Electrostatics

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Transcript Electrostatics

Electrostatics
July 21, 2015
iClicker
An object is placed on the axis of a converging thin lens
of focal length 2 cm, at a distance of 8 cm from the
lens. The distance between the image and the lens is
most nearly
a.
b.
c.
d.
e.
0.4 cm
0.8 cm
1.6 cm
2.0 cm
2.7 cm
Atomic Charges

The atom has positive charge in the nucleus,
located in the protons.


The positive charge cannot move from the atom
unless there is a nuclear reaction.
The atom has negative charge in the electron
cloud on the outside of the atom.

Electrons can move from atom to atom without all
that much difficulty.
Charge

Charge comes in two forms, which Ben Franklin designated as positive (+) and
negative(–).

Charge is quantized

It only comes in packets

The smallest possible stable charge, which we designate as e, is the magnitude of
the charge on 1 electron or 1 proton.

e is referred to as the “elementary” charge.


e = 1.602 × 10-19 Coulombs.
We say a proton has charge of e, and an electron has a charge of –e.
Problem

A certain static discharge delivers -0.5 Coulombs of electrical
charge. How many electrons are in this discharge?
Problem

The total charge of a system composed of 1800 particles, all of
which are protons or electrons, is 31x10-18 C. How many protons
are in the system? How many electrons are in the system?
Coulomb’s Law and
Electrical Force
Electric Force

Charges exert forces on each other.

Like charges (two positives, or two negatives)
repel each other, resulting in a repulsive force.

Opposite charges (a positive and a negative)
attract each other, resulting in an attractive
force.
Coulomb’s Law

Coulomb’s law tells us how the magnitude of the force between two
particles varies with their charge and with the distance between them.





k = 9.0 × 109 N m2 / C2
q1, q2 are charges (C)
r is distance between the charges (m)
F is force (N)
Coulomb’s law applies directly only to spherically symmetric
charges.
Problem

A point charge of positive 12.0 μC experiences an attractive
force of 51 mN when it is placed 15 cm from another point
charge. What is the other charge?
iClicker
Two isolated charges, + q and - 2q, are 2 centimeters
apart. If F is the magnitude of the force acting on
charge -2q, what are the magnitude and direction of
the force acting on charge + q ?
Magnitude
(A) (1/2) F
(B) 2 F
(C) F
(D) F
(E) 2F
Direction
Toward charge - 2q
Away from charge -2q
Toward charge - 2q
Away from charge - 2q
Toward charge - 2q
Superposition

Electrical force, like all forces, is a vector
quantity.

If a charge is subjected to forces from more
than one other charge, vector addition must be
performed.

Vector addition to find the resultant vector is
sometimes called superposition.
Problem
Electric Fields
The Electric Field

The presence of + or – charge modifies empty space. This
enables the electrical force to act on charged particles without
actually touching them.

We say that an “electric field” is created in the space around a
charged particle or a configuration of charges.

If a charged particle is placed in an electric field created by other
charges, it will experience a force as a result of the field.

We can easily calculate the electric force from the electric field.
Electric Fields

Forces exist only when two or more particles are
present.

Fields exist even if no force is present.
Field Lines Around Charges

The arrows in a field are not vectors, they are
“lines of force”.

The lines of force indicate the direction of the
force on a positive charge placed in the field.

Negative charges experience a force in the
opposite direction.
Field Line Rules
Single Charges
Double Charges
Plate Charges
Field Vectors from Field Lines

The electric field at a given point is not the field
line itself, but can be determined from the field
line.

The electric field vectors, and thus force vector,
is always tangent to the line of force at that
point.
Force due to Electric Field

The force on a charged particle placed in an
electric field is easily calculated.

F = Eq
F: Force (N)
 E: Electric Field (N/C)
 q: Charge (C)

Problem

The electric field in a given region is 4000 N/C pointed toward
the north. What is the force exerted on a 400 μg Styrofoam bead
bearing 600 excess electrons when placed in the field? Ignore
gravitational effects.
iClicker
An electron e and a proton p are simultaneously released from rest in a uniform
electric field E. Assume that the particles are sufficiently far apart so that the
only force acting on each particle after it is released is that due to the electric
field. At a later time when the particles are still in the field, the electron and
the proton will have the same
a.
b.
c.
d.
e.
direction of motion
speed
displacement
magnitude of acceleration
magnitude of force acting on them
Problem

A 400 mg Styrofoam bead has 600 excess electrons on its
surface. What is the magnitude and direction of the electric field
that will suspend the bead in midair?
Problem

A proton traveling at 440 m/s in the +x direction enters an
electric field of magnitude 5400 N/C directed in the +y
direction. Find the acceleration.
Electric Fields

The Electric Field surrounding a point charge or
a spherical charge can be calculated by:

E = kq / r2
E: Electric Field (N/C)
 k: 9 x 109 N m2/C2
 q: Charge (C)
 r: distance from center of charge q (m)

iClicker
A point P is 0.50 meter from a point charge of 5.0 X 10-8
coulomb. The intensity of the electric field at point P is
most nearly
(A) 2.5 x 10-8 N/C
(B) 2.5 x 101 N/C
(C) 9.0 x 102 N/ C
(D) 1.8 x 103 N/C
(E) 7.5 x 108 N/C
iClicker
Charges + Q and - 4Q are situated as shown above. The
net electric field is zero nearest which point?
(A) A
(B) B
(C) C
(D) D
(E) E
Superposition of Fields

When more than one charge contributes to the
electric field, the resultant electric field is the
vector sum of the electric fields produced by the
various charges.

