phys113 electricity and electromagnetism semester 2 2002 l.
Download
Skip this Video
Loading SlideShow in 5 Seconds..
PHYS113 Electricity and Electromagnetism Semester 2; 2002 PowerPoint Presentation
Download Presentation
PHYS113 Electricity and Electromagnetism Semester 2; 2002

Loading in 2 Seconds...

play fullscreen
1 / 65

PHYS113 Electricity and Electromagnetism Semester 2; 2002 - PowerPoint PPT Presentation


  • 194 Views
  • Uploaded on

PHYS113 Electricity and Electromagnetism Semester 2; 2002. Professor B. J. Fraser. 1. Electric Charge. What is charge? 700 BC - Greeks write of effects of rubbing amber ( Electrum ) 1600’s - Gilbert shows electrification is a general phenomenon

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about 'PHYS113 Electricity and Electromagnetism Semester 2; 2002' - darci


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
1 electric charge
1. Electric Charge
  • What is charge?
    • 700 BC - Greeks write of effects of rubbing amber (Electrum)
    • 1600’s - Gilbert shows electrification is a general phenomenon
    • 1730 - C. Dufay concludes “there are 2 distinct Electricities”
    • 1750 - Ben Franklin shows +ve & -ve charges
  • Electrostatics involves the forces between stationary charges.
  • Charge is a basic atomic property
    • forces between electrons & nuclei
    • unlike charges attract
    • like charges repel
transfer of charge
Transfer of Charge
  • Charge transfer
  • touching  charge sharing (conduction)
  • only the electrons move
  • Can appear as though positive charge has moved
  • Unit of charge: Coulomb, C
  • 1 Coulomb = 1 Ampere  second
  • Electronic charge, e = 1.602 x 10-19 C
  • i.e. 1 C = 6.3 x 1018 electrons
  • a small number!!

+

+

+

+

+

+

+

+

+

conservation quantisation
Conservation & Quantisation
  • Charge is always conserved
    • it cannot be created or destroyed
  • Charge only comes in fixed packets
    • the packet size is ± e
    • It cannot wear off
    • The light from distant quasars (billions of years old) shows evidence of exactly the same atomic charge.
forces between charges
Forces Between Charges
  • Coulomb’s Law
    • 1785: Coulomb experimentally determines force law between 2 charged point sources, q1 and q2.
    • Thus:
    • where k = 8.99 x 109 N m2 C-2
  • Electric force has direction (vector)
  • Hence Coulomb’s Law is:
    • F12 is the force on q1 due to q2
    • r12 is a unit vector from q2 to q1 along the line that joins them.
hints for problem solving

q3

-3 nC

2.0 m

q1

q2

+2 nC

+2 nC

2.0 m

Hints for Problem Solving
  • Draw a clear diagram
  • Forces are vectors
    • include coordinates
    • i.e. Fx and Fy or i and j components
    • add vectorially
  • Shortcuts due to symmetry?

Example: Electric Forces in a Plane

  • Calculate the forces on q1 and q3
solution forces in a plane

j

i

Solution: Forces in a plane

Force on q1

  • This is due to q2 and q3

q3

F1

F13

F12

q1

q2

solution forces in a plane8

j

i

Solution: Forces in a plane

Force on q3

  • This is due to q2 and q3
  • Find magnitude as before

q3

F31

F32

F3

q

q1

q2

2 the concept of the electric field
2.The Concept of the Electric Field
  • Why is there a force between charged particles?
  • How does each particle know that the other one is there?
  • What happens in space between charged particles?
  • This is an example of an action-at-a-distance force.
  • E.g. Gravitation, Magnetism
  • These forces are described in terms of a field in space surrounding the particle or object.
electric field strength

q0

Electric Field Region

+

+

Q

+

F

+

+

+

+

+

+

+

+

+

+

Test Charge

Charged Object

Electric Field Strength
  • Test an invisible force field?
    • See if a test object experiences a force!
  • Test for an electric field by measuring force experienced by a positive test charge.
  • We know that E F and: E 1/q0
    • where q0 = charge of test charge
    • E = electric field, N C-1
  • Hence: and since:
  • Then:
  • Electric field seen by q0 due to q.
electric field lines
Electric Field Lines
  • Electric field strength is a vector quantity.
  • Much easier to represent using vectors pointing in field direction - electric field lines.
  • Concept due to M. Faraday
  • “lines of force”
  • Electric Field lines point away from positive charges
  • Field lines point in the direction of the force or electric field
  • Density (spacing) of field lines depends upon magnitude of E.
  • Field lines never intersect.
electric fields in nature
Electric Fields in Nature
  • All charges (fixed & moving) produce an electric field that carries energy through space at the speed of light.
field due to point charges

