16.4 induced charge - department of physics &...
TRANSCRIPT
1
1. Text book
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Physics: Principles with Applications, 7th ed, Vol 2. by Giancoli.
Available in printed form or as an eBook with Mastering Physics
Electric Charge• Its Conservation• In the Atom• Insulators and Conductors• Induced Charge • Electroscope
Electric ForceCoulomb’s Law
Electric Field• Field Lines• E fields and conductors• Gauss’s Law
16.1 Static Electricity; Electric Charge and Its Conservation
-+
Charge comes in two types: + and -
Objects can be
charged by rubbing
(which moves electrons from one object to the other)
Human skin
Rabbit's fur
Glass
Human hair
Nylon
Wool
Lead
Cat's fur
Silk
Aluminium
Paper (Small positive charge)
Cotton (No charge)
Steel (No charge)
Wood (Small negative charge)
Amber
Polystyrene
Rubber balloon
Brass
Gold
Synthetic rubber
Styrofoam
Plastic wrap
Scotch tape
PVC
rubber
Triboelectric series
Most
positively
charged
Most
negatively
charged
0
+
-
Electric charge is an intrinsic property
of some elementary particles
A negatively negatively negatively negatively charged object has an excess of - charge
A positively positively positively positively charged object has lost some of its - charge
Net chargeNet chargeNet chargeNet charge = sum of all + and - charges in an object
FFFFree charges ree charges ree charges ree charges are loosely bound electrons that can move freely around a conductor
A neutralneutralneutralneutral object has equal amounts of + and - charge
[not for the exam]
Static Electricity; Electric Charge and Its Conservation
Like charges repel
i.e. the net charge
produced in any
process is zero
Opposite charges attract.
+---
Electric charge is conserved
In a conductor (e.g. a metal), charge flows freely
Almost no charge flows in an insulator
Semiconductors (e.g. Si) have an intermediate
conductivity
16.4 Induced Charge
or by induction.
Charge is induced at each
end of the metal rod.
We can see induction inconductors because it gives riseto a net attractive interaction
+
+ -
Metal objects can be charged by conduction,
Neutral metal rod acquires a charge when
it is touched by a charged metal object.
2
To get a permanent charge on the metal rod by inductiona path must be provided for charge to flow …
Induced Charge
Cut the wire with the charged object nearby.
The rod is now permanentlycharged. It has the opposite charge
to the original charged object.
electrophorus
Ground or Earth
= reservoir where charge moves freely to/from
Induced Charge
Charge separation can also be induced inside the atoms or molecules of a nonconductor where it again gives rise to a
… so balloons stick to walls.
weak attractive interaction
Atom or molecule
The Electroscope
A charged electroscope can be used to determine the sign of an unknown charge.
Same chargeLeaves open more
Opposite chargeLeaves close
Then bring the unknown charge
close to the electroscope.First charge it by
induction or conduction.
A metal ball hangs from the ceilingby an insulating thread. The ball isattracted to a positive-charged rodheld near the ball.
The charge of the ball must be:
1. Only positive
2. Only negative
3. Only neutral
4. Positive or neutral
5. Negative or neutral
Question
16.5 Coulomb’s Law
The magnitude of the electric force between two point charges is
proportional to the product of the charges and inversely proportional to the square of the distance between them.
Units = Coulomb
(16-1)
k = 8.988 × 109 Nm2/C2
Insignificant volume
Units of charge = Coulomb, C
k can also be written in terms of εεεε0, the permittivity of free permittivity of free permittivity of free permittivity of free spacespacespacespace
The electric force is typically much stronger than the gravitational force
aka Electric constantElectric constantElectric constantElectric constant
�� =1
4��= 8.85 × 10 !" #"/%&"
Coulomb’s Law
Force acts along the line connecting the charges.
repel
attract
Same charges
Opposite charges
force on 1 due to 2
force on 2 due to 1
Which charge feels the
bigger force?
+ 1C + 2 C
Question
3
Coulomb’s Law
All electrons have the same charge:
Smallest isolated charge found in nature
Electric charge is quantized in units of the electron charge, e.
