Bright Storm on Electric Field
(Start to minute 6:18)
Introduction to Electric Field Khan
Academy
(Start to 7:03)
Read and take notes on pgs: 505-507
Electrical Field
• Faraday developed an approach to discussing
fields.
• An electric field is said to exist in the region of
space around a charged object.
– When another charged object enters this electric
field, the field exerts a force on the second
charged object.
Section 15.4
Electric Field, Cont.
• A charged particle, with
charge Q, produces an
electric field in the region of
space around it.
• A small test charge, qo,
placed in the field, will
experience a force.
Section 15.4
Electric Field
• Mathematically,
•
•
•
•
SI unit: N / C
Use this for the magnitude of the field
The electric field is a vector quantity
The direction of the field is defined to be the
direction of the electric force that would be exerted
on a small positive test charge placed at that point.
Section 15.4
Direction of Electric Field
• The electric field
produced by a negative
charge is directed
toward the charge.
– A positive test charge
would be attracted to
the negative source
charge.
Section 15.4
Direction of Electric Field, Cont.
• The electric field
produced by a positive
charge is directed away
from the charge.
– A positive test charge
would be repelled from
the positive source
charge.
Section 15.4
More About a Test Charge and The
Electric Field
• The test charge is required to be a small charge.
– It can cause no rearrangement of the charges on the
source charge.
– Mathematically, the size of the test charge makes no
difference.
• Using qo = 1 C is convenient
• The electric field exists whether or not there is a test
charge present.
Section 15.4
Electric Field, Direction Summary
Section 15.4
EXAMPLE 15.4 Electrified Oil
Goal Use electric forces and fields together with Newton's second law in a onedimensional problem.
Problem Tiny droplets of oil acquire a small negative charge while dropping
through a vacuum (pressure = 0) in an experiment. An electric field of magnitude
5.92 104 N/C points straight down. (a) One particular droplet is observed to
remain suspended against gravity. If the mass of the droplet is 2.93 10-15 kg,
find the charge carried by the droplet. (b) Another droplet of the same mass falls
10.3 cm from rest in 0.250 s, again moving through a vacuum. Find the charge
carried by the droplet.
Strategy We use Newton's second law with both gravitational and electric
forces. In both parts the electric field is pointing down, taken as the negative ydirection, as usual. In part (a) the acceleration is equal to zero. In part (b) the
acceleration is uniform, so the kinematic equations yield the acceleration.
Newton's law can then be solved for q.
SOLUTION
(a) Find the charge on the suspended droplet.
Apply Newton's second law to the droplet in the vertical direction.
(1)
may = ΣFy = -mg + Eyq
points downward, hence Ey is negative. Set ay = 0 in Equation (1) and solve for
q.
mg (2.93 10-15 kg)(9.80 m/s2)
-19
=
=
-4.85
10
C
q=
4
-5.92 10 N/C
Ey
(b) Find the charge on the falling droplet.
Use the kinematic displacement equation to find the acceleration:
Δy = ½ayt2 + v0t
Substitute Δy = -0.103 m, t = 0.250 s, and v0 = 0:
-0.103 m = ½ay(0.250 s)2 → ay = -3.30 m/s2
Solve Equation (1) for q and substitute.
m(ay + g) (2.93 10-15 kg)(-3.30 m/s2 + 9.80 m/s2)
=
q=
-5.92 104 N/C
Ey
= -3.22 10-19 C
LEARN MORE
Remarks This example exhibits features similar to the Millikan Oil-Drop
experiment which determined the value of the fundamental electric charge e.
Notice that in both parts of the example, the charge is very nearly a multiple of e.
Question What would be the acceleration of the oil droplet in part (a) if the
electric field suddenly reversed direction without changing in magnitude? (Select
all that apply.)
The acceleration would be downward.
upward.
The acceleration would be
The magnitude of the acceleration would be 0 m/s2.
magnitude of the acceleration would be 9.80 m/s2.
acceleration would be 19.60 m/s2.
The
The magnitude of the
Read and take notes on pgs: 508-509
Electric Fields and Superposition Principle
• The superposition principle holds when
calculating the electric field due to a group of
charges.
– Find the fields due to the individual charges.
– Add them as vectors.
– Use symmetry whenever possible to simplify the
problem.
Section 15.4
Problem Solving Strategy
• Draw a diagram of the charges in the problem.
• Identify the charge of interest.
– You may want to circle it
• Units – Convert all units to SI.
– Need to be consistent with ke
Section 15.4
Problem Solving Strategy, Cont.
• Apply Coulomb’s Law.
– For each charge, find the force on the charge of interest.
– Determine the direction of the force.
– The direction is always along the line of the two charges.
• Sum all the x- and y- components.
– This gives the x- and y-components of the resultant force
• Find the resultant force by using the Pythagorean
theorem and trigonometry.
Section 15.4
Problem Solving Strategy, Electric
Fields
• Calculate Electric Fields of point charges.
– Use the equation to find the electric field due to
the individual charges.
– The direction is given by the direction of the force
on a positive test charge.
– The Superposition Principle can be applied if more
than one charge is present.
Section 15.4
EXAMPLE 15.5 Electric Field Due to Two Point Charges
The resultant electric field
at P equals the vector sum
field due to the positive charge q1 and
2
1
+
, where
2
1
is the
is the field due to the negative charge q2.
Goal Use the superposition principle to calculate the
electric field due to two point charges.