j

i

6 cm

4 cm

+

-

A

q1 = +12 nC

q2 = -12 nC

Field Due to Point Charges
  • Electric fields add vectorially:

E = E1 + E2 + E3 + …

  • Thus:

Worked Example

  • Find the electric field at point A for the dipole shown

Field at A due to q1.

field due to point charges14

EA1

A

+q1

- q2

EA2

Field Due to Point Charges

Field at A due to q2

  • Total electric field at A:
  • No component in the j direction
  • Example of an electric dipole
  • Often found in nature (e.g. molecules)
  • For more: See Section 21.11
field due to a line segment

j

i

^

r

dy

dE

L

y

R

q

O

P

-L

r

dE

dQ

Field due to a line segment
  • Charge, Q, distributed uniformly along length, L, with charge density:

l = Q/L

Worked Example

  • What is the electric field at a distance R from a rod of length 2L carrying a uniform charge density, l?
  • Consider an infinite collection of charge elements, dQ.
field due to line segment contd
Field due to line segment (contd)

But l = Q/L and thus dQ = l dy

  • Centre rod at origin
  • For every charge at + y, there is another corresponding charge at -y
  • Thus, fields in j component add to 0.

cosq = R/r

r2 =(R2 + y2)

Can you do this integral?

solution to field due to a line segment
Solution to Field Due to a Line Segment
  • The solution to the field due to a line segment is:
  • So, what is the big deal?
  • Well, what happens if L >> R?

Thus, the field from a line charge is proportional to 1/R and not 1/R2.

field due to a surface

P

^

^

k

j

r

R

^

i

q

L

x

dx

Field Due to a Surface
  • Consider a charge Q uniformly distributed across surface of area A
  • Surface charge density is: s = Q/A

Worked Example

  • Find the electric field at distance R from an infinite plane sheet with surface charge density s.

Divide the sheet into an infinite collection of line segments, L, long and, dx, wide

field due to a surface19
Field Due to a Surface
  • Charge on each strip:

dQ = s dA = s L dx

  • Charge per unit length:

l = dQ/L = s dx

  • From previous example, each strip sets up electric field:

E = 2kl/r = 2 k s dx/r

  • Summing for all the strips:

But i components sum to 0

Can you do this integral?

solution to field due to a surface
Solution to Field Due to a Surface
  • The solution to the field due to a surface is:
  • So, what’s the big deal this time?
  • How does the field vary with R?

Thus, the field from a surface in independent of the distance R!

particles in an electric field

screen

+ + + + + + + + +

e-

- - - - - - - - -

Particles in an Electric Field
  • A particle of charge, q, in an electric field, E, experiences a force:

E = F/q

F = qE = ma

  • The particle accelerates at a
  • +ve particle moves in direction of E
  • I.e. from +ve to -ve charge regions.
  • Thus an electron will be deflected toward a +ve charged plate as its moving past it.
  • Examples: operation of CRT’s, TV tubes, etc.
particles in an electric field22
Particles in an Electric Field

Worked Example

  • An electron in near-Earth space is accelerated Earthward by an electric field of 0.01 NC-1. Find its speed when it strikes air molecules in the atmosphere after travelling 3 Earth radii (19 000 km).
  • The electron experiences a force:

F = ma = qE

a = qE/m

  • For motion at constant acceleration:

v2 = u2 + 2as = 2as

v = 2.6 x 108 ms-1

  • i.e. 0.8 x speed of light
fishnets and flux the gaussian surface

A

q

Fishnets and Flux: The Gaussian Surface
  • Consider a fishnet with water flowing through it.
  • The rate of flow through net is the flux.

