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General Physics (PHY 2140)

Lecture 6Lecture 6 Electrostatics and electrodynamics

Capacitance and capacitorscapacitors with dielectrics

Electric current current and drift speed resistance and Ohm’s law

Chapter 16-17

http://www.physics.wayne.edu/~apetrov/PHY2140/

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Lightning ReviewLightning Review

Last lecture:

1.1. Capacitance and capacitorsCapacitance and capacitors

Parallel-plate capacitorParallel-plate capacitor Combinations of capacitorsCombinations of capacitors

ParallelParallel SeriesSeries

Energy stored in a capacitorEnergy stored in a capacitor

Review Problem: Consider an isolated simple parallel-plate capacitor whose plates are given equal and opposite charges and are separated by a distance d. Suppose the plates are pulled apart until they are separated by a distance D>d. The electrostatic energy stored in a capacitor is

a. greater thenb. the same asc. smaller then before the plates were pulled apart.

QC

V

1 2

1 1 1...

eqC C C

1 2 ...eqC C C 0

AC

d

221 1

2 2 2

QU QV CV

C

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Q QQ Q

V0 V

16.10 Capacitors with dielectrics16.10 Capacitors with dielectrics

A dielectrics is an insulating material (rubber, glass, etc.)Consider an insolated, charged capacitor

Notice that the potential difference decreases (k = V0/V)Since charge stayed the same (Q=Q0) → capacitance increases

dielectric constant: k = C/C0

Dielectric constant is a material property

0 0 00

0 0

Q Q QC C

V V V

Insert a dielectric

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Capacitors with dielectrics - notesCapacitors with dielectrics - notes

Capacitance is multiplied by a factor Capacitance is multiplied by a factor kk when the when the dielectric fills the region between the plates dielectric fills the region between the plates completelycompletely

E.g., for a parallel-plate capacitorE.g., for a parallel-plate capacitor

The capacitance is limited from above by the electric The capacitance is limited from above by the electric discharge that can occur through the dielectric material discharge that can occur through the dielectric material separating the platesseparating the plates

In other words, there exists a maximum of the electric In other words, there exists a maximum of the electric field, sometimes called field, sometimes called dielectric strengthdielectric strength, that can be , that can be produced in the dielectric before it breaks downproduced in the dielectric before it breaks down

0

AC

d

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Dielectric constants and dielectric Dielectric constants and dielectric strengths of various materials at room strengths of various materials at room temperaturetemperatureMaterialMaterial Dielectric Dielectric

constant, kconstant, kDielectric Dielectric strength (V/m)strength (V/m)

VacuumVacuum 1.001.00 -- --

AirAir 1.000591.00059 3 3 101066

WaterWater 8080 ----

Fused quartzFused quartz 3.783.78 9 9 101066

For a more complete list, see Table 16.1

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Take a parallel plate capacitor whose plates have an area of 2.0 m2 and are separated by a distance of 1mm. The capacitor is charged to an initial voltage of 3 kV and then disconnected from the charging source. An insulating material is placed between the plates, completely filling the space, resulting in a decrease in the capacitors voltage to 1 kV. Determine the original and new capacitance, the charge on the capacitor, and the dielectric constant of the material.

ExampleExample

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Take a parallel plate capacitor whose plates have an area of 2 m2 and are separated by a distance of 1mm. The capacitor is charged to an initial voltage of 3 kV and then disconnected from the charging source. An insulating material is placed between the plates, completely filling the space, resulting in a decrease in the capacitors voltage to 1 kV. Determine the original and new capacitance, the charge on the capacitor, and the dielectric constant of the material.

Given:

V1=3,000 VV2=1,000 VA = 2.00 m2

d = 0.01 m

Find:

C=?C0=?Q=?k=?

Since we are dealing with the parallel-plate capacitor, the original capacitance can be found as

2

12 2 20 0 3

2.008.85 10 18

1.00 10

A mC C N m nF

d m

9 50 18 10 3000 5.4 10Q C V F V C

The charge on the capacitor can be found to be

The dielectric constant and the new capacitance are

10 0

2

3 18 54V

C C C nF nFV

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How does an insulating dielectric material reduce electric fields by producing effective surface charge densities?