~ 1013 electrons
Charges produced by rubbing are
typically around a microcoulomb: ~ 10-6 C
Magnitude of resultant force is,above the
+x axis' = '(" + '*
"+ = ,-. !
'*
'(
+
+
140 N 850 N−−−− x
y
10o
40o
'( = 850 cos 10 − 140 cos(50)
'* = 850 sin 10 + 140 sin(50)And the direction is given by,
θ
Take care when working in other quadrants
To calculate the Net Force on a charge…
Example: What is the
net force acting on the negative charge?
Force is a vector quantity
1. Draw a picture and roughly determine the net force2. Calc. the components of force acting on a charge due to all other charges
3. Add the x and y components of the forces separately:
16.7 The Electric Field
Electric field is defined to be the force felt by a smallpositive ‘test’ charge, q, divided by that charge:
(16-3)
Units N/CSmall enough not to disturb charges that created E
… it depends on the charges that produce the electric field.Note, E does not depend on q …
If q is positive, E and F point
in the same direction
If q is negative, E and F point
in opposite directions
Near a Uranium nucleus92 protons
3 x 1021 N/C
5 x 1011 N/C
3,000,000 N/C+
+
+
+ +++
++
++
+
++
+
++
+
++
++
+
+proton
electron
+
~150N/C
Near a comb
~1000 N/C
Inside a
hydrogen atom
-
Near the Earth’s surface
Electrical
breakdown in air
How big are the Electric fields around us?
Electric Field
Electric field (like force) is a vector.
Components can be added to give the net electric field at a point
+ -
E is a vector that points away from (+) charges...
… and towards (-) charges
For a point charge:
(16-4a)
[magnitude]
A small −−−− 4.0 µµµµC charge sits in auniform electric field and feels adownward (electric) force of 6.0 N.
F = 6.0 N
- 4.0 µµµµC
1. 1.5 x 106 N/C UP
2. 1.5 x 106 N/C DOWN
3. 6.7 x 10-7 N/C UP
4. 6.7 x 10-7 N/C DOWN
5. E = kQ/r2
What is the electric field at this point (ignore gravity) ?
Question
4
The Electric Field
Problem solving in electrostatics:
Electric forces and electric fields
(show all charges with signs, and fields and forces with directions)
2. Calculate all components of forces or fields.
' =23435
65; or 7 =
23
65 for point charges
3. Add forces or fields vector components to get resultant.
1. Draw a diagram.
Example. Calculate the electric field near a dipole
Field at B: E1= (9x109)(50x10-6)/.3972 = 2.86x106 N/C
x y
E1 2.86x106cos49 2.86x106sin49
E2 2.86x106cos49 - 2.86x106sin49
Add x and y components:
E2 = (9x109)(50x10-6)/.3972 = 2.86x106 N/C
30 cm
21
+50 µC -50 µC
B
26 cm
49o
49o
7 = ( 3.75 × 10: " + 0")
= 3.75 × 10: %/#
From geometry, and are 49o
above and below the horizontal
7!
7"
7! 7"
In the +x direction.
7
E 3.75x106 0
30 cm
21
+50 µC -50 µC
B
Example. Calculate the electric field near a dipole
26 cm
49o
49o
OR - Use the symmetry:
E = 2 E1 cos49
Generally,
Sum of any 2 vectors of equal
magnitude is
θθ
and have same magnitude
7!
7! 7"
7"
So we can immediately write down the answer…
= 3.75 × 10: %/#
In the +x direction.
E = 2 E1 cosθ
7!
7"
½ the angle
between the
2 vectors
7
Is there a point on the line where E = 0?
-Q +2Q
What happens if a 3rd charge is placed at that point?
10 cm
What if the two charge have the same sign?
Question
16.8 Field Lines
The electric field can be represented by field lines.
E.g. Electric dipole(two opposite charges)
Electric field is stronger when the
field lines are closer together.
Field lines start on (+) charges
and end on (-) charges.
The field at a point is a vector tangential to the field
line there.
Field lines
never cross
Field Lines: near point charges
Two charges of the same magnitude and sign.
