Electric Fields, Voltage, Electric Current, and Ohm’s Law ISAT 241 Fall 2003 David J. Lawrence

Electric Fields, Voltage, Electric

Current, and Ohm’s Law

ISAT 241

Fall 2003

David J. Lawrence

Properties of Electric Charges Two kinds of charges. Unlike charges

attract, while like charges repel each other. The force between charges varies as the

inverse square of their separation: F 1/r2.

Charge is conserved. It is neither created nor destroyed, but is transferred.

Charge is quantized. It exists in discrete “packets”: q = / N e, where N is some integer.

Serway & Jewett, Principles of Physics, 3rd ed.

Figure 19.2


Figure 19.2

Properties of Electric Charges “Electric charge is conserved” means

that objects become “charged” when charges (usually electrons) move from one neutral object to another.

This movement results in a Net Positive charge on one object,

and a Net Negative charge on the other

object.

Properties of Electric Charges

Neutral, uncharged matter contains as many positive charges as negative charges.

Net charge is caused by an excess (or shortage) of charged particles of one sign.

These particles are protons and electrons.

Properties of Electric Charges Charge of an electron = e = 1.6 10-19 C

Charge of a proton = e = 1.6 10-19 C

“C” is the Coulomb. Charge is Quantized!

• Total Charge = N e = N 1.6 10-19 C where N is the number of positive charges minus the number of negative charges.

• But, for large enough N, quantization is not evident.

Electrical Properties of Materials

Conductors: materials in which electric charges move freely, e.g., metals.

Insulators: materials that do not readily transport charge, e.g., most plastics, glasses, and ceramics.

Electrical Properties of Materials

Semiconductors: have properties somewhere between those of insulators and conductors, e.g., silicon, germanium, gallium arsenide, zinc oxide.

Superconductors: “perfect” conductors in which there is no “resistance” to the movement of charge, e.g., some metals and ceramics at low temperatures: tin, indium, YBa2Cu3O7

Coulomb’s Law The electric force between two charges is given by:

2

21

r

qqkFF eee

(newtons, N)

Attractive if q1 and q2 have opposite sign.

Repulsive if q1 and q2 have same sign. r = separation between the two charged

particles. ke = 9.0 x 109 Nm2/C2 = Coulomb Constant.

= electric force exerted by q1 on q2

r12 = unit vector directed from q1 to q2

Coulomb’s Law

Force is a vector quantity.

12221

1212 ˆF rr

qqkF e


Figure 19.9

Gravitational Field Consider the uniform gravitational field near the

surface of the earth If we have a = small “test mass” mo , the force on

that mass is Fg = mo g

g

y mo

We define the gravitational field to be

gF

mg

o

Recall that

g = | g | = 9.8 m/s2

The Electric Field The electric field vector E at a point in

space is defined as the electric force FE acting on a positive “Test Charge” placed at that point, divided by the magnitude of the test charge qo.

q >> qo qo

FEq

The Electric Field

q >> qo qo

FEq

EF

qE

o

Units:

~newtons/coulomb, N/C


See Figure 19.11

The Electric Field

In general, the electric force on a charge qo in an electric field E is given by

EqF oE

+ FEFE

E

The Electric Field

E is the electric field produced by q, not the field produced by qo.

Direction of E = direction of FE (qo > 0).

qo << |q|

We say that an electric field exists at some point if a test charge placed there experiences an electric force.

The Electric Field For this situation, Coulomb’s law gives:

FE = |FE| = ke (|q||qo|/r2) Therefore, the electric field at the position of

qo due to the charge q is given by: E = |E| = |FE|/qo = ke (|q|/r2)

q >> qo

qoE

q

|q| >> qo

qo E

q

Gravitational Field Lines Consider the uniform gravitational field near the

surface of the earth = g If we have a small “test mass” mo , the force on that

mass is Fg = mo g We can use gravitational field lines as an aid for

visualizing gravitational field patterns.

g

y mo

Recall that

g = | g | = 9.8 m/s2

Electric Field Lines

An aid for visualizing electric field patterns.

Point in the same direction as the electric field vector, E, at any point.

E is large when the field lines are close together, E is small when the lines are far apart.

Electric Field Lines The lines begin on positive charges and

terminate on negative charges, or at infinity in the case of excess charge.

The number of lines leaving a positive charge or approaching a negative charge is proportional to the magnitude of the charge.

No two field lines can cross. E is in the direction that a positive test

charge will tend to go.


The lines begin on positive charges and terminate on negative charges, or at infinity in the case of excess charge.

+


The lines terminate on negative charges. -

Electric Field Lines More examples

Field lines cannot cross!


Figure 19.17


Figure 19.18


Figure 19.19


Figure 19.20

See the discussion about this figure on page 683 in your book.


Figure 19.21

See Example 19.6 on page 684 in your book.


