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General Physics (PHY 2140)

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The electric field ... First find the field at point P due to charge q1 and q2. ... Produces large fields (and force) near sharp edges. 18. 7/31/09. Remarks ... – PowerPoint PPT presentation

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Title: General Physics (PHY 2140)


1
General Physics (PHY 2140)
Lecture 3
  • Electrostatics
  • Electric field (cont.)
  • Conductors in electrostatic equilibrium
  • The oscilloscope
  • Electric flux and Gausss law

http//www.physics.wayne.edu/apetrov/PHY2140/
Chapter 15
2
Lightning Review
  • Last lecture
  • Coulombs law
  • the superposition principle
  • The electric field
  • Review Problem A free electron and free
    proton are placed in an identical electric field.
    Compare the electric force on each particle.
    Compare their accelerations.

3
Example Electric Field Due to Two Point Charges
  • Question
  • Charges q14.9 mC and charge q2-2.8 mC are
    placed as shown, 2 cm and 4 cm from the origin.
    Find the electric field at point P, which has
    coordinates (0, 0.015) m.

q1 (0, 0.02 m)
P (0, 0.015 m)
0
q2 (0, 0.04 m)
4
Question Charges q14.9 mC and charge q2-2.8
mC are placed as shown, 2 cm and 4 cm from the
origin. Find the electric field at point P, which
has coordinates (0, 0.015) m.
  • Observations
  • First find the field at point P due to charge q1
    and q2.
  • Field E1 at P due to q1 is directed away from q1.
  • Field E2 at due to q2 is directed towards q2(to
    the right).
  • The net field at point P is the vector sum of E1
    and E2.
  • The magnitude is obtained with

5
Question Charges q14.9 mC and charge q2-2.8
mC are placed as shown, 2 cm and 4 cm from the
origin. Find the electric field at point P, which
has coordinates (0, 0.015) m.
Given q1 4.9 mC q2 -2.8 mC d1 0.02 m d2
0.04 m d3 0.015 m Find E12 ?
6
15.5 Electric Field Lines
  • A convenient way to visualize field patterns is
    to draw lines in the direction of the electric
    field.
  • Such lines are called field lines.
  • Remarks
  • Electric field vector, E, is tangent to the
    electric field lines at each point in space.
  • The number of lines per unit area through a
    surface perpendicular to the lines is
    proportional to the strength of the electric
    field in a given region.
  • E is large when the field lines are close
    together and small when far apart.

7
15.5 Electric Field Lines (2)
  • Electric field lines of single positive (a) and
    (b) negative charges.

a)
b)

-
q
q
8
15.5 Electric Field Lines (3)
  • Rules for drawing electric field lines for any
    charge distribution.
  • Lines must begin on positive charges (or at
    infinity) and must terminate on negative charges
    or in the case of excess charge at infinity.
  • The number of lines drawn leaving a positive
    charge or approaching a negative charge is
    proportional to the magnitude of the charge.
  • No two field lines can cross each other.

9
15.5 Electric Field Lines (4)
  • Electric field lines of a dipole.


-
10
Application Measurement of the atmospheric
electric field
  • The electric field near the surface of the Earth
    is about 100 N/C downward. Under a thundercloud,
    the electric field can be as large as 20000 N/C.
  • How can such a (large) field be measured?

A
11
(No Transcript)
12
15.6 Conductors in Electrostatic Equilibrium
  • Good conductors (e.g. copper, gold) contain
    charges (electron) that are not bound to a
    particular atom, and are free to move within the
    material.
  • When no net motion of these electrons occur the
    conductor is said to be in electro-static
    equilibrium.

13
15.6 Conductors in Electrostatic Equilibrium
  • Properties of an isolated conductor (insulated
    from the ground).
  • Electric field is zero everywhere within the
    conductor.
  • Any excess charge field on an isolated conductor
    resides entirely on its surface.
  • The electric field just outside a charged
    conductor is perpendicular to the conductors
    surface.
  • On an irregular shaped conductor, the charge
    tends to accumulate at locations where the radius
    of curvature of the surface is smallest at
    sharp points.

14
Electric field is zero everywhere within the
conductor.
  • If this was not true, the field inside would be
    finite.
  • Free charge there would move under the influence
    of the field.
  • A current would be induced.
  • The conductor would not be in an electrostatic
    state.

15
Any excess charge field on an isolated conductor
resides entirely on its surface.
  • This property is a direct result of the 1/r2
    repulsion between like charges.
  • If an excess of charge is placed within the
    volume, the repulsive force pushes them as far
    apart as they can go.
  • They thus migrate to the surface.

16
The electric field just outside a charged
conductor is perpendicular to the conductors
surface.
  • If not true, the field would have components
    parallel to the surface of the conductor.
  • This field component would cause free charges of
    the conductor to move.
  • A current would be created.
  • There would no longer be a electro-static
    equilibrium.

