Transcript Gauss`s law

Electric Charge
An intrinsic characteristic of a fundamental particle, electrons
& protons that accompanies it wherever they exist
Proton = positive
Electron = negative
Each has the same value of charge
Electrically neutral – most objects have equal
amounts of protons & electrons & therefore no net
charge
Electrically charged objects
•an object with excess protons will have a positive charge
•an object with excess electrons will have a negative charge
Melectron = 9.11x10-31 kg
Mproton = 1.67x10-27 kg
Conductors & Insulators
• Fix the next few slides…check notes
Semiconductors:
intermediate between conductors & insulators. Silicon are
germanium are examples & these have been used to
revolutionize electronics.
Grounding (not what happens when you get busted)
– giving the charge on an object a free pathway to the Earth.
(which is so large it can absorb an infinite amount of charge &
remain electrically neutral) Known as discharging an object. It
is easier to do to conductors than insulators
Charge is Conserved
Conservation of charge states that the net sum of total
charge in the universe cannot change.
Just like with other conservation laws, charge is exchanged
between the objects
In the production of charged particles, equal amount of
positive & negative charges are always produced (ex: when
an electron is created, a positron is also created – same
mass but opposite charge; when a proton is created, an antiproton is also created, & so on – the same when charged
particles are destroyed & turned into energy)
Methods of Charging an object
Friction – rubbing two neutral objects together giving them
equal amounts of opposite charges
Contact – touching a neutral object with a charged object
giving each object the same sign charge & the total charge on
the two objects will add up to the original charge (but does not
always split evenly)
Induction – bringing a charged object near a neutral object.
The charges in the neutral object will separate, making one end
act as a positive charge & the other end as a negative charge.
Polarization
Involves the separation of charge, it
is not a method of charge.
Electrostatic
Photocopier
Coulomb’s Law
Coulomb’s Law states that the electrostatic force between two
object is proportional to the product of the forces & inversely
proportional to the square of the distance between them.
Unit of charge:
k is Coulomb’s constant
Charge is Quantized
Charge is not continuous, but made up of multiples of the
elementary charge which is that found on an electron or
proton
e = 1.6x10-19 C
q = ne
n  1,2,3,...
When a physical quantity can only have discrete
values instead of any value, we say that the quantity
is quantized
Example
Imagine 3 charges, separated in an equilateral triangle as
shown with L = 2.0 cm, q = 1.0 nC. What is the magnitude
and direction of the force felt by the upper charge?
Example2
The charges and coordinates of two charged particles held
fixed in the xy plane are: q1 = +3.0μC, x1 = 3.5cm, y1 =
0.50cm, and q2 = −4.0μC, x2 = −2.0cm, y2 = 1.5cm.
How far away from
q2 should a third
charge q3 = +4.0μC
be placed such that
the net electrostatic
force on q2 is zero?
ELECTRIC FIELD
The cause of electric force between charges results
from the ELECTRIC FIELD that is established.
ELECTRIC FIELD is a space or region where
electric charges experience a electrical force. Each
charge creates its own E-field. This is very similar
to the g-field that surrounds the Earth.
E-FIELD is a VECTOR
and is defined as:
E-field Lines
This is where the test
charge qo comes into play
Based on the diagram, what is the sign of the charge
at C and D?
Based on the diagram, what is the charge ratio of
C/D?
Typical Electric Fields
APPLICATION
The motion of electrons in a radio transmitting antenna
creates an electric field. The effect of this electric field can
be experienced for many miles around the transmitting
antenna. When a radio in the vicinity is tuned to that
particular station, the electrons in the receiving antenna
experience a force that, with the help of the receiver’s
circuits, results in the sounds produced
Example
Point charges q1 =8.0nC
and q2 = -5.0nC are on the
x-axis located at x = -5.0 cm
and x = 5.0 cm.
Find the NET electric
field at a point on the yaxis, 10.0cm above
x=0.
Example 2 Two point charges of +3.6uC and -8.7uC are
located 3.5m apart. Determine the position where
the E-field would be zero other than infinity.
Conceptually, where along the
axis would this be possible?
a) To the left of q1
b) To the right of q2
c) In between q1 and q2 ?
Example: A charged cork ball is suspended on a light string in
the presence of a uniform electric field as shown. The ball is in
equilibrium. E = (3.0i + 5.0j) x 10^5 N/C.
a) Draw an FBD for the ball
b) Find the sign & magnitude of charge on the ball.
c) Find the tension in the string.
Example: Ring of Charge
A nonconducting ring of radius R lies in the yz-plane and
carries a uniformly distributed positive charge Q
a) Determine an expression for the E-field at a
point along the x-axis some distance x from origin.
b) Determine the value of x for which Ex is a maximum.
c) Determine the maximum E-field, Ex max.
d) On the axes, sketch Ex vs x from x = -2R to x = +2R.
e) Describe the motion of an electron placed at x = R/2 and
released from rest.
Parallel Plate – Uniform E
A negatively charged particle is moving in
the +x-direction when it enters a region with
a uniform electric field pointing in the +xdirection. Which graph gives its position as a
function of time correctly? (Its initial
position is x = 0 at t = 0.)
Lightning Rods are
pointed
Electric Flux
E
A
area A
We define the electric flux ,
of the electric field E,
through the surface A, as:
Electric Flux
Consider a cylinder immersed in a uniform E-field as shown
E
Determine total flux through cylinder.
Consider a positive point charge outside a spherical
container. Determine the net flux,Ф, for the sphere.
+q
r
R
If the surface is irregular, we can
integrate over the surface by using tiny
pieces of area, dA
Gauss’s Law
Carl Friedrich Gauss 1777-1855
Gauss law relates E on a closed surface, any
closed surface, to net charge inside the surface.
A closed surface encloses a volume of space
such that there is an inside and an outside.
Interior of a circle is not a closed surface.
Gauss states that you can tell how much charge
you have inside the ‘box’ without actually looking
inside the ‘box’. One can just look at field lines
entering or exiting.
Gauss’ Law is a restatement of Coulomb’s Law (for
E-field, kq/r2 )
Consider a point charge, +q, surrounded by a
spherical Gaussian surface, a distance r from +q.
The Gaussian surface
is fictitious but could
represent the closed
surface of any object.
We want to see how
the enclosed charge
is related to the Efield at the surface of
our Gaussian
surface.
E-field at surface is given by:
Flux through surface is given by:
Gauss’s Law
The total flux within
a closed surface …
… is proportional to
the enclosed charge.
It gives us great insight into the electric fields in
and on conductors and within voids inside metals.
It doesn’t matter where or how charge is
distributed within surface. Gaussian Surface
(GS) is an imaginary surface we fit around
charge where E is constant to make integration
easier (reason for symmetry). If E isn’t going to
be constant at all points, then you must use
Coulomb’s Law (more difficult) and look all
contributions from each charge.
For which of these closed surfaces (a, b, c, d),
will the flux of the electric field, produced by the
charge +2q, be zero?
Calculate the flux of the electric field
for each of the closed surfaces a, b, c, and d
Surface a, a =
Surface b, b =
Surface c, c =
Surface d, d =
Consider 2 identical point charges, one positive, the other negative.
Surround the charges with a Gaussian
surface
What is the net charge enclosed?
What is the net flux through the surface?
What is E at the surface?
Applying Gauss’s Law
Gauss’s law is useful only when the electric field is constant on a
given surface
Procedure to follow for all cases of high degree
of symmetry:
1. Select Gaussian surface
Spherical, cylindrical, or planar
2. Setup integral using appropriate surface area
3. Solve for E
Example 1: Thin spherical hollow conductor
of radius R with uniformly distributed charge
+Q residing on all over surface.