Figure 19.22


Work Done by a Constant Force (Review)

Fluffy exerts a constant force of 12N to drag her dinner a distance of 3m across the kitchen floor.

How much work does Fluffy do?

Work Done by a Constant Force (Review)

Ingeborg exerts a constant force of 12N to drag her dinner a distance of 3 m across the kitchen floor.

= 30o

How much work does Ingeborg do?

Similar to Serway & Jewett, Principles of Physics, 3rd ed.

Figure 6.1

See page 179 in your book.

Work Done by a Force (Review) Is there a general expression that will give us

the work done, whether the force is constant or not?

Yes! Assume that the object that is being moved is

displaced along the x-axis from xi to xf. Refer to Figure 6.7 and Equation 6.11 on p. 184.

W F dxxx

x

i

f

= area under graph of Fx from xi to xf


Figure 6.7

Gravitational Field Consider the uniform gravitational field

near the surface of the earth = g

Recall that g = | g | = 9.8 m/s2

g

yb

ya

y

b mo

a mo

dSuppose we allow a “test mass” mo to fall from a to b, a distance d.

Gravitational Field

How much work is done by the gravitational field when the test mass falls?

dgmsFW ogba

g

yb

ya

y

b mo

a mo

dSuppose we allow a “test mass” mo to fall from a to b, a distance d .

Electric Field

A uniform electric field can be produced in the space between two parallel metal plates.

The plates are connected to a battery.

E


Figure 20.3

Electric Field

How much work is done by the electric field in moving a positive test charge (qo) from a to b?

E

d

a

qo

b

qo

Wa b ?

Electric Field

Recall that FE = qo E

Magnitude of displacement = d

E

d

a

qo

b

qo

dEqsFW oEba

Potential Difference = Voltage Definition

The Potential Difference or Voltage between points a and b is always given by

VW

qab

a b

o

= (work done by E to move test chg. from a to

b)(test charge)

This definition is true whether E is uniform or not.

Potential Difference = Voltage For the special case of parallel metal plates

connected to a battery -- The Potential Difference between points a and

b is given by

o

baab q

WV

This is also called the Voltage between points a and b.

Remember, E is assumed to be uniform.

dEq

dEq

o

o

Potential Difference = Voltage We need units! Potential Difference between points a & b

Voltage between points a & b

VW

qab

a b

o

~Joules

Coulomb

J

CVolt V

Potential Difference = Voltage More units! Recall that for a uniform electric field

VdEVab so

d

VE ab

C

N

m

V

meter

Volts~

In your book’s notation: dEV

Where d is positive when the displacement is in the same direction as the field lines are pointing.

Potential Difference = Voltage In the general case

b

ao sdEq

= a “path integral” or “line integral”Therefore

VW

qE dsab

a b

o a

b

b

a Eba sdFW

Potential Difference = Voltage If E, FE , and the displacement are all along

the x-axis, this doesn’t look quite so imposing!

W F dx q E dxa ba

b

x oa

b

x So

VW

qE dxab

a b

o a

b

x

Potential Difference = Voltage What about the uniform E case?

E

d

a

qo

b

qo

VW

qE dxab

a b

o a

b

x ?


Figure 20.3


Voltage & Electric FieldB

A

12V

d

If the separation between the plates is

d = 0.3 cm, find the magnitude of the electric field between the plates.

E

Voltage & Electric Field Solution: Recall that for a uniform electric field,

the voltage (or electric potential difference) between two points is given by VAVBVAB Ed

where d is the distance between the two points. E VAB / d

In our case, we know that the voltage between the two plates is just the battery voltage, so VAB = 12 V. The two plates are separated by d = 0.3cm

E 12V/(0.3cm x 10-2 m/cm) = 4000 V/m

Voltage and Electric Potential Sometimes, physicists talk about the “electric

potential” at some location, e.g.,

VA = “electric potential at point A”

VB = “electric potential at point B” Electric potential really needs to be measured

with respect to some reference point. For example, the reference could be “ground” (the earth) or some distant point in space (“at ”), so to be precise we could say

VA = electric potential difference (voltage) between point A and ground, etc.

Voltage and Electric Potential If VA = electric potential difference (voltage)

between point A and ground, and

VB = electric potential difference (voltage) between point B and ground,

Then the voltage (electric potential difference) between points A & B can be written

VAB = VA VB and VBA = VB VA If A is more + than B VAB > 0

VBA< 0 If B is more + than A VBA > 0

VAB< 0

Electric Field Lines and Electric Potential

Electric field lines always point in the direction of decreasing electric potential.

Page 713

In your book’s notation: dEV

Where d is positive when the displacement is in the same direction as the field lines are pointing.

So d is negative when the displacement is in the opposite direction as the field lines are pointing.

Gravitational Field -- P.E. Consider the uniform gravitational field near

the surface of the earth = g Recall that g = | g | = 9.8 m/s2

g

yb

ya

y

b m

a m

d

Suppose we lift a “test mass” m from a to b, a distance d, against the field g .

