We saw in the previous tutorials that a **Capacitor** consists of two parallel conductive plates (usually a metal) which are prevented from touching each other (separated) by an insulating material called the “dielectric”.

We also saw that when a voltage is applied to these plates an electrical current flows charging up one plate with a positive charge with respect to the supply voltage and the other plate with an equal and opposite negative charge.

Then, a capacitor has the ability of being able to store an electrical charge Q (units in **Coulombs**) of electrons. When a capacitor is fully charged there is a potential difference, p.d. between its plates, and the larger the area of the plates and/or the smaller the distance between them (known as separation) the greater will be the charge that the capacitor can hold and the greater will be its **Capacitance**.

The capacitors ability to store this electrical charge ( Q ) between its plates is proportional to the applied voltage, V for a capacitor of known capacitance in Farads. Capacitance C is always positive and never negative. The greater the applied voltage the greater will be the charge stored on the plates of the capacitor. Likewise, the smaller the applied voltage the smaller the charge. Therefore, the actual charge Q on the plates of the capacitor and can be calculated as:

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Where: Q (Charge, in Coulombs) = C (Capacitance, in Farads) x V (Voltage, in Volts)

It is sometimes easier to remember this relationship by using pictures. Here the three quantities of Q, C and V have been superimposed into a triangle giving charge at the top with capacitance and voltage at the bottom. This arrangement represents the actual position of each quantity in the *Capacitor Charge* formulas.

and transposing the above equation gives us the following combinations of the same equation:

__Units of:__ Q measured in Coulombs, V in volts and C in Farads.

Then from above we can define the unit of **Capacitance** as being a constant of proportionality being equal to the coulomb/volt which is also called a **Farad**, unit F.

As capacitance represents the capacitors ability (capacity) to store an electrical charge on its plates we can define one Farad as the “*capacitance of a capacitor which requires a charge of one coulomb to establish a potential difference of one volt between its plates*” as firstly described by Michael Faraday. So the larger the capacitance, the higher is the amount of charge stored on a capacitor for the same amount of voltage.

The ability of a capacitor to store a charge on its conductive plates gives it its **Capacitance** value. Capacitance can also be determined from the dimensions or area, A of the plates and the properties of the dielectric material between the plates. A measure of the dielectric material is given by the permittivity, ( ε ), or the dielectric constant. So another way of expressing the capacitance of a capacitor is;

where A is the area of the plates in square metres, m^{2} with the larger the area, the more charge the capacitor can store. d is the distance or separation between the two plates. The smaller is this distance, the higher is the ability of the plates to store charge, since the -ve charge on the -Q charged plate has a greater effect on the +Q charged plate, resulting in more electrons being repelled off of the +Q charged plate, and thus increasing the overall charge.

ε_{0} (epsilon) is the value of the permittivity for air which is 8.84 x 10^{-12} F/m, and ε_{r} is the permittivity of the dielectric medium used between the two plates.

We have said previously that the capacitance of a parallel plate capacitor is proportional to the surface area A and inversely proportional to the distance, d between the two plates and this is true for dielectric medium of air. However, the capacitance value of a capacitor can be increased by inserting a solid medium in between the conductive plates which has a dielectric constant greater than that of air.

Typical values of epsilon ε for various commonly used dielectric materials are: Air = 1.0, Paper = 2.5 – 3.5, Glass = 3 – 10, Mica = 5 – 7 etc.

The factor by which the dielectric material, or insulator, increases the capacitance of the capacitor compared to air is known as the **Dielectric Constant**, (**k**). “k” is the ratio of the permittivity of the dielectric medium being used to the permittivity of free space otherwise known as a vacuum.

Therefore, all the capacitance values are related to the permittivity of vacuum. A dielectric material with a high dielectric constant is a better insulator than a dielectric material with a lower dielectric constant. Dielectric constant is a dimensionless quantity since it is relative to free space.

A parallel plate capacitor consists of two plates with a total surface area of 100 cm^{2}. What will be the capacitance in pico-Farads, (pF) of the capacitor if the plate separation is 0.2 cm, and the dielectric medium used is air.

then the value of the capacitor is 44pF.