Reorientation of polar molecules

Induced polarization of non-polar molecules

Dielectric Breakdown: breaking of molecular bonds/ionization of molecules.

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17.1 Electric Current17.1 Electric Current

Whenever charges of like signs move in a given Whenever charges of like signs move in a given direction, a current is said to exist.direction, a current is said to exist.Consider charges are moving perpendicularly to a Consider charges are moving perpendicularly to a surface of area A.surface of area A.Definition: Definition: the current is the rate at which charge the current is the rate at which charge flows through this surfaceflows through this surface..

A

+

+

+

+

+

I

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17.1 Electric Current - Definition 17.1 Electric Current - Definition

Given an amount of charge, Given an amount of charge, Q, passing through the area A in a Q, passing through the area A in a time interval time interval t, the current is the ratio of the charge to the time t, the current is the ratio of the charge to the time interval.interval.

A

+

+

+

+

+

I

QI

t

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17.1 Electric Current - Units17.1 Electric Current - Units

The SI units of current is The SI units of current is the ampere (A)the ampere (A).. 1 A = 1 C/s1 A = 1 C/s 1 A of current is equivalent to 1 C of charge passing through the 1 A of current is equivalent to 1 C of charge passing through the

area in a time interval of 1 s.area in a time interval of 1 s.

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17.1 Electric Current – Remark 117.1 Electric Current – Remark 1

Currents may be carried by the motion of Currents may be carried by the motion of positive or positive or negative chargesnegative charges..

It is conventional to give the current the same It is conventional to give the current the same direction as the flow of direction as the flow of positivepositive charge. charge.

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17.1 Electric Current – Remark 217.1 Electric Current – Remark 2

In a metal conductor such as copper, the current is due In a metal conductor such as copper, the current is due to the motion of the electrons (negatively charged). to the motion of the electrons (negatively charged).

The direction of the current in copper is thus The direction of the current in copper is thus oppositeopposite the the direction of the electronsdirection of the electrons..

--

-- v

I

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17.1 Electric Current – Remark 317.1 Electric Current – Remark 3

In a beam of protons at a particle In a beam of protons at a particle accelerator (such as RHIC at accelerator (such as RHIC at Brookhaven national laboratory), the Brookhaven national laboratory), the current is the same direction as the current is the same direction as the motion of the protons.motion of the protons.

In gases and electrolytes (e.g. Car In gases and electrolytes (e.g. Car batteries), the current is the flow of batteries), the current is the flow of both positive and negative charges.both positive and negative charges.

                      

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17.1 Electric Current – Remark 417.1 Electric Current – Remark 4

It is common to refer to a moving charge as a It is common to refer to a moving charge as a mobile mobile charge carriercharge carrier..

In a metal the charge carriers are electrons.In a metal the charge carriers are electrons.

In other conditions or materials, they may be positive or In other conditions or materials, they may be positive or negative ions.negative ions.

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17.1 Electric Current – Example 17.1 Electric Current – Example Current in a light bulbCurrent in a light bulb

The amount of charge that passes The amount of charge that passes through the filament of a certain through the filament of a certain light bulb in 2.00 s is 1.67 c. Find.light bulb in 2.00 s is 1.67 c. Find.

(A) the current in the light bulb.(A) the current in the light bulb.

(B) the number of electrons that pass (B) the number of electrons that pass through the filament in 1 second.through the filament in 1 second.

                                        

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The amount of charge that passes through the filament The amount of charge that passes through the filament of a certain light bulb in 2.00 s is 1.67 c. Find.of a certain light bulb in 2.00 s is 1.67 c. Find.(A) the current in the light bulb.(A) the current in the light bulb.