Gravitational Field -- P.E. Gravitational Potential Energy: Ug = mgy

where y is the height.

The change in the gravitational P.E. as we

lift the mass is: P.E.g = Ug = Ugb Uga

= mgyb mgya = mgd +++ positive

If instead we let the mass fall from b to a:

P.E.g = Ug = Uga Ugb

= mgya mgyb = mgd --- negative

Electric Potential Energy

If a particle with charge q moves through a potential difference V = Vfinal Vinitial , then the change in electric potential energy of the particle is given by

P.E.E = UE = q Vor

UE final UE initial = q Vfinal Vinitial)

Electric Potential Energy

Repeating

VqUE

Note that:

Electric Potential Electric Potential Energy

but they are related (by the above equation)

Electric Potential Energy Consider a uniform electric field = E in an

environment without gravity. Vba > 0 “point b is more positive than point a”

E

yb

ya

y

b qo

a qo

dSuppose we move a “test charge” qo from a to b, a distance d, against the field

E (qo is positive).

Electric Potential Energy The change in electric P.E. of the test

charge when we move it is:

P.E.E = UE = qoV

UEb UEa = qoVba

Vba= E d

UEb UEa = qoE d =

= the work we do in moving

the charge

Energy -- Units Recall that the SI unit of energy is the Joule

(J). Another common unit of energy is the electron

volt (eV), which is the energy that an electron (or proton) gains or loses by moving through a potential difference of 1 V.

1 eV = 1.60210-19 J Example: electron in beam of CRT has speed

of 5107 m/s KE = 0.5mv2 = 1.110-15 J = 7.1103 eV electron must be accelerated from rest through potl. diff. of 7.1103 V in order to reach this speed.

Electric Current Consider a bar of material in which positive

charges are moving from left to right:

imaginary surface

I

Electric current is the rate at which charge passes through the surface, Iavg = Q/t, and the instantaneous current is I = dQ/dt.

Electric Current

IdQ

dt

SI unit of charge: Coulomb (C) SI unit of current: Ampere (1A= 1C/s) A current of 1 ampere is equivalent to 1

Coulomb of charge passing through the surface each second.

Electric Current By definition, the direction of the current is in the

direction that positive charges would tend to move if free to do so, i.e., to the right in this example.

In ionic solutions (e.g., salt water) positive charges (Na+ ions) really do move. In metals the moving charges are negative, so their motion is opposite to the conventional current.

In either case, the direction of the current is in the direction of the electric field.

Electric Current Na+ ions moving through salt water

Electrons moving through copper wire

E I

E I

Electric Current The electric current in a conductor is given

by

wheren = number of mobile charged particles (“carriers”) per unit volumeq = charge on each carriervd = “drift speed” (average speed) of each carrierA = cross-sectional area of conductor

In a metal, the carriers have charge q e.

I nqv Ad

Electric Current

The average velocity of electrons moving through a wire is ordinarily very small ~ 10-4 m/s.

It takes over one hour for an electron to travel 1 m!!!

E I

Ohm’s Law For metals, when a voltage (potential

difference) Vba is applied across the ends of a bar, the current through the bar is frequently proportional to the voltage.

area

A

Vb VaE

I

The voltage across the bar is denoted:

Vba = VbVa .

Ohm’s Law

IR

V or V R Iba ba 1

This relationship is called Ohm’s Law.

The quantity R is called the resistance of the conductor.

R has SI units of volts per ampere. One volt per ampere is defined as the Ohm (. 1=1V/A.

Ohm’s Law is not always valid!!

Ohm’s Law The resistance can be expressed as

where

is the length of the bar (m)

A is the cross-sectional area of the bar (m2)

, “Rho”, is a property of the material called the

resistivity. SI units of ohm-meters (-m).

RA

area

A

Vb VaE

I

Ohm’s Law

The inverse of resistivity is called conductivity:

So we can write

1 1

RV

I A A

Resistance and Temperature The resistivity of a conductor varies with

temperature (approximately linearly) as

where resistivity at temperature T (oC)

oresistivity at some reference temperature To (usually 20oC)

“temperature coefficient of resistivity”.

Variation of resistance with T is given by

o oT T1 ( )

R R T To o 1 ( )

Electrical Power

The power transferred to any device carrying current I (amperes) and having a voltage (potential difference) V (volts) across it is

P = VI Recall that power is the rate at which energy

is transferred or the rate at which work is done.

Units: W (Watt) = J/s

Electrical Power

Since a resistor obeys Ohm’s Law

V = IR , we can express the power dissipated in a resistor in several alternative ways:

P VI I RV

R 2

2

Documents

Electric Fields, Voltage, Electric Current, and Ohm’s Law ISAT 241 Fall 2003 David J. Lawrence