Consider the following circuit.

Assume that the capacitor is fully discharged and the switch connected to the capacitor has just been moved to position A. The voltage across the 100uf capacitor is zero at this point and a charging current ( i ) begins to flow charging up the capacitor until the voltage across the plates is equal to the 12v supply voltage. The charging current stops flowing and the capacitor is said to be “fully-charged”. Then, Vc = Vs = 12v.

Once the capacitor is “fully-charged” in theory it will maintain its state of voltage charge even when the supply voltage has been disconnected as they act as a sort of temporary storage device. However, while this may be true of an “ideal” capacitor, a real capacitor will slowly discharge itself over a long period of time due to the internal leakage currents flowing through the dielectric. This is an important point to remember as large value capacitors connected across high voltage supplies can still maintain a significant amount of charge even when the supply voltage is switched “OFF”.

If the switch was disconnected at this point, the capacitor would maintain its charge indefinitely, but due to internal leakage currents flowing across its dielectric the capacitor would very slowly begin to discharge itself as the electrons passed through the dielectric. The time taken for the capacitor to discharge down to 37% of its supply voltage is known as its Time Constant.

If the switch is now moved from position A to position B, the fully charged capacitor would start to discharge through the lamp now connected across it, illuminating the lamp until the capacitor was fully discharged as the element of the lamp has a resistive value. The brightness of the lamp and the duration of illumination would ultimately depend upon the capacitance value of the capacitor and the resistance of the lamp (t = C x R). The larger the value of the capacitor the brighter and longer will be the illumination of the lamp as it could store more charge.

Calculate the charge in the above capacitor circuit.

then the charge on the capacitor is 1.2 millicoulombs.

Electrical current can not actually flow through a capacitor as it does a resistor or inductor due to the insulating properties of the dielectric material between the two plates. However, the charging and discharging of the two plates gives the effect that current is flowing.

The current that flows through a capacitor is directly related to the charge on the plates as current is the rate of flow of charge with respect to time. As the capacitors ability to store charge (Q) between its plates is proportional to the applied voltage (V), the relationship between the current and the voltage that is applied to the plates of a capacitor becomes:

As the voltage across the plates increases (or decreases) over time, the current flowing through the capacitance deposits (or removes) charge from its plates with the amount of charge being proportional to the applied voltage. Then both the current and voltage applied to a capacitance are functions of time and are denoted by the symbols, i_{(t)} and v_{(t)}.

However, from the above equation we can also see that if the voltage remains constant, the charge will become constant and therefore the current will be zero!. In other words, no change in voltage, no movement of charge and no flow of current. This is why a capacitor appears to “block” current flow when connected to a steady state DC voltage.

We now know that the ability of a capacitor to store a charge gives it its capacitance value C, which has the unit of the **Farad, F**. But the farad is an extremely large unit on its own making it impractical to use, so submultiple’s or fractions of the standard Farad unit are used instead.

To get an idea of how big a Farad really is, the surface area of the plates required to produce a capacitor with a value of just one Farad with a reasonable plate separation of just say 1mm operating in a vacuum. If we rearranging the equation for capacitance above this would give us a plate area of:

A = Cd ÷ 8.85pF/m = (1 x 0.001) ÷ 8.85×10^{-12} = 112,994,350 m^{2}

or 113 million m^{2} which would be equivalent to a plate of more than 10 kilometres x 10 kilometres (over 6 miles) square. That’s huge.

Capacitors which have a value of one Farad or more tend to have a solid dielectric and as “One Farad” is such a large unit to use, prefixes are used instead in electronic formulas with capacitor values given in micro-Farads (μF), nano-Farads (nF) and the pico-Farads (pF). For example:

Convert the following capacitance values from a) **22nF** to **uF**, b) **0.2uF** to **nF**, c) **550pF** to **uF**.

a) 22nF = 0.022uF

b) 0.2uF = 200nF

c) 550pF = 0.00055uF

While one Farad is a large value on its own, capacitors are now commonly available with capacitance values of many hundreds of Farads and have names to reflect this of “Super-capacitors” or “Ultra-capacitors”.