Again, as with force vectors, this is referred to as
superposition.
iClicker
The diagram above shows an isolated, positive charge Q. Point (B) is twice as far
away from Q as point A. The ratio of the electric field strength at point A
to the electric field strength at point B is
(A) 8 to 1
(B) 4 to 1
(C) 2 to 1
(D) 1 to 1
(E) 1 to 2
Electric Field Lines

Electric field lines are NOT VECTORS, but
may be used to derive the direction of electric
field vectors at given points.

The resulting vector gives the direction of the
electric force on a positive charge placed in the
field.
Problem

A particle bearing -5.0 μC is placed at -2.0 cm, and a particle
bearing 5.0 μC is placed at 2.0 cm. What is the field at the origin?
iClicker
An electron is accelerated from rest for a time of 10-9 second by a
uniform electric field that exerts a force of 8.0 x 10-15 Newton on
the electron.
What is the magnitude of the electric field?
(A) 8.0 x 10-24 N/C
(B) 9.1 x 10-22 N/C
(C) 8.0 x 10-6 N/C
(D) 2.0 x 10-5 N/C
(E) 5.0 x 104 N/C
iClicker
An electron is accelerated from rest for a time of 10-9 second by a
uniform electric field that exerts a force of 8.0 x 10-15 Newton
on the electron.
The speed of the electron after it has accelerated for the 10-9 second
is most nearly
(A) 101 m/s
(B) 103 m/s
(C) 105 m/s
(D) 107 m/s
(E) 109 m/s
Problem

A particle bearing 10.0 mC is placed at the origin, and a particle
bearing 5.0 mC is placed at 1.0 m. Where is the field zero?
Electric Potential
Electric Potential Energy

Electrical potential energy is the energy
contained in a configuration of charges.

Like all potential energies, when it goes up the
configuration is less stable; when it goes down,
the configuration is more stable.

The unit is the Joule.
Electric Potential Energy

Electrical potential energy increases when
charges are brought into less favorable
configurations
+ -
-
Electric Potential Energy

Electrical potential energy decreases when
charges are brought into more favorable
configurations.
+ +
+
-
Work

Work must be done on the charge to increase
the electric potential energy.
Electric Potential

Electric potential is hard to understand, but easy to
measure.

We commonly call it “voltage”, and its unit is the Volt.

1 V = 1 J/C

Electric potential is easily related to both the electric
potential energy, and to the electric field.
Electric Potential

The change in potential energy is directly related to the
change in voltage.

ΔU = qΔV




Δ U: change in electrical potential energy (J)
q: charge moved (C)
Δ V: potential difference (V)
All charges will spontaneously go to lower potential
energies if they are allowed to move.
Electric Potential

Since all charges try to decrease U, and ΔU = q ΔV,
this means that spontaneous movement of charges result
in negative ΔU.

V=U/q

Positive charges like to DECREASE their potential
(ΔV < 0)

Negative charges like to INCREASE their potential.
(ΔV > 0)
Problem

A 3.0 μC charge is moved through a potential
difference of 640 V. What is its potential energy
change?
E-field & Potential

The electric potential is related in a simple way
to a uniform electric field.

ΔV = -Ed
ΔV: change in electrical potential (V)
 E: Constant electric field strength (N/C or V/m)
 d: distance moved (m)

Problem

An electric field is parallel to the x-axis. What is
its magnitude and direction of the electric field if
the potential difference between x =1.0 m and x
= 2.5 m is found to be +900 V?
More Electric Field
Stuff
iClicker
Two large, flat, parallel, conducting plates are 0.04 m apart, as shown above. The
lower plate is at a potential of 2 V with respect to ground. The upper plate is
at a potential of 10 V with respect to ground. Point P is located 0.01 m above
the lower plate. The electric potential at point P is
a.
b.
c.
d.
e.
10 V
8V
6V
4V
2V
iClicker
The electron volt is a measure of
(A) Charge
(B) Energy
(C) Impulse
(D) Momentum
(E) velocity
iClicker
A point P is 0.50 meter from a point charge of 5.0 X 10-8
coulomb. The electric potential at point P is most
nearly
(A) 2.5 x l0-8 V
(B) 2.5 x 101 V
(C) 9.0 x 102 V
(D) 1.8x 103 V
(E) 7.5x 103 V
iClicker
The figure above shows two particles, each with a charge of +Q,
that are located at the opposite corners of a square of side d.
What is the direction of the net electric field at point P ?
(A) Up and to the Right
(B) Up and to the Left
(C) Down and to the Right
(D) Down and to the Left
(E) Down
Charge Location

Electric field lines are more dense near a sharp point, indicating
the electric field is more intense in such regions.