1.67

2.000.835

Q CI

t sA

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The amount of charge that passes through the filament of a certain The amount of charge that passes through the filament of a certain light bulb in 2.00 s is 1.67 c. Find.light bulb in 2.00 s is 1.67 c. Find.

(b) the number of electrons that pass through the filament in 1 (b) the number of electrons that pass through the filament in 1 second.second.

Reasoning: Reasoning:

In 1 s, 0.835 C of charge passes the cross-sectional In 1 s, 0.835 C of charge passes the cross-sectional area of the filament. area of the filament.

This total charge per second is equal to the number This total charge per second is equal to the number of electrons, N, times the charge on a single electron.of electrons, N, times the charge on a single electron.

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The amount of charge that passes through the filament of a certain The amount of charge that passes through the filament of a certain light bulb in 2.00 s is 1.67 c. Find.light bulb in 2.00 s is 1.67 c. Find.(b) the number of electrons that pass through the filament in 1 (b) the number of electrons that pass through the filament in 1 second.second.

Solution:Solution:

1

18

19

9

1.60 10 / 0.835

0.835

1

5.22 10

.60 10 /

qN N C electron C

CN

C e

N electron

le o

s

ctr n

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15.2 Current and Drift Speed15.2 Current and Drift Speed

Consider the current on a conductor of cross-sectional Consider the current on a conductor of cross-sectional area A. area A.

Avdq

vdt

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15.2 Current and Drift Speed (2)15.2 Current and Drift Speed (2)

Volume of an element of length Volume of an element of length x is : x is : V = A V = A x.x.

Let n be the number of carriers per unit of volume.Let n be the number of carriers per unit of volume.

The total number of carriers in The total number of carriers in V is: n A V is: n A x.x.

The charge in this volume is: The charge in this volume is: Q = (n A Q = (n A x)q.x)q.

Distance traveled at Distance traveled at drift speed drift speed vvdd by carrier in time by carrier in time t: t: x = x = vvd d t.t.

Hence: Hence: Q = (n A Q = (n A vvd d t)q.t)q.

The current through the conductor: The current through the conductor:

I = I = Q/ Q/ t = n A t = n A vvd d qq..

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15.2 Current and Drift Speed (3)

• In an isolated conductor, charge carriers move randomly in all directions.

• When an external potential is applied across the conductor, it creates an electric field inside which produces a force on the electron.

• Electrons however still have quite a random path.

• As they travel through the material, electrons collide with other electrons, and nuclei, thereby losing or gaining energy.

• The work done by the field exceeds the loss by collisions.

• The electrons then tend to drift preferentially in one direction.

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15.2 Current and Drift Speed - Example15.2 Current and Drift Speed - Example

Question:Question:

A copper wire of cross-sectional area 3.00x10A copper wire of cross-sectional area 3.00x10-6-6 m m22 carries a current of carries a current of 10. A. Assuming that each copper atom contributes one free electron to 10. A. Assuming that each copper atom contributes one free electron to the metal, find the drift speed of the electron in this wire. The density of the metal, find the drift speed of the electron in this wire. The density of copper is 8.95 g/cmcopper is 8.95 g/cm33..

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Question:

A copper wire of cross-sectional area 3.00x10-6 m2 carries a current of 10 A. Assuming that each copper atom contributes one free electron to the metal, find the drift speed of the electron in this wire. The density of copper is 8.95 g/cm3.

Reasoning: We know:• A = 3.00x10-6 m2 • I = 10 A.• = 8.95 g/cm3.• q = 1.6 x 10-19 C.• n = 6.02x1023 atom/mol x 8.95 g/cm3 x ( 63.5 g/mol)-1

• n = 8.48 x 1022 electrons/ cm3.

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Question:

A copper wire of cross-sectional area 3.00x10-6 m2 carries a current of 10 A. Assuming that each copper atom contributes one free electron to the metal, find the drift speed of the electron in this wire. The density of copper is 8.95 g/cm3.

Ingredients:

A = 3.00x10-6 m2 ; I = 10 A.; = 8.95 g/cm3.; q = 1.6 x 10-19 C.

n = 8.48 x 1022 electrons/ cm3.

