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• Early History

• Applications

• Related equations

• Capacitor Networks

Ch.26: Capacitance and Dielectrics.

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Opposite charges insulated from one another form a CAPACITOR.

Charges in space set-up Electric Field.

Fields can do work. Why? __________

________________

E = F/q , F = qE E = kQ/d2 , W = qEd , W/q = Ed , V = Ed , V = kQ/d , and W = qV

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Early History of Capacitors.

Galvanism: In 1800, produced electrical current from the contact of two different metals in a moist environment. Luigi Galvani

(1737-1798)  Frog-leg experiment.

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Alessandro Giuseppe Volta (1745-1827)

The Voltaic Pile.

In 1799, metal discs (zinc with copper or silver) , separated with paperboard discs soaked in saline solution. This stack was the first electric battery.

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Sir Humphry Davy

(1778-1829) Founder of Electrochemistry.                                                 

In 1810, Davy uses the Voltaic Pile to begin electrochemistry. Isolated elemental K, Ca, Ba, Na, Sr, Mg, B, and Si.

His lab assistant was Michael Faraday! Considered his greatest discovery.

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Michael Faraday (1791-1867) discovered variable capacitor.                                   

Later, the SI unit used for capacitance was named a Farad (F) in his honor. 1F = 1 C / V In words: one Farad is equal to one Coulomb per Volt

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Capacitance

•Capacitance is the ability of a conductor to store energy in the form of electrically separated charges.

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Capacitance

•Capacitance is the ratio of charge to potential difference.

C = ΔV

Q

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A parallel plate capacitor before charging. There is no net charge on the plates.

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During charging, the negative charges will move to one plate. Now, each plate has a small charge.

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After charging, each plate has a larger net charge, which it can hold even after the battery is gone.

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Farads

• Capacitance is measured in units called farads (F).

• 1 farad = 1 coulomb/1 volt

• Common measures are in microfarads (F) and picofarads (pF = 1 x 10-12 F).

• Named for Michael Faraday (1791-1867), a prominent British physicist.

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Capacitance depends on:

•Size and shape of capacitor.

•Material between the plates.

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Size and shape of the capacitor:

If no material is between the plates of a capacitor (has to be in a vacuum), then the formula for capacitance is

C = εo d

A

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Permittivity of free space:

The Greek letter epsilon, represents a material-dependent constant called the permittivity of the material. When it is followed by a subscripted o, it is called the permittivity of free space, and has a value of 8.85 x 10-12 C2/N. m2

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Capacitance Example

How much charge is on a 1 F capacitor which has a potential difference of 110 Volts?

Q = CV

How much energy is stored in this capacitor?

Ecap = QV/2

= (1)(110) = 110 Coulombs

= (110) (110) / 2 = 6,050 Joules!

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Parallel Plate CapacitorExample C = 0A/d

Calculate the capacitance of a parallel plate capacitor made from two large square metal sheets 1.3 m on a side, separated by 0.1 m.

A

d

A

F 105.1 10

)1.0()3.1)(3.1)(1085.8( 12C

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(a) Two capacitors in series, (b) Same CHARGE different VOLTAGE.

1 / Ceq = 1 / C1 + 1 / C2

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(a) Capacitors in parallel, (b) Same VOLTAGE different CHARGE.

Ceq = C1 + C2 + C3 ··· Cn

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Dielectric• Placing a dielectric

between the plates increases the capacitance.

• C = C0

Capacitance with dielectric

Dielectric constant (kappa) ( > 1)

Capacitance without dielectric

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Dielectric Constant, Strength• Large capacitance (energy storage)

– Small gap (d)

• Breakdown voltage is material dependent

– Large area (A)

• Would like to keep the device size tolerable

– Large dielectric constant ()

• Large dielectric strength

– Dielectric constant/strength of material

Material Constant Strength (V/m)

Vacuum 1.00000 -

Air 1.00059 3 x 106

Paper 3.7 16 x 106

Teflon 2.1 60 x 106

C

0Ad

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Capacitors can store charge and ENERGY U = q V, and the potential V increases as the charge is placed on the plates (V = Q / C). Since the V changes as the Q is increased, we have to integrate over all the little charges “dq” being added to a plate: U = q V leads to U = V dq = q/c dq = 1/C q dq = Q2 / 2C. And using Q = C V, we get U = Q2 / 2C = C V2 / 2 = Q V / 2So the energy stored in a capacitor can be thought of as the potential energy stored in the system of positive charges that are separated from the negative charges, much like a stretched spring has potential energy associated with it.

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ELECTRIC FIELD ENERGY

Here's another way to think of the energy stored in a charged capacitor: If we consider the space between the plates to contain the energy (equal to 1/2 C V2) we can calculate an energy DENSITY (Joules per volume). The volume between the plates is area x plate separation, or A d. Then the energy density u is

u = 1/2 C V2 / A d = o E2 / 2

Recall C = o A / d and V =E d.

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Energy density: u = o E2 / 2

This is an important result because it tells us that empty space contains energy if there is an electric field (E) in the "empty" space.

If we can get an electric field to travel (or propagate) we can send or transmit energy and information through empty space!!!