fw = vA

    • v = velocity of flow
    • A = area of net
  • If the net is angled at q to the flow:

fw = vA cosq

  • In vector form:

fw = vA

  • where the direction of A is normal to net
defining electric flux
Defining Electric Flux
  • For an irregular shape, area A is sum of infinitesimal elements dA.
  • Thus, summing over 2-D surface S:
  • Now, replace water with electric field, i.e. there is no physical motion.
  • The electric flux through a surface of area A is:
  • The electric flux through a surface is proportional to the number of field lines passing through a surface.
  • If the fishnet is formed into a closed shape (e.g. lobster pot) its called a Gaussian surface.
the gaussian surface
The Gaussian Surface
  • The total electric flux (number of field lines) passing through this surface is:
  • where A points perpendicularly away from each element dA.
  • If the flux in one side is the same as that out then the total flux is zero.
  • If there is no net charge inside a Gaussian surface the electric flux through it adds to zero.
  • Gaussian surfaces are imaginary constructions!

dA

E

3 gauss law

Flux lines

Gaussian surface

3. Gauss’ Law
  • Consider a point charge surrounded by a Gaussian sphere.
  • The electric field is:
  • where e0 = permittivity of free space

= 8.85 x 10-12 C2 N-1 m-2

  • The electric flux through the surface is then:

Radial field lines are always normal to sphere

gauss law in general
Gauss’ Law in General
  • Gauss’ Law states that the electric flux through any closed surface enclosing a point charge Q is proportional to Q.
  • The surface need not be centred on Q and can be any shape.

Example: Coulomb’s Law from Gauss’ Law

  • What is the electric field due to a point charge?
  • Consider a Gaussian sphere of radius r centred on a charge q.
  • Only interested in radial field direction.
  • All fields in other directions cancel.
coulomb from gauss
Coulomb from Gauss
  • Consider surface elements dA
  • If E is along dA then:

E.dA = E dA cos(0º) = E dA

  • Hence:
  • From Gauss’ Law:
  • Rearranging:
  • Which, since F = qE, gives Coulomb’s Law, where we put E radially outward from the charge q.

E

+q

r

dA

Gaussian surface

applications of gauss law

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

P

Applications of Gauss’ Law
  • Use Gauss’ law to find electric flux or field in a symmetrical situation.
  • Shape of the Gaussiansurface is dictated by the symmetry of the problem.
  • Worked Example
  • Find the electric field due to an infintely long rod, positively charged, of constant charge density, l.
electric field of long rod
Electric Field of Long Rod

dA

h

+ + + + + + + + + +

dA

dA

r

  • Consider motion of a test charge.
  • Only field lines radially away from the rod are important.
  • Consider a Gaussian cylinder around part of the rod, radius r, height, h.
  • Total flux through cylinder is:
  • But, @ top & bottom EdA

E.dA=0

  • For the side E is parallel to dA

E.dA=E dA

field due to a long rod
Field due to a Long Rod

Gauss’ Law

Rod Charge Density

Compare this with our previous result.

E varies as 1/R

charged spherical shell

r

R

E

Charged Spherical Shell

Worked Example

  • E-field inside & outside a charged spherical shell (e.g. plane, car)

Outside the shell

  • Use a Gaussian sphere of radius r centred on the shell. Then:

E.dA = E dA (since E ||dA)

inside a charged spherical shell

E

Inside a Charged Spherical Shell

Inside the shell

  • r < R so consider a Gaussian sphere inside the shell.
  • no net charge enclosed by sphere
  • Qencl = 0, so
  • Inside the shell the field is zero: a physically important result.

No field inside the shell

Faraday Cage!!

solid polarisable sphere

+

Solid Polarisable Sphere

Worked Example

  • What is the electric field outside & inside a solid nonconducting sphere of radius R containing uniformly distributed charge Q.

Outside the sphere:

  • r > R
  • consider spherical Gaussian surface

As before

inside the solid sphere

r

R

Inside the Solid Sphere

Inside the sphere

  • r < R
  • Charge enclosed by a Gaussian sphere of radius r<R is:
field inside solid charged sphere
Field Inside Solid Charged Sphere
  • Hence, from Gauss’ Law:
  • The same behaviour is found for other forces, e.g. gravity.