22 3 19 6 2

6

10.0 /

8.48 10 1.6 10 3.00 10

2.46 10 /

d

I C sv

nqA electrons m C m

m s

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15.2 Current and Drift Speed - Comments15.2 Current and Drift Speed - Comments

Drift speeds are usually very small.Drift speeds are usually very small.

Drift speed much smaller than the average speed Drift speed much smaller than the average speed between collisions. between collisions.

Electrons traveling at 2.46x10Electrons traveling at 2.46x10-6-6 m/s would would take 68 min to m/s would would take 68 min to travel 1m.travel 1m.

So why does light turn on so quickly when one flips a So why does light turn on so quickly when one flips a switch?switch?

The info travels at roughly 10The info travels at roughly 1088 m/s… m/s…

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Mini-quizMini-quiz

Consider a wire has a long conical shape. How does the Consider a wire has a long conical shape. How does the velocity of the electrons vary along the wire?velocity of the electrons vary along the wire?

Every portion of the wire carries the same current: as the cross Every portion of the wire carries the same current: as the cross sectional area decreases, the drift velocity must increase to sectional area decreases, the drift velocity must increase to carry the same value of current. This is dues to the electrical carry the same value of current. This is dues to the electrical field lines being compressed into a smaller area, thereby field lines being compressed into a smaller area, thereby increasing the strength of the electric field.increasing the strength of the electric field.

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17.3 Resistance and Ohm’s Law - Intro17.3 Resistance and Ohm’s Law - Intro

When a voltage (potential difference) is applied across When a voltage (potential difference) is applied across the ends of a metallic conductor, the current is found to the ends of a metallic conductor, the current is found to be proportional to the applied voltage.be proportional to the applied voltage.

I V V

I

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17.3 Definition of Resistance17.3 Definition of Resistance

In situations where the proportionality is exact, one can In situations where the proportionality is exact, one can write.write.

V IR • The proportionality constant R is called resistance

of the conductor.• The resistance is defined as the ratio.

VR

I

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17.3 Resistance - Units17.3 Resistance - Units

In SI, resistance is expressed in volts per ampere.In SI, resistance is expressed in volts per ampere.

A special name is given: A special name is given: ohms (ohms ().).

Example: if a potential difference of 10 V applied across Example: if a potential difference of 10 V applied across a conductor produces a 0.2 A current, then one a conductor produces a 0.2 A current, then one concludes the conductors has a resistance of 10 V/0.2 a concludes the conductors has a resistance of 10 V/0.2 a = = 50 50 ..

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17.3 Ohm’s Law17.3 Ohm’s Law

Resistance in a conductor arises because of collisions Resistance in a conductor arises because of collisions between electrons and fixed charges within the material.between electrons and fixed charges within the material.

In many materials, including most metals, the resistance is In many materials, including most metals, the resistance is constant over a wide range of applied voltages.constant over a wide range of applied voltages.

This is a statement of Ohm’s law.This is a statement of Ohm’s law.

Georg Simon Ohm(1787-1854)

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I

V

I

V

Linear or Ohmic MaterialNon-Linear or Non-Ohmic Material

Semiconductorse.g. diodes

Most metals, ceramics

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V IR

Ohm’s Law

R understood to be independent of V.

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Definition:

Resistor: a conductor that provides a specified resistance in an electric circuit.

The symbol for a resistor in circuit diagrams.

       

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Example: Example: Resistance of a Steam IronResistance of a Steam Iron

All household electric devices are required to have a All household electric devices are required to have a specified resistance (as well as many other specified resistance (as well as many other characteristics…). Consider that the plate of a certain characteristics…). Consider that the plate of a certain steam iron states the iron carries a current of 7.40 A when steam iron states the iron carries a current of 7.40 A when connected to a 120 V source. What is the resistance of the connected to a 120 V source. What is the resistance of the steam iron?steam iron?

12016.2

7.40

V VR

I A