how to find potential difference across a capacitor

Capacitor

Capacitance

A capacitor is a device for storing separated charge.  No unmarried electronic component plays a more of import function today than the capacitor.  This device is used to store information in estimator memories, to regulate voltages in power supplies, to establish electrical fields, to shop electric energy, to observe and produce electromagnetic waves, and to mensurate fourth dimension.  Whatever two conductors separated by an insulating medium form a capacitor.

A parallel plate capacitor consists of two plates separated by a sparse insulating textile known as a dielectric.   In a parallel plate capacitor electrons are transferred from one parallel plate to another.

We have already shown that the electrical field between the plates is constant with magnitude E = σ/ε₀ and points from the positive towards the negative plate.

The potential difference between the negative and positive plate therefore is given by

∆U = U_pos - U_neg = -q ∫_neg ^pos E·dr = q E d.

When integrating, dr points from the negative to the positive plate in the opposite direction from Due east.  Therefore E·dr = -Edr, and the minus signs cancel.
The positive plate is at a higher potential than the negative plate.

Field lines and equipotential lines for a abiding field betwixt two charged plates are shown on the correct.  One plate of the capacitor holds a positive charge Q, while the other holds a negative charge -Q.  The accuse Q on the plates is proportional to the potential deviation Five across the ii plates.  The capacitance C is the proportional constant,

Q = CV,  C = Q/5.

C depends on the capacitor's geometry and on the type of dielectric material used. The capacitance of a parallel plate capacitor with two plates of expanse A separated by a distance d and no dielectric material between the plates is

C = ε₀A/d.

(The electric field is E = σ/ε₀. The voltage is V = Ed = σd/ε₀.  The accuse is Q = σA. Therefore Q/Five = σAε₀/σd = Aε₀/d.)
The SI unit of capacitance is Coulomb/Volt = Farad (F).
Typical capacitors have capacitances in the picoFarad to microFarad range.

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The capacitance tells u.s.a. how much accuse the device stores for a given voltage.  A dielectric between the conductors increases the capacitance of a capacitor.  The molecules of the dielectric material are polarized in the field between the two conductors.  The entire negative and positive charge of the dielectric is displaced by a small amount with respect to each other.  This results in an effective positive surface charge on one side of the dielectric and a negative surface charge on the other side of the dielectric.  These constructive surface charges on the dielectric produce an electrical field, which opposes the field produced by the surface charges on the conductors, and thus reduces the voltage betwixt the conductors.  To keep the voltage up, more accuse must be put onto the conductors.  The capacitor thus stores more than charge for a given voltage.  The dielectric constant κ is the ratio of the voltage Five₀ between the conductors without the dielectric to the voltage 5 with the dielectric, κ = V₀/5, for a given amount of charge Q on the conductors.

In the diagram in a higher place, the same amount of accuse Q on the conductors results in a smaller field between the plates of the capacitor with the dielectric.  The college the dielectric constant κ, the more charge a capacitor can shop for a given voltage. For a parallel-plate capacitor with a dielectric between the plates, the capacitance is

C = Q/Five = κQ/V₀ = κε₀A/d = εA/d,

where ε = κε₀.  The static dielectric constant of any material is always greater than i.

Typical dielectric constants

Cloth Dielectric Constant Air one.00059 Aluminum Silicate five.3 to five.5 Bakelite 3.7 Beeswax (yellow) 2.7 Butyl Rubber ii.4 Formica 20 4.00 Germanium 16 Drinking glass four to 10 Gutta-percha two.6 Halowax oil 4.viii Kel-F ii.6 Lucite 2.8 Mica four to eight Micarta 254 3.four to v.4 Mylar 3.ane Neoprene rubber 6.7 Nylon 3.00

Material Dielectric Abiding Paper 1.5 to 3 Alkane series 2 to 3 Plexiglass 3.4 Polyethylene 2.2 Polystyrene 2.56 Porcelain 5 to 7 Pyrex drinking glass 5.half dozen Quartz 3.7 to 4.five Silicone oil two.5 Steatite 5.3 to six.v Strontium titanate 233 Teflon two.1 Tenite 2.9 to iv.v Vacuum 1.00000 Vaseline 2.sixteen Water (distilled) 76.7 to 78.two Wood 1.2 to 2.ane

If a dielectric with dielectric constant κ is inserted between the plates of a parallel-plate of a capacitor, and the voltage is held constant past a battery, the charge Q on the plates increases by a factor of κ.  The battery moves more electrons from the positive to the negative plate.  The magnitude of the electric field between the plates, East = V/d stays the aforementioned.

If a dielectric is inserted betwixt the plates of a parallel-plate of a capacitor, and the accuse on the plates stays the same because the capacitor is asunder from the battery, then the voltage V decreases by a factor of κ, and the electric field between the plate, E = V/d, decreases past a gene of κ.

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Module 5: Question 2:

(a)  A parallel-plate capacitor initially has a voltage of 12 V and stays continued to the battery. If the plate spacing is now doubled, what happens?
(b)  A parallel-plate capacitor initially is continued to a battery and the plates hold charge ±Q. The battery is and so asunder. If the plate spacing is now doubled, what happens?

Hint:  The battery is a charge pump. It tin pump accuse from i plate to the other to maintain a constant potential deviation.
No battery <--> no charge pump.  Accuse cannot move from 1 plate to the other.

Talk over this with your fellow students in the discussion forum!

