01 Capacitance.pptx

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Objectives:

After completing this module,

you should be able to:

• Define capacitance in terms of charge and voltage, and calculate the capacitance for a parallel plate capacitor given separation and area of the plates.

• Define dielectric constant and apply to calculations of voltage, electric field

intensity, and capacitance.

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-Capacitor

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Capacitor

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-+ + + + + + + + +

Q V∝

-+ + + + + + + + + + + + + + + + + + + + + +

Q

∝

V

Q V

∝

Q = C·V

C = Q

V

C = Capacitance

U = (qE)d U/q = Ed

V = Ed Work = Fd

More Charge, More Lines in a given Area, Larger Electric Field, More Voltage. Voltage (V) is directly proportional to Charge (Q).

0

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Capacitance in Farads

One farad (F) is the capacitance C of a conductor that holds one coulomb of charge for each volt of potential.

Example: When 40 mC of charge are placed on a conductor, the potential is 8 V. What is the capacitance?

C = 5 mF

(C) ; (F)

(V)

Q coulomb

C farad

V volt

 

40 C

8 V

Q

C

V

m

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𝐼 =𝐼 ₀∙ 𝑒

(

− 𝑡 𝜏

)

𝐼 =𝐼₀∙ 𝑒

(

−𝑡 𝑅𝐶

)

.012= 1

𝑅𝐶

𝑅𝐶=83.3

𝑅(.9412 𝐹)=83.3

𝑅=88.5 Ω

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4.70 nF

𝑉

=

𝑉

₀

∙

𝑒

(

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100 nF

𝑉

=

𝑉

₀

∙

𝑒

(

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2,200,000 nF

𝑉

=

𝑉

₀

∙

𝑒

(

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Parallel Plate Capacitance

d

Area A +Q

-Q

You will recall from Gauss’ law that E is also: For these two parallel plates:

Q is charge on either plate. A is area of plate.

And and Q V C E V d   0 0 Q E A





  0

V

Q

E

d



A



C

Q

₀

A

V



d

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PERMITIVITY OF FREE SPACE - LAB

Use the parallel plate capacitor and a capacitance meter to make a graph that will allow you to find the

permittivity of free space, ε_0.

0

A

C

d



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1 m m

0.5m m

1.5m m

13.9 => 63.9

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Example 3. The plates of a parallel plate capacitor

have an area of 0.4 m

2

and are 3 mm apart in air.

What is the capacitance?

3 mm

d

A

0.4 m2

C = 1.18 nF

0

Q

A

C

V



d



2 2

-12 _C 2 Nm

(8.85 x 10 )(0.4 m ) (0.003 m)

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0

Q

A

C

V



d

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Applications of Capacitors

d

Changing d

Microphone

A microphone converts sound waves into an

electrical signal (varying voltage) by changing d.

0

A

C

d





Q

V

C

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Applications of Capacitors

Changing Area

+ + + +

-Variable Capacitor

The tuner in a radio is a variable capacitor. The changing area A alters capacitance until desired signal is obtained.

0

A

C

d





V

Q

C

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Q = .9412 V C = .9412 F Q = C·V

V ≡ U_E/q U_E = q·V

U_c = ½ q·V U_c = ½ Q·V

U_c = ½ (C·V)V U_c = ½ C·V2

U_c = ½·.9412·4.52

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Example 6: In Ex-4, we found capacitance to be 11.1 nF, the voltage 200 V, and the charge 2.22 mC. Find the potential energy U.

U = 222 mJ

Verify your answer from the other formulas for P.E.

C = 11.1 nF

200 V

Q = 2.22 mC U = ?

Capacitor of Example 5.

2 1

2

(11.1 nF)(200 V)

U



2 1

2

U



CV

2 1

2

;

2

Q

U

QV

U

C

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Capacitance of Spherical Conductor

+Q r

E and V at surface. At surface of sphere:

Recall: And: Capacitance: Capacitance, C 2 ; kQ kQ E V r r   0

1

4

k





0

4

kQ

Q

V

r



r



C

Q

V



0

4

Q

C

V

Q



r

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--

-C

=

Q

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Find the total

charge

and

energy

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Example 1: What is the capacitance of a metal sphere

of radius 8 cm?

r = 0.08 m Capacitance, C

+Q r

Capacitance: C = 4_or

C = 8.90 x 10-12 F

Note: The capacitance depends only on physical para- meters (the radius r) and is not determined by either charge or potential. This is true for all capacitors.

2

-12 C

N m

4 (8.85 x 10

)(0.08 m)

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Q = 3.56 nC

Q = 3.56 nC Total Charge on Conductor:

Example 1 (Cont.):

What charge Q is needed to give

a potential of 400 V?

r = 0.08 m Capacitance, C

+Q r

C = 8.90 x 10-12 F

Note: The farad (F) and the coulomb (C) are

extremely large units for static electricity. The SI prefixes micro m, nano n, and pico p are often used.

