Download Chapter 18 Notes - Valdosta State University

CHAPTER 18 - ELECTRIC FORCES AND FIELDS
Electric charge is produced when electrons are separated from
neutral atoms so that one object has more electrons than protons
and another has more protons than electrons.
The charge on the electron is negative so an excess of electrons
gives an object a negative charge.
An object that has lost electrons so that it has fewer electrons than
protons is considered to have a positive charge.
The basic unit of charge in the metric system is the Coulomb and
corresponds to the amount of charge associated with 6.25 x 1018
electrons or protons.
The charge on one electron is -e = - 1.6 x 10-19 C.
The charge on one proton is +e = + 1.6 x 10-19 C.
Above is an illustration of the positive nucleus surrounded by the
negative electrons. If the number of electrons is the same as the
number of protons the atom is said to be neutral.
Example
A metal plate has a charge of - 3.0 μC and a rod has a charge of
+ 2.0 μC. How many electrons must be transferred from the plate
to the rod so that they have the same charge?
1 Coulomb is 6.25 x 1018 electrons.
The law of conservation of electric charge says that during any
process, the net electric charge of an isolated system remains
constant. Chemical reactions, electric circuits, and radioactive
decay provide examples of this law.
Fundamental characteristic of electric charges says like charges
repel and unlike charges attract.
Electric conductors and insulators differ much in the same way
heat conductors and insulators differ.
If a substance has electrons that are not tightly bound in molecules,
these relatively free electrons can transmit electric energy as well
as heat energy. In substances where these free electrons are not
present, neither electric or heat energy is transmitted as well.
Two methods of placing an electric charge on an object are
conduction and induction.
In charging by conduction, a charged object touches an uncharged
object. Charge flows from the charged object to the uncharged
object and they are both left with the same type of charge.
In charging by induction, the charged object is brought near the
uncharged object causing electrons to be driven to or drawn from
the ground. Once the grounding connection is removed and the
original charged object is removed, a charge remains on the
originally uncharged object opposite in sign to the original charge.
Coulomb's Law
Coulomb's law states that the electric force that exists between two
point charges is directly proportional to the product of the
magnitudes of the charges and inversely proportional to the square
of the distance between them. The equation is:
F = (kq1q2)/r2
where q1 and q2 are the two charges expressed in Coulombs, r is
the distance between them expressed in meters and k is Coulomb's
constant with a value of 8.99 x 109 Nm2/C2 in empty space. The
direction of the electrostatic force is along the line connecting the
two charges. It is repulsive if they are alike and attractive if they
are opposite.
Example
Two tiny spheres have the same mass of 2.0 x 10-6 kg and carry
charges of equal magnitude. The gravitational force each exerts on
the other is balanced by the electric force. Are the charges both
positive, both negative, or one positive and one negative? Find the
magnitude of the two charges. G = 6.67 x 10-11Nm2/kg2.
The permittivity of free space is another way to express Coulomb's
constant and is ε0 = 1/4πk or 8.85 x 10-12 C2/Nm2. This will be
handy when we look at electric flux.
When more than two point charges are involved, we must find the
vector sum of all of the electric forces acting on the point charge of
interest.
If they are arranged in a linear fashion, we have a one dimensional
problem and the forces can be in the positive direction or the
negative direction
If the charges are arranged in a plane so that they do not form a
line, we must consider X and Y components or use trigonometry to
solve parts of a triangle.
Example
Three point charges are fixed in place as in the diagram. q1 has a
charge of +8.00 μC, q2 has a charge of -5.00 μC and q3 has a
charge of +5.00 μC. (a) Determine the net force exerted on q1 by
the other two charges and (b) if q1 had a mass of 1.50 g and it were
free to move, what would be its acceleration?
An electric field is defined as a region of space in which an electric
force will be experienced by a charged particle that is placed there.
The SI unit of electric field strength is the newton per
coulomb(N/C). The formula used to calculate electric field strength
is:
ξ = F/q
Where ξ is the electric field strength, F is the force experienced by
the charge at that location, and q is the magnitude of the charge
placed in the field. Electric fields are vectors and must be treated
as such in calculation.