All lightning rods take advantage of this by having a sharply
pointed tip.

During an electrical storm, the electric field at the tip becomes so
intense that charge is given off into the atmosphere, discharging
the area near a house at a steady rate and preventing a sudden
blast of lightning.
Excess Charges

Excess charges reside on the surface of a
charged conductor.

If excess charges were found inside a conductor,
they would repel one another until the charges
were as far from each other as possible, thus the
surface.
E-Field of a Charged Sphere

The electric field inside a conductor must be
zero.
+
+
+
+
+
+
E=0
+
+
+
+
+
+
+
Conductor in an E-Field

The electric field inside a conductor must be
zero.
-
-
+
-
+
+
E=0
-
+
+
-
+
+
iClicker
A positive charge of 10-6 coulomb is placed on an insulated solid conducting sphere.
Which of the following is true?
(A) The charge resides uniformly throughout the sphere
(B) The electric field inside the sphere is constant in magnitude, but not zero.
(C) The electric field in the region surrounding the sphere increases with increasing
distance from the sphere.
(D) An insulated metal object acquires a net positive charge when brought near to, but not
in contact with, the sphere.
(E) When a second conducting sphere is connected by a conducting wire to the first
sphere, charge is transferred until the electric potentials of the two spheres are equal.
Electric Potential
Energy
iClicker
Of the following phenomena, which provides the best evidence that
particles can have wave properties?
(A) The absorption of photons by electrons in an atom
(B) The alpha-decay of radioactive nuclei
(C) The interference pattern produced by neutrons incident on a
crystal
(D) The production of x-rays by electrons striking a metal target
(E) The scattering of photons by electrons at rest
Conservation of Energy

In a conservative system, energy changes from one
form of mechanical energy to another.

When only the conservative electrostatic force is
involved, a charged particle released from rest in an
electric field will move so as to lose potential energy
and gain an equivalent amount of kinetic energy.

The change in electrical potential energy can be
calculated by

ΔU = qΔV.
Problem

If a proton is accelerated through a potential
difference of -2,000 V, what is its change in
potential energy?

How fast will this proton be moving if it started
at rest?
iClicker
Two parallel conducting plates are connected to a constant voltage
source. The magnitude of the electric field between the plates is
2,000 N/C. If the voltage is doubled and the distance between
the plates is reduced to 1/5 the original distance, the magnitude
of the new electric field is
(A) 800 N/C
(B) 1,600 N/C
(C) 2,400 N/C
(D) 5,000 N/C
(E) 20,000 N/C
Problem

A proton at rest is released in a uniform electric
field. How fast is it moving after it travels
through a potential difference of -1200 V? How
far has it moved?
Electric Potential Energy

Electric potential energy is a scalar, like all forms of
energy.

U = kq1q2/r





U: electrical potential energy (J)
k: 9 × 109 N m2 / C2
q1, q2 : charges (C)
r: distance between centers (m)
This formula only works for spherical charges or
point charges.
Electric Potential

For a spherical or point charge, the electric
potential can be calculated by the following
formula

V = kq/r
V: potential (V)
 k: 9 × 109 N m2 / C2
 q: charge (C)
 r: distance from the charge (m)

Problem

How far must the point charges q1 = +7.22 μC
and q2 = -26.1 μC be separated for the electric
potential energy of the system to be -126 J?
Problem

The electric potential 1.5 m from a point charge
q is 2.8 x 104 V. What is the value of q?
iClicker
The hollow metal sphere shown above is positively charged. Point C is the center
of the sphere and point P is any other point within the sphere. Which of the
following is true of the electric field at these points?
a.
b.
c.
d.
e.
It is zero at both points.
It is zero at C, but at P it is not zero and is directed inward.
It is zero at C, but at P it is not zero and is directed outward.
It is zero at P, but at C it is not zero.
It is not zero at either point.
E-Field and Potential
Electric Field and Potential

E=-V/d

Two things about E and V
The electric field points in the direction of
decreasing electric potential.
 The electric field is always perpendicular to the
equipotential surface.

E-Field and Potential

The electric field points in the direction of
decreasing electric potential.
E-Field is Perpendicular

Zero work is done when a charge is moved
perpendicular to an electric field

W = Fd cos = 0

If zero work is done, and V = - W/q, then there
is no change in potential.

Thus, the potential is constant in a direction
perpendicular to the electric field.
Problem

Draw a negative point charge of -Q and its
associated electric field.

Draw 4 equipotential surfaces such that ΔV is
the same between the surfaces, and draw them
at the correct relative locations.

What do you observe about the spacing between
the equipotential surfaces?