These capacitors are electrochemical energy storage devices which utilise a high surface area of their carbon dielectric to deliver much higher energy densities than conventional capacitors and as capacitance is proportional to the surface area of the carbon, the thicker the carbon the more capacitance it has.

Low voltage (from about 3.5V to 5.5V) super-capacitors are capable of storing large amounts of charge due to their high capacitance values as the energy stored in a capacitor is equal to 1/2(C x V^{2}).

Low voltage super-capacitors are commonly used in portable hand held devices to replace large, expensive and heavy lithium type batteries as they give battery-like storage and discharge characteristics making them ideal for use as an alternative power source or for memory backup. Super-capacitors used in hand held devices are usually charged using solar cells fitted to the device.

Ultra-capacitor are being developed for use in hybrid electric cars and alternative energy applications to replace large conventional batteries as well as DC smoothing applications in vehicle audio and video systems. Ultra-capacitors can be recharged quickly and have very high energy storage densities making them ideal for use in electric vehicle applications.

When a capacitor charges up from the power supply connected to it, an electrostatic field is established which stores energy in the capacitor. The amount of energy in **Joules** that is stored in this electrostatic field is equal to the energy the voltage supply exerts to maintain the charge on the plates of the capacitor and is given by the formula:

so the energy stored in the 100uF capacitor circuit above is calculated as:

The next tutorial in our section about Capacitors, we look at Capacitor Colour Codes and see the different ways that the capacitance and voltage values of the capacitor are marked onto its body.

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1. If supposing a capacitor of capacitance 1 farad is connected to a voltage supply of two volt, will the capacitor store an energy of 2 joule?

2. If a capacitor is connected is connected to a voltage of 1 volt and can 1 coulomb of charge will be stored in it? Will this be linear? i.e 2 volt , 2 coulomb so on?

I am experimenting with ultra capacitors in automotive applications and am curious about their construction. How much surface area do the plates of a 2.5V rated 2,700Farad Maxwell super capacitor have (2.5v 2600F BCAP0010 A08) Approximate dimensions 60mm DIA x 150mm length cylindrical shape

Please consult the manufacturers datasheet.

I just need to know the name of any simulation that allow me to know the behavior and the motion of the charges under the sun with different angles >

About the capacitors working

can we charge a capacitor with positive charges only or not and why ?

No, for the same reason a battery can not have only one terminal. Charges come in pairs.

What happens when a charge passes from battery to capacitor . Whether the charge is same as flown or half of it.

Some losses will happen ESR Equivalent series resistance of component

I have a question. A d.c of 4A flows into a previously uncharged 20 micro farads for 3ms determine the p.d between the plates

i = C(dV/dt) re-arranging gives: dV = i/C.dt = (1/C)i.dt

Therefore, V = (1/20uF) x 4A x 3ms = 600V

Question:

Assuming that a capacitor is disconnected from the charging battery,explain how the 1)capacitance 2) p.d. across the plates and 3) energy stored in parallel plate capacitor change ,when a medium of dielectric constant k is introduced.

Wow! what an awesome website! I have a question about capacitance.

there are two conducing rods of which radii are a and b with the separation distance of 2d from center to center.

picture like this …

O (rod 1) o.(origin) O(rod 2)

V=V0 V=0

radius a radius b

-d d

↑y axis

↙ →x axis

Z axis

(1) calculate potential equation describing this system

(2) calculate capacitance from the relationship between potential and charge.

(3) calculate the capacitance if the permittivity of x>0 region is ε1 and that of x<0 region is ε2.

Please solve my problem

The capacitance of a parallel plate capacitor is with a certain dielectric medium between the plates of capacitor. Find the capacitance of the capacitor if

1) the distance between two plates is double; and

2) the area of plates is halved.

plz send reply fast

Capacitance is inversely proportional to the distance between the plates of the capacitor so with the increase in the distance there will be a decrease in capacitance.

In the case of the area, it is directly proportional to capacitance, therefore the smaller the area the lesser the capacitance there will be.