Example
A tiny ball with a mass of 0.012 kg carries a charge of -18.0 μC.
What magnitude and direction electric field is needed to cause the
ball to float above the ground?(g = 9.8 m/s2)
The electric field produced by a point charge is symmetrical
around the location of the point with field lines radiating outward
from a positive charge and radiating inward towards a negative
charge.
The magnitude of the electric field strength produced by a point
charge varies inversely as the square of the distance from the
charge's location. The formula is:
ξ = kq/r2
where q is the magnitude of the point charge causing the field to
exist. This equation indicates that there will be shells surrounding
the point charge on which the electric force will not change.
Movement from one shell to another will change the electric force
and require a change in energy. Movement on a shell will change
neither the electric force or electric potential energy.
Example
A 3.0 μC point charge is placed in an external uniform electric
field with magnitude 1.6 x 104 N/C. Where relative to the point
charge is the electric field zero?
The case of two point charges relatively near each other and with
opposite charges of the same magnitude is called an electric dipole.
The product of the magnitude of one of the charges and the
distance between them is called the electric dipole moment. Polar
molecules such as H2O and HCl have dipole moments.
Electric dipoles produce curved electric field lines. The electric
field vector direction is tangent to the electric field line at any
point.
The parallel plate capacitor is a device which consists of two metal
plates arranged so that the relatively large, flat sides face each
other. The electric field strength within the capacitor can be
calculated using the formula:
ξ = q/ε0A = σ/ ε0
where A is the area of one plate and σ is the charge density on the
plate. Near the edges of the plates, the field strength will be
different, but over most of the volume between the plates this
equation can be used to find the field strength.
Example
A small plastic ball with a mass of 6.50 x 10-3kg and a charge of
+0.150 μC hangs in equilibrium between the plates of a capacitor
as in the diagram. If the area of each plate is 0.0150 m2, find the
magnitude of the charge on each plate.
Shielding and the electric field inside a conductor
If a net charge exists within a solid conductor, all of the individual
charges would repel each other. This net repulsive force would
cause them to move apart until they reached the outer boundary of
the conductor. This gives us the statement under equilibrium
conditions, all charge resides on the outside of a conductor.
The charges arrange themselves so that the electric field within the
conductor is zero.
This is the basis for electric shielding. Any charge inside the
conductor will experience no electric force due to any electric field
outside the conductor.
Gauss's Law
Consider a positive point charge surrounded by a spherical surface
of radius r.
This closed sphere is called a Gaussian surface
Since the equation for the electric field strength at this surface is:
ξ = kq/r2
and can be rewritten as:
ξ = q/ε04πr2
since k = 1/4πεo. Also A = 4πr2.
Then
ξ = q/ε0A
and
ξA= q/ε0
The product ξA is called electric flux and is represented by the
symbol Φ. So the equation becomes:
Φ = ξA= q/ε0
If the surface is not a sphere, we must find the component of the
electric field that is perpendicular to the surface since the electric
flux depends only on the component of the electric field that passes
through the surface at right angles to the surface. A field
component parallel to the surface does not actually pass through it.
The component of the electric field passing through the surface is
ξcosφ where φ is the angle between the direction of the electric
field and the normal to the surface.
The equation for electric flux becomes:
Φ = ξcosφA= q/ε0
for a flat surface.
If the surface is not spherical or flat, we must add the flux through
all of the individual surfaces to get the total flux. The equation
becomes:
Φ = ΣξcosφΔA= q/ε0
It can be shown that this equation is true for any shape surface. The
SI unit of electric flux is the Nm2/C.
Example
Two planar surfaces intersect and are perpendicular. Surface 1 has
an area of 1.7 m2 and surface 2 has an area of 3.2 m2. The electric
field is uniform and has a magnitude of 250 N/C. Find the electric
flux through (a) surface 1, (b) surface 2, and (c) the total electric
flux through both surfaces.
P 566 Questions 1, 5, 6, 7, 8, 11, 12, 14, 15, 18
P 567 Problems 1, 2, 5, 7, 9, 13, 17, 18, 20, 22, 23, 27, 33, 41, 42,
43, 47, 48