a) Determine E outside the conductor

b) Determine E inside the conductor
Conductors
Why does E = 0 inside a conductor?
Conductors are full of free electrons, roughly one
per cubic Angstrom (1 angstrom = 1x10-10). These
are free to move. If E is nonzero in some region,
then the electrons in that region ‘feel’ a force (qE)
and start to move.
A charge, +q, inside a neutral conducting
shell. Find E for r = a, b, & c
Spherical
cavity
r =a:
a
c
b
r = b:
r = c:
c
What would E be like inside cavity if we placed the
pt charge outside of the uncharged conductor?
Excess charge is deposited on a
conductor. An imaginary Gaussian
surface is drawn completely inside
the conductor as shown. There can be
no electric field on the imaginary
surface because it is inside the
conductor. By Gauss's Law we must
have Qencl = 0. We conclude that all the
excess charges must reside on the
outer surface of the solid conductor.
G.S.
A section of conductor is removed . How
will that change charge distribution?
What happens when we move the inner charge
off-center?
Properties of Conductors in electrostatic equilibrium
In a conductor there are large number of electrons free to move.
Any Excess charge placed on a conductor moves to the exterior
surface of the conductor where equilibrium is reached.
The electric field inside a conductor (metal) is zero when charges
are at rest.
*A conductor shields a cavity within it from external electric
Fields but if there is charge within cavity, E can be present.
Electric field lines contact conductor surfaces at right angles.
There can be no net horizontal component or charges would move.
A conductor can be charged by contact or induction
Connecting a conductor to ground is referred to as grounding
The ground can accept of give up an unlimited number of electrons
Example: A very long line of charge
Determine E-field at a distance, r, from
the line of charge with linear charge
density λ.
r
Example: A very long cylindrical insulator has a
uniformly distributed +Q per unit length. Find E outside
and inside the cylinder. Insulator has radius R.
Inside insulator:
Very large non-conducting
sheet of surface charge density,
+σ. Find E outside of sheet.
Very large Conducting sheet of
surface charge density, +σ.
Find E outside of sheet.
A positive charge distribution exists within a nonconducting
spherical region of radius a. The volume charge density is
NOT uniform but varies with the distance r from the center of
the spherical charge distribution, according to ρ = βr for 0 ≤ r ≤
a, where β is a positive constant and ρ = 0 for r > a.
a) Show that total charge Q in the spherical region of radius ‘a’
is βπa4.
b) In terms of β, a, r, and constants, determine the
magnitude of the electric field at a point a distance r
from the center of the spherical charge distribution
for each of the following cases.
i) r > a
ii) r = a
iii) 0 < r < a