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Q = CV . size 12{Q= ital "CV"} {}

This equation expresses the two major factors affecting the amount of charge stored. Those factors are the physical characteristics of the capacitor, C size 12{C} {} , and the voltage, V . Rearranging the equation, we see that capacitance C size 12{C} {} is the amount of charge stored per volt, or

C = Q V . size 12{C=Q/V} {}

Capacitance

Capacitance C size 12{C} {} is the amount of charge stored per volt, or

C = Q V . size 12{C=Q/V} {}

The unit of capacitance is the farad (F), named for Michael Faraday (1791–1867), an English scientist who contributed to the fields of electromagnetism and electrochemistry. Since capacitance is charge per unit voltage, we see that a farad is a coulomb per volt, or

1 F = 1 C 1 V . size 12{F= { {"1 C"} over {"1 V"} } } {}

A 1-farad capacitor would be able to store 1 coulomb (a very large amount of charge) with the application of only 1 volt. One farad is, thus, a very large capacitance. Typical capacitors range from fractions of a picofarad 1 pF = 10 –12 F size 12{ left (1" pF"="10" rSup { size 8{-"12"} } " F" right )} {} to millifarads 1 mF = 10 –3 F size 12{ left (1" mF"="10" rSup { size 8{-3} } " F" right )} {} .

[link] shows some common capacitors. Capacitors are primarily made of ceramic, glass, or plastic, depending upon purpose and size. Insulating materials, called dielectrics, are commonly used in their construction, as discussed below.

There are various types of capacitors with varying shapes and color. Some are cylindrical in shape, some circular in shape, some rectangular in shape, with two strands of wire coming out of each.
Some typical capacitors. Size and value of capacitance are not necessarily related. (credit: Windell Oskay)

Parallel plate capacitor

The parallel plate capacitor shown in [link] has two identical conducting plates, each having a surface area A size 12{A} {} , separated by a distance d size 12{d} {} (with no material between the plates). When a voltage V size 12{V} {} is applied to the capacitor, it stores a charge Q size 12{Q} {} , as shown. We can see how its capacitance depends on A size 12{A} {} and d size 12{d} {} by considering the characteristics of the Coulomb force. We know that like charges repel, unlike charges attract, and the force between charges decreases with distance. So it seems quite reasonable that the bigger the plates are, the more charge they can store—because the charges can spread out more. Thus C size 12{C} {} should be greater for larger A size 12{A} {} . Similarly, the closer the plates are together, the greater the attraction of the opposite charges on them. So C size 12{C} {} should be greater for smaller d size 12{d} {} .

Two parallel plates are placed facing each other. The area of each plate is A, and the distance between the plates is d. The plate on the left is connected to the positive terminal of the battery, and the plate on the right is connected to the negative terminal of the battery.
Parallel plate capacitor with plates separated by a distance d size 12{d} {} . Each plate has an area A size 12{A} {} .

It can be shown that for a parallel plate capacitor there are only two factors ( A size 12{A} {} and d size 12{d} {} ) that affect its capacitance C size 12{C} {} . The capacitance of a parallel plate capacitor in equation form is given by

C = ε 0 A d . size 12{C=e rSub { size 8{0} } A/d} {}

Capacitance of a parallel plate capacitor

C = ε 0 A d size 12{C=e rSub { size 8{0} } A/d} {}

A size 12{A} {} is the area of one plate in square meters, and d is the distance between the plates in meters. The constant ε 0 is the permittivity of free space; its numerical value in SI units is ε 0 = 8.85 × 10 12 F/m . The units of F/m are equivalent to C 2 /N · m 2 size 12{ left (C rSup { size 8{2} } "/N" cdot m rSup { size 8{2} } right )} {} . The small numerical value of ε 0 size 12{e rSub { size 8{0} } } {} is related to the large size of the farad. A parallel plate capacitor must have a large area to have a capacitance approaching a farad. (Note that the above equation is valid when the parallel plates are separated by air or free space. When another material is placed between the plates, the equation is modified, as discussed below.)

Capacitance and charge stored in a parallel plate capacitor

(a) What is the capacitance of a parallel plate capacitor with metal plates, each of area 1.00 m 2 size 12{m rSup { size 8{2} } } {} , separated by 1.00 mm? (b) What charge is stored in this capacitor if a voltage of 3.00 × 10 3 V is applied to it?

Strategy

Finding the capacitance C size 12{C} {} is a straightforward application of the equation C = ε 0 A / d size 12{C=e rSub { size 8{0} } A/d} {} . Once C size 12{C} {} is found, the charge stored can be found using the equation Q = CV size 12{Q= ital "CV"} {} .

Solution for (a)

Entering the given values into the equation for the capacitance of a parallel plate capacitor yields

C = ε 0 A d = 8.85 × 10 –12 F m 1.00 m 2 1.00 × 10 –3 m = 8.85 × 10 –9 F = 8.85 nF.

Discussion for (a)

This small value for the capacitance indicates how difficult it is to make a device with a large capacitance. Special techniques help, such as using very large area thin foils placed close together.

Solution for (b)

The charge stored in any capacitor is given by the equation Q = CV size 12{Q= ital "CV"} {} . Entering the known values into this equation gives

Q = CV = 8.85 × 10 –9 F 3.00 × 10 3 V = 26.6 μC. alignl { stack { size 12{Q= ital "CV"= left (8 "." "85"´"10" rSup { size 8{-9} } " F" right ) left (3 "." "00"´"10" rSup { size 8{3} } " V" right )} {} #="26" "." 6 µC "." {} } } {}

Discussion for (b)

This charge is only slightly greater than those found in typical static electricity. Since air breaks down at about 3 . 00 × 10 6 V/m size 12{3 "." "00" times "10" rSup { size 8{6} } } {} , more charge cannot be stored on this capacitor by increasing the voltage.

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Questions & Answers

a15kg powerexerted by the foresafter 3second
Firdos Reply
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Xolani Reply
movement in a direction
Jason
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Hosea
Explain why magnetic damping might not be effective on an object made of several thin conducting layers separated by insulation? can someone please explain this i need it for my final exam
anas Reply
Hi
saeid
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Yimam
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Oluwakayode Reply
while
Hosea
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thomas Reply
this is the mass of an atom of an element in ratio with the mass of carbon-atom
Chukwuka
show me how to get the accuracies of the values of the resistors for the two circuits i.e for series and parallel sides
Jesuovie Reply
Explain why it is difficult to have an ideal machine in real life situations.
Isaac Reply
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Promise
what's the s . i unit for couple?
Promise
its s.i unit is Nm
Covenant
Force×perpendicular distance N×m=Nm
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Oluwakayode
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yes
haider
what affects fluid
Doreen Reply
pressure
Oluwakayode
Dimension for force MLT-2
Promise Reply
what is the dimensions of Force?
Osueke Reply
how do you calculate the 5% uncertainty of 4cm?
melia Reply
4cm/100×5= 0.2cm
haider
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melia Reply
= 200g±(5%)10g
haider
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melia
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haider
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haider
the relationship between the applied force and the deflection
melia
sorry wrong question i meant the 5% uncertainty of 4cm?
melia
its 0.2 cm or 2mm
haider
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melia
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Practice Key Terms 6

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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