Note: The farad (F) and the coulomb (C) are

extremely large units for static electricity. The SI prefixes micro m, nano n, and pico p are often used.

(8.90 pF)(400 V)

Q



;

Q

C

Q CV

V

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Dielectric

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Dielectric Materials

Most capacitors have a dielectric material between their plates to provide greater dielectric strength

and less probability for electrical discharge.

The separation of dielectric charge allows more charge to be placed on the plates—greater capacitance C > C_o.

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Smaller plate separation without contact. Increases capacitance of a capacitor.

Higher voltages can be used without breakdown. Often it allows for greater mechanical strength.

Smaller plate separation without contact. Increases capacitance of a capacitor.

Higher voltages can be used without breakdown. Often it allows for greater mechanical strength.

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Dielectric Constant

K K

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Dielectric Constant, K

The dielectric constant K for a material is the ratio of the capacitance C with this material as compared with the capacitance C_o in a vacuum.

Dielectric constant: K = 1 for Air

K can also be given in terms of voltage V, electric field intensity E, or permittivity :

0

C K

C



0 0

0

V

E

K

V

E



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The Permittivity of a Medium

The capacitance of a parallel plate capacitor with a dielectric can be found from:

The constant  is the permittivity of the medium which relates to the density of field lines.

0

or

0

or

A

C KC

C K

C

d





2 2 -12 C 0

;

0

8.85 x 10

_Nm

K

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Example 4: Find the capacitance C and the charge Q if

connected to 200-V battery. Assume the dielectric constant is K = 5.0.

2 mm

d

A

0.5 m2

  K₀

  K₀ 5(8.85 x 10-12C/Nm2)

  44.25 x 10-12 C/Nm2

C = 11.1 nF

Q if connected to V = 200 V?

Q = CV = (11.1 nF)(200 V) Q _Q= 2.22 _{= 2.22}m_mC_C

2 2

-12 _C 2

Nm

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Example 5: A capacitor has a capacitance of 6mF with air as the dielectric. A battery charges the capacitor to 400 V and is then disconnected. What is the new voltage if a sheet of

mica (K = 5) is inserted? What is new capacitance C ?

V = 80.0 V

C = Kc_o = 5(6 mF)

C = 30 mF

V_o = 400 V

Mica, K = 5

Air dielectric Mica dielectric 0 0 0 ; V V C K V

C V K

  

400 V ; 5

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Example 5 (Cont.): If the 400-V battery is reconnected after insertion of the mica, what additional charge will be added to the plates due to the increased C?

Q₀ = C₀V₀ = (6 mF)(400 V)

DQ = 9.60 mC

V_o = 400 V

Mica, K = 5

Air C_o = 6 mF

Mica C = 30 mF Q₀ = 2400 mC

Q = CV = (30 mF)(400 V) Q = 12,000 mC

DQ = 12,000 mC – 2400 mC

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Dielectric Strength

The dielectric strength of a material is that electric FIELD intensity E_m for which the

material becomes a conductor. (Charge leakage.)

r Q

Dielectric

E_m varies considerably with physical and environmental conditions such as pressure, humidity, and surfaces.

For air: E_m = 3 x 106 N/C for spherical surfaces

and as low as 0.8 x 106 N/C for sharp points.

For air: E_m = 3 x 106 N/C for spherical surfaces

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Example 2: What is the maximum charge that can be

placed on a spherical surface one meter in diameter?

(R = 0.50 m)

r Q

E_m = 3 x 106 N/C

Maximum Q

Air

Maximum charge in air: Q_Q_m_m = ₌83.3_83.3m_mC_C

This illustrates the large size of the coulomb as a unit of charge in electrostatic applications.

2 2

;

m m

E r

kQ

E

Q

r

k



2 2

6 _N 2

C

9 Nm C

(3 x 10 )(0.50 m)

9 x 10

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Capacitance and Shapes

The charge density on a surface is significantly affected by the curvature. The density of charge is greatest where the curvature is greatest.

+ + + + + ++ + + + + +++ + ++ +++++++ + + + + +

Leakage (called corona discharge) often occurs at sharp points where curvature is greatest.

(r is least)

Leakage (called corona discharge) often occurs at sharp points where curvature is greatest.

(r is least)

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--

-

--

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-+

+ +

+

+ + + ₊

+ + +

+

+ +

+ + +

+

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Summary of Formulas

(C) ; (F)

(V) Q coulomb C farad V volt   0

4

C





r

0

Q

A

C

K

V



d



0 0 0 0

V

E

C

K

C

V

E





2 2 1 1

2

;

2

;

2

Q

U

QV

U

CV

U

C