Link:  PhET Capacitor Lab (Bones)

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Energy stored in a capacitor

The energy U stored in a capacitor is equal to the work W done in separating the charges on the conductors.  The more than charge is already stored on the plates, the more work must be washed to divide additional charges, because of the strong repulsion betwixt like charges.  At a given voltage, it takes an minute amount of work ∆W = Five∆Q to separate an additional infinitesimal amount of charge ∆Q.
(The voltage V is the corporeality of work per unit of measurement accuse.)
Since Five = Q/C, V increases linear with Q.  The total work done in charging the capacitor is

W = ∫₀ ^Qf VdQ =  ∫₀ ^Qf (Q/C)dQ = ½(Q_f ^two/C) = ½VQ_F = V_averageQ_f
Using Q = CV nosotros can also write U = ½(Qⁱⁱ/C)  or  U = ½CV².

Problem:

Each memory prison cell in a reckoner contains a capacitor to store charge.  Charge being stored or not beingness stored corresponds to the binary digits 1 and 0.  To pack the cells more densely, trench capacitors are frequently used in which the plates of a capacitor are mounted vertically forth the walls of a trench etched into a silicon chip.  If nosotros take a capacitance of l femtoFarad = 50*10^-15 F and each plate has an area of 20*10^-12 yard² (micron-sized trenches), what is the plate separation?

Solution:

• Reasoning:
  The capacitance of a parallel plate capacitor with two plates of area A separated past a altitude d and no dielectric material betwixt the plates is C = ε₀A/d.
• Details of the calculation:
  C = ε₀A/d, d = ε₀A/C = (8.85*10^-12*20*10^-12/(l*ten^-15)) m = 3.54*ten^-nine m.
  Typical atomic dimensions are on the gild of 0.1 nm, so the trench is on the order of 30 atoms broad.

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For whatever insulator, there is a maximum electric field that can be maintained without ionizing the molecules.  For a capacitor this means that there is a maximum commanded voltage that that can be placed across the conductors.  This maximum voltage depends the dielectric in the capacitor.  The corresponding maximum field E_b is called the dielectric strength of the material.  For stronger fields, the capacitor 'breaks down' (similar to a corona discharge) and is usually destroyed.  Almost capacitors used in electrical circuits carry both a capacitance and a voltage rating.  This breakup voltage Five_b is related to the dielectric strength Due east_b.  For a parallel plate capacitor nosotros have V_b = E_bd.

Material	Dielectric Strength (5/m)
Air	3*106
Bakelite	24*10six
Neoprene rubber	12*10six
Nylon	14*10vi
Paper	16*10six
Polystyrene	24*tenhalf dozen
Pyrex drinking glass	14*xhalf-dozen
Quartz	viii*106
Silicone oil	15*106
Strontium titanate	8*ten6
Teflon	60*106

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Capacitors in serial or parallel

A capacitor is a device for storing separated charge and therefore storing electrostatic potential free energy.  Circuits often contain more one capacitor.

Consider ii capacitors in parallel as shown on the right

When the bombardment is continued, electrons will catamenia until the potential of point A is the aforementioned every bit the potential of the positive terminal of the battery and the potential of point B is equal to that of the negative terminal of the battery.  Thus, the potential difference betwixt the plates of both capacitors is V_A - V_B = V_bat.  We have C₁ = Q₁/V_bat and C₂ = Q_two/5_bat, where Q_i is the charge on capacitor C_i, and Q₂ is the accuse on capacitor C_ii.  Allow C be the equivalent capacitance of the 2 capacitors in parallel, i.due east. C = Q/Five_bat, where Q = Q₁ + Q₂.  Then C = (Q₁ + Q₂)/V_bat = C_ane + C_ii.

For capacitors in parallel, the capacitances add.

For more than ii capacitors we accept

C = C_one + C₂ + C_iii + C₄ + ... .

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Consider two capacitors in series every bit shown on the right.
Let Q correspond the total charge on the top plate of C₁, which and so induces a accuse -Q on its bottom plate.  The charge on the bottom plate of C₂ volition be -Q, which in plow induces a charge +Q on its acme plate as shown.
Permit V₁ and 5_two represent the potential differences between plates of capacitors C₁ and C₂, respectively.
Then V₁ + 5₂ = 5_bat, or (Q/C₁) + (Q/C₂) = Q/C, or (ane/C₁) + (1/C₂) = 1/C.

For more than than 2 capacitors in serial we have
1/C = 1/C₁ + ane/C₂ + 1/C₃ + 1/C₄ + ... .
where C is equivalent capacitance of the two capacitors.

For capacitors in series the reciprocal of their equivalent capacitance equals the sum of the reciprocals of their individual capacitances.

Problem:

What full capacitances can yous make by connecting a 5 μF and an 8 μF capacitor together?

Solution:

• Reasoning:
  We can connect the capacitors either in series or in parallel.
  To obtain the largest capacitance, nosotros have to connect the capacitors in parallel.
  To obtain the smallest capacitance, nosotros have to connect the capacitors in series.
• Details of the calculation:
  Connecting the capacitors in parallel:
  C_largest = (five + 8) μF = 13 μF.
  Connecting the capacitors in series.
  1/C_smallest = (1/5+ 1/viii) (μF)^-1 = 13/(40 μF) = 0.325/μF.
  C_smallest = twoscore/xiii μF = 3.077 μF.

Source: http://labman.phys.utk.edu/phys136core/modules/m5/capacitors.html

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