E

r

R

behaviour of charges fields near conductors
Behaviour of Charges & Fields Near Conductors
  • The electric field is zero everywhere inside a conductor.
  • Electrons move to create an E field which opposes any external field.
  • Free charges move to the outside surfaces of conductors
  • A result of Gauss’ law.
  • The electric field near a conductor is perpendicular to its surface.
  • A parallel component would move charges and establish an electric field inside.
why doesn t my radio work

+

+

+

+

+

+

+

+

+

E = 0

+

+

En

+

+

+

+

+

+

+

+

dA

+

Why Doesn’t My Radio Work
  • The electric field outside

a charged conductor is:

  • where:

Proof

  • Consider a Gaussian cylinder straddling the conductor’s surface.
  • Closed hollow conductors admit no electric field
  • EM shielding  “Faraday Cages”
  • Car Radios and biomagnetics
importance and tests of gauss law
Importance and Tests of Gauss’ Law
  • Coulomb’s law  experimental evidence of Gauss’s law
  • 1/r2 law is the key prediction
  • Gauss’ law is so basic that its essential to test its validity
  • Tests of F  1/r2±d
4 electric potential technology can t live without it
4. Electric Potential: Technology Can’t Live Without It!
  • Technology relies on using energy associated with electrical interactions
  • Work is done when Coulomb forces move a charged particle in an electric field.
  • This work is expressed in terms of electric potential (energy)
  • Electric potential is measured in Volts.
  • Basic to the operation of all electric machines and circuits.
mechanical analogue
Mechanical Analogue

In mechanics

  • Work done in moving from point a  b
  • results in a change in potential energy:

W a  b= Ua - Ub

  • When W a  b > 0

Ua > Ub

  • e.g. a mass falling under gravity
what is electric potential
What is Electric Potential?

In electricity

  • Consider a test charge q0 moving with respect to a charge, q, fixed at the origin.
  • The work done is:
  • When integrated along the path and thus:
  • This is the change in electric potential energy, for a charge q0 moving from a  b.
electric potential energy
Electric Potential Energy
  • Since:
  • By definition, a charge infinitely far away has zero potential energy.
  • The electric potential energy between 2 charges is then:
  • Since this is a scalar the total potential energy for a system of charges is:
uranium nucleus example
Uranium Nucleus Example

Worked Example

  • Calculate the electrostatic potential energy between 2 protons in a Uranium nucleus separated by 2 x 10-15 m.
electric potential
Electric Potential

Definition:

  • Electric potential is potential energy per unit charge:
  • where U(r) is the potential energy of test charge q0 due to a charge distribution.
  • V(r) is a property of the charges producing it, not q0.
  • Volt = unit of electric potential

1 V = 1 volt = 1 J/C

  • Note also that 1 V/m = 1 N/C
potential charge distribution
Potential & Charge Distribution
  • For a single point charge; q, a distance r away, the electric potential is:
  • Potential is zero if r = 
  • For a collection of charges:
  • For a charge distribution:
electric potential difference
Electric Potential Difference
  • Difference in electric potential for a charge q between points a and b.
  • i.e potential difference can be expressed as a path-independent integral over an electric field.
  • All charge distributions have an electric potential
  • The potential difference Va - Vb is the work/unit charge needed to move a test charge from a  b without changing its kinetic energy.

For a uniform field, d || E

the electron volt
The electron volt
  • For the definition of volt, 1J of work is needed to move 1 C of charge through a potential difference of 1V
  • A more convenient unit at atomic scales is the electron-volt:
  • The energy gained by an electron (or proton) moving through a potential difference of 1 volt:

1 eV = (1.6 x 10-19 C)(1 V)

= 1.6 x 10-19 J

  • Not an SI unit but a very useful one!

Worked Example

  • In a hydrogen atom the e- revolves around the p+ at a distance of 5.3 x 10-11 m. Find the electric potential at the e- due to the p+, and the electrostatic potential energy between them.
worked examples

e -

p+

r

Worked Examples

A very simplistic picture

  • Electric potential due to proton:
  • Electrostatic p.e. is given by:
forces on charged particles
Forces on Charged Particles

Worked Example

  • In a CRT an electron moves 0.2 m in a straight line (from rest) driven by an electric field of 8 x 103 V/m. Find:

(a) The force on the electron.

(b) The work done on it by the E-field.

(c) Its potential difference from start to finish.

(d) Its change in potential energy.