Two charges of different magnitude and opposite sign.
Where is E = 0 in these figures?
Electric dipole
Magnitude of point charge
∝∝∝∝ Number of field linesstarting/ending on that charge.
5
Field Lines: inside a capacitor
The electric field between twoclosely spaced, oppositely chargedparallel plates is UNIFORM
(same value everywhere)
Can be derived using Gauss’s Law.
(each plate carries charge Q and has area A)
Field lines are
equally spaced
7 =;
<��
= =>.?,-.,
If it were not, the charges inside the conductor would
feel a force and move around in order to make E zero.
Static electric field inside a solid conductor is zero
Any net charge in a conductor distributes itself on the surface.
E = 0
----
--------
------ - - - - -
Solid metal sphere with -Q
net charge
16.9 Electric Fields and Conductors
Electric field is perpendicular tothe outer surface of a conductor
E
– if it were not, charges would move to make it perpendicular.
E = 0inside
+Q
Inner surface -Q
Outer surface +Q
Neutral
conducting shell
Electric Fields and Conductors
What are some of the consequences of this ?
From the outside it looks as though the shell is not there
+Q
Neutral
conducting shell
If a point charge Q is placed at the center of a NEUTRAL hollow
conducting shell, the E field induces a separation of charge in the shell.
Electric Fields and Conductors
+3Q
−3Q appears on inner surface
+2Q appears on outer surface
Field lines radiate out from the central +3Q charge.
-Q
→
→
%@,=A-BC@>.?A@DD = ;inner + ;outer−; = −3; + ;outer
Now let the shell carry some charge..
E field outside shell is identical to one that would be produced
by a +2Q point charge, (= 3Q −Q) at the center of the shell
Shell has net charge of -Q.
A +3Q point charge is at the center of
hollow conducting spherical shell.
What is the charge on the
outer surface of the shell?
E field outside any charged spherical conductor (solid or hollow) is the same as if all the charge were concentrated at the center of the sphere.
Electric Fields and Conductors
Application: Faraday cage – used to shield people and equipment from external electric fields
16.12 Gauss’s Law
Electric flux
(16-7)
Electric flux through an area is proportional to the number of field lines crossing the area.
Introduce a new concept..
Gauss’s Law can be used to calculate E
Between 7 and area normal
6
Gauss’s Law
Total Flux through (an imaginary) closed surface,
Double charge enclosed → double the flux through
the closed surface
+
Net flux flows outof closed surface+
Zero flux flows through closed
surface
++
Flux in (left) = - Flux out (right)
Gauss’s Law
(16-9)
Gauss’s Law can be used to find the electric field in situations with a high degree of symmetry…
Total electric flux through closed surface
charge enclosed by closed surface
By convention, flux is positive if
it’s coming out of a closed surface
Gauss’s
Law
Total electric flux passing through a closed surface is proportional to the charge enclosed by that surface.
Ex.16-12. Calc. E field using Gauss’s Law for a charged metal shell of radius r0 with net charge Q.
so 7 = ;/4EF0B2
Charge enclosed by <2 = 0
So 7 = 0 0B<B01
E
rr0
~ 1/r2
Outside the shell: imagine a spherical closed surface, <1 0B H B01
0B H B01
Charge enclosed by <1= E Σ ∆A = E(4πr2)
A2
rr
A1
What if the shell was a solid sphere instead?
[there will be no derivations using Gauss’s Law in the Exam]
Inside the shell: imagine surface <2 0B I B01
Σ E⊥∆A = Q/ε0
E field outside a charged spherical shellis the same as if all the charge were
concentrated at the center of the shell.
So,
• Two kinds of electric charge – positive and negative
• Charge is conserved
• Charge on electron:
• Charge is quantized in units of e
• Conductors: electrons free to move
Summary of Chapter 16
• Objects can be charged by conduction or induction
• Coulomb’s law: Electric field is force per
unit charge:
• Electric field of a point charge, Q:
• Electric field can be represented by electric field lines
• Static electric field inside conductor is zero; surface field is perpendicular to surface
• Electric flux: Gauss’s law: