suspension protects the car (and passengers) from damage by absorbing some of the energy of the vertical movement of the car. If a wheel is pushed up very quickly by going over a large stone, the rest of the car is less affected thanks to the suspension, which absorbs energy then releases it slowly by pushing the car up. In the same way, a decoupling capacitor can "smooth out" electrical signals or pulses. Analogous to the stone, sometimes the nature of electrical generation, or malfunctions, can cause voltage "spikes". Even very short voltage spikes can cause severe damage to some equipment. The decoupling capacitor is able to absorb this "shock" and reduce the chance of damage occuring. In addition, it can be used to produce smoother transitions between states - like a light fading out, even after the switch has been turned off.

Hope that helps. Sorry if it's a bit verbose, but I aim to be comprehensive.

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    \$\begingroup\$ "Say I charge a capacitor with a constant voltage" - if you do that it charges immediately - it does not exponentially rise as you have said. It charges exponentially when a resistor is in series with it and attains 63.2% percent of its charge in time = RC. Caps charge linearly with a constant current applied. It is WRONG to say they charge exponentially in the general way you have. A lot of folk "see" the mechanical model of a cap like a flywheel/mass and reserve a spring for the model of inductance - this naturally equates current to force which I think is easier to understand. \$\endgroup\$
    – Andy aka
    Commented Jul 8, 2013 at 7:57
  • \$\begingroup\$ You're right, I should have been more specific. I normally go by the real world assumption that there will always be some sort of resistance and therefore one can see the exponential charge, but I forgot to stop that for the purposes of a theoretical model. \$\endgroup\$ Commented Jul 10, 2013 at 14:36

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