(e) Its final speed.

worked examples51
Worked Examples

(a) Force is in opposite direction to the E-field, magnitude:

(b) Work done by force:

(c) Potential difference is defined as work/unit charge:

Alternatively (e- opposite to p+):

worked examples52
Worked Examples

(d) Change in potential energy:

(e) Loss of PE = gain in KE = ½mv2

worked examples53
Worked Examples

Worked Example

  • A proton is accelerated across a potential difference of 600 V. Find its change in K.E. and its final velocity.
  • By definition, 1 eV = 1.6 x 10-19 J.
  • Acceleration across 600 V
  • Proton gains 600 eV.

DK.E. = 600(1.6 x 10-19) = 9.6 x 10-17 J

  • Final velocity is:
  • If it started from rest
equipotentials
Equipotentials
  • Regions of equal electric potential may be joined by contour lines.
  • These are equipotentials.
  • In 3-D these can form equipotential surfaces where the potential is the same at each point on the surface.
  • Field lines and equipotentials are always perpendicular.
  • No work is done in moving a charge along an equipotential surface because there is no change in potential.
  • The surface of a conductor is an equipotential since charge is uniformly distributed across the surface of conductors.
obtaining e from the electric potential
Obtaining E from the Electric Potential
  • Recall:
  • If the direction s is parallel to E for infinitesimal elements ds from a to b
  • Electric field is the rate of change of potential V in the direction ds.
  • In 3-D space we use x, y & z components to express in terms of partial derivatives.
vector notation
Vector Notation
  • In vector notation:
  • Where  is the gradient operator.
vector notation example
Vector Notation Example

Worked Example

  • A potential distribution in space is described by:

V = Axy2 - Byz

where A and B are constants. Find the electric field.

potential due to charge distributions
Potential Due to Charge Distributions
  • If E is known, use:
  • If E is not known, use:
  • for continuous charge distributions:
parallel plates

ds

0V

0.25V

0.1V

0.2V

x =0

x =0.5m

Parallel Plates

Worked Example

  • Two parallel metal plates have an area A = 225 cm2 and are l=0.5 cm apart, with a p.d. of 0.25 V between them. Calculate the electric field.
  • This is obvious from the definition of units of electric field = V/m.
uniformly charged disc
Uniformly Charged Disc

Worked Example

  • Find the electric potential and electric field along the axis of a uniformly charged disc of radius R and total charge Q.
  • Consider the disc divided into rings of radius,r, width, dr.

R

(y2 + x2)½

y

x

P

dy

uniformly charged disc contd
Uniformly Charged Disc (contd)
  • For the ring shown:
  • For the total potential we integrate over all rings:
  • By definition of charge density:
  • For the ring: dq = s 2p y dy
uniformly charged disc62
Uniformly Charged Disc
  • The field is only in the x direction.
why sparks occur at pointed tips
Why Sparks Occur at Pointed Tips
  • Recall: Conducting objects contain zero electric field.
  • Charge resides on outer surface
  • This surface is an equipotential.
  • Equipotential surfaces outside the conductor are parallel to its surface.
  • For curved conductors, surface charge density:
  • Hence: (radius of curvature)
  • Small radius implies s and E are large
  • E.g. at points and tips
st elmo s fire
St. Elmo’s Fire
  • Regions of strong E-field
    • Ionisation of air
  • Corona discharge
    • greenish glow (St. Elmo’s Fire)
    • E > 3 x 106 V/m
  • Ionisation
    • Current flow
    • Carry away excess charge

Lightning conductors

  • Do not attract lightning
  • Introduce a lower potential difference region close to clouds.

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

uses in technology
Uses in Technology

Accelerators (1929)

(Giancoli Section 44.2, p1115)

  • Van der Graaf HV Accelerator
  • Works because E-field inside Gaussian sphere is zero
  • 1m sphere  3 x 106 V
  • Up to 20 MV produced

Precipitators (See Figure shown)

  • Remove dust and particles from coal combustion
  • -ve wire @ 40 - 100 kV
  • E-field  particles to wall
  • > 99% effective.

Photocopiers (1940)

(Giancoli Example 21.5, p555)

  • Image on +ve photoconductive drum
  • Charge pattern  -ve toner pattern
  • Heat fixing  +ve paper.