17
On an irregular shaped conductor, the charge
tends to accumulate at locations where the radius
of curvature of the surface is smallest at
sharp points.
  • Consider, for instance, a conductor fairly flat
    at one end and relatively pointed at the other.
  • Excess of charge move to the surface.
  • Forces between charges on the flat surface, tend
    to be parallel to the surface.
  • Those charges move apart until repulsion from
    other charges creates an equilibrium.
  • At the sharp ends, the forces are predominantly
    directed away from the surface.
  • There is less of tendency for charges located at
    sharp edges to move away from one another.
  • Produces large fields (and force) near sharp
    edges.

-
-
-
-
18
Remarks
  • Property 4 is the basis for the use of lightning
    rods near houses and buildings. (Very important
    application)
  • Most of any charge on the house will pass through
    the sharp point of the lightning rod.
  • First developed by B. Franklin.

19
Faradays ice-pail experiment
Demonstrates that the charge resides on the
surface of a conductor.
20
Mini-quiz
  • Question
  • Suppose a point charge Q is in empty space.
    Wearing rubber gloves, we sneak up and surround
    the charge with a spherical conducting shell.
    What effect does this have on the field lines of
    the charge?

?


q
21
  • Question
  • Suppose a point charge Q is in empty space.
    Wearing rubber gloves, we sneak up and surround
    the charge with a spherical conducting shell.
    What effect does this have on the field lines of
    the charge?
  • Answer
  • Negative charge will build up on the inside of
    the shell.
  • Positive charge will build up on the outside of
    the shell.
  • There will be no field lines inside the conductor
    but the field lines will remain outside the shell.




-
-
-


-
-
-



-
q

-
-


-
-
-



22
Mini-Quiz
  • Question
  • Is it safe to stay inside an automobile during a
    lightning storm? Why?

23
Question Is it safe to stay inside an automobile
during a lightning storm? Why? Answer Yes. It
is. The metal body of the car carries the excess
charges on its external surface. Occupants
touching the inner surface are in no danger.
SAFE
24
15.8 The Van De Graaff Generator
  • Read Textbook Discuss in Lab.

25
15.9 The oscilloscope
  • Changing E field applied on the deflection plate
    (electrodes) moves the electron beam.

V2
?
d
L
26
Oscilloscope deflection angle (additional)
V2
?
d
L
27
15.10 Electric Flux and Gausss Law
  • Discuss a technique introduced by Karl F. Gauss
    (1777-1855) to calculate electric fields.
  • Requires symmetric charge distributions.
  • Technique based on the notion of electrical flux.

28
15.10 Electric Flux
  • To introduce the notion of flux, consider a
    situation where the electric field is uniform in
    magnitude and direction.
  • Consider also that the field lines cross a
    surface of area A which is perpendicular to the
    field.
  • The number of field lines per unit of area is
    constant.
  • The flux, F, is defined as the product of the
    field magnitude by the area crossed by the field
    lines.

AreaA
E
29
15.10 Electric Flux
  • Units Nm2/C in SI units.
  • Find the electric flux through the area A 2 m2,
    which is perpendicular to an electric field E22
    N/C

Answer F 44 Nm2/C.
30
15.10 Electric Flux
  • If the surface is not perpendicular to the field,
    the expression of the field becomes
  • Where q is the angle between the field and a
    normal to the surface.

N
q
q
31
15.10 Electric Flux
  • Remark
  • When an area is constructed such that a closed
    surface is formed, we shall adopt the convention
    that the flux lines passing into the interior of
    the volume are negative and those passing out of
    the interior of the volume are positive.

32
Example
  • Question
  • Calculate the flux of a constant E field (along
    x) through a cube of side L.

y
1
2
E
x
z
33
  • Question
  • Calculate the flux of a constant E field (along
    x) through a cube of side L.
  • Reasoning
  • Dealing with a composite, closed surface.
  • Sum of the fluxes through all surfaces.
  • Flux of field going in is negative
  • Flux of field going out is positive.
  • E is parallel to all surfaces except surfaces
    labeled 1 and 2.
  • So only those surface contribute to the flux.

34
  • Question
  • Calculate the flux of a constant E field (along
    x) through a cube of side L.
  • Reasoning
  • Dealing with a composite, closed surface.
  • Sum of the fluxes through all surfaces.
  • Flux of field going in is negative
  • Flux of field going out is positive.
  • E is parallel to all surfaces except surfaces
    labeled 1 and 2.
  • So only those surface contribute to the flux.
  • Solution

35
15.10 Gausss Law
  • The net flux passing through a closed surface
    surrounding a charge Q is proportional to the
    magnitude of Q
  • In free space, the constant of proportionality is
    1/eo where eo is called the permittivity of of
    free space.

36
15.10 Gausss Law
  • The net flux passing through any closed surface
    is equal to the net charge inside the surface
    divided by eo.
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