Problem Charge q1 = 7.00 µC is at the origin, and charge
q2 = -5.00 µC is on the x-axis, 0.300 m from the origin. (a)
Find the magnitude and direction of the electric field at
point P, which has coordinates (0, 0.400) m. (b) Find the
force on a charge of 2.00 10-8 C placed at P.
Strategy Follow the problem-solving strategy, finding the electric field at point
P due to each individual charge in terms of x- and y-components, then adding the
components of each type to get the x- and y-components of the resultant electric
field at P. The magnitude of the force in part (b) can be found by simply
multiplying the magnitude of the electric field by the charge.
SOLUTION
(a) Calculate the electric field at P.
Find the magnitude of 1:
-6
(7.00
10
C)
|q1|
9
2
2
E1 = ke 2 = (8.99 10 N · m /C )
(0.400 m)2
r1
= 3.93 105 N/C
The vector 1 is vertical, making an angle of 90° with respect to the positive xaxis. Use this fact to find its components:
E1x = E1 cos (90°) = 0
E1y = E1 sin (90°) = 3.93 105 N/C
Next, find the magnitude of 2:
-6
(5.00
10
C)
|q2|
9
2
2
E2 = ke 2 = (8.99 10 N · m /C )
(0.500 m)2
r2
= 1.80 105 N/C
Obtain the y-component in the same way, but a minus sign has to be provided for sin θ
because this component is directed downwards:
opp 0.400
sin θ =
=
= 0.800
hyp 0.500
E2y = E2 sin θ = (1.80 105 N/C)(-0.800) =-1.44 105 N/C
Sum the x-components to get the x-component of the resultant vector:
Ex = E1x + E2x = 0 + 1.08 105 N/C = 1.08 105 N/C
Sum the y-components to get the y-component of the resultant vector:
Ey = E1y + E2y = 3.93 105 N/C - 1.44 105 N/C
Ey = 2.49 105 N/C
Use the Pythagorean theorem to find the magnitude of the resultant vector:
E = √Ex2 + Ey2 = 2.71 105 N/C
The inverse tangent function yields the direction of the resultant vector:
2.49 105 N/C
Ey
= tan-1 E = tan-1 1.08 105 N/C = 66.6°
( )
x
(
)
(b) Find the force on a charge of 2.00 10-8 C placed at P.
Calculate the magnitude of the force (the direction is the same as that of because the charge
is positive):
F = Eq = (2.71 105 N/C)(2.00 10-8 C) = 5.42 10-3 N
LEARN MORE
Remarks There were numerous steps to this problem, but each was very short.
When attacking such problems, it's important to focus on one small step at a time.
The solution comes not from a leap of genius, but from the assembly of a number
of relatively easy parts.
Question Suppose q2 were moved slowly to the right. What would happen to the
angle ?
It would increase.
It would decrease.
It would first increase and then decrease.
increase.
It would remain the same.
It would first decrease and then
Introduction to Electric Field Khan
Academy
(Minute 7:03 to End)
Read and take notes on pgs: 510-516
Electric Field Lines
• A convenient aid for visualizing electric field
patterns is to draw lines pointing in the
direction of the field vector at any point.
• These lines are called electric field lines and
were introduced by Michael Faraday.
Section 15.5
Electric Field Lines, Cont.
• The field lines are related to the field in the
following manners:
– The electric field vector, , is tangent to the
electric field lines at each point.
– 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.
Section 15.5
Electric Field Line Patterns
• Point charge
• The lines radiate equally
in all directions.
• For a positive source
charge, the lines will
radiate outward.
Section 15.5
Electric Field Line Patterns
• For a negative source
charge, the lines will
point inward.
Section 15.5
Rules for Drawing Electric Field Lines
• The lines for a group of charges must begin on
positive charges and end on negative charges.
– In the case of an excess of charge, some lines will begin or
end infinitely far away.
• The number of lines drawn leaving a positive charge
or ending on a negative charge is proportional to the
magnitude of the charge.
• No two field lines can cross each other.
Section 15.5
Electric Field Line Patterns
• An electric dipole
consists of two equal
and opposite charges.
• The high density of lines
between the charges
indicates the strong
electric field in this
region.
Section 15.5
Electric Field Line Patterns
• Two equal but like point charges
• At a great distance from the
charges, the field would be
approximately that of a single
charge of 2q
• The bulging out of the field lines
between the charges indicates
the repulsion between the
charges.
• The low field lines between the
charges indicates a weak field in
this region.
Section 15.5
Electric Field Patterns
• Unequal and unlike
charges
• Note that two lines
leave the +2q charge for
each line that
terminates on –q
• At a great distance from
the charges, the field
would be equivalent to
that of a single charge
+q
Section 15.5
Electric Field Lines, Final
• The electric field lines are not material
objects.
• They are used only as a pictorial
representation of the electric field at various
locations.
• They generally do not represent the path of a
charged particle released in the electric field.
Section 15.5
Conductors in Electrostatic Equilibrium
• When no net motion of charge occurs within a
conductor, the conductor is said to be in
electrostatic equilibrium.
• An isolated conductor has the following
properties:
– The electric field is zero everywhere inside the
conducting material.
– Any excess charge on an isolated conductor
resides entirely on its surface.
Section 15.6
Properties, Cont.
• The electric field just outside a charged
conductor is perpendicular to the conductor’s
surface.
• On an irregularly shaped conductor, the
charge accumulates at locations where the
radius of curvature of the surface is smallest
(that is, at sharp points).
Section 15.6
Property 1
• The electric field is zero everywhere inside the
conducting material.
– Consider if this were not true
• If there were an electric field inside the conductor, the
free charge there would move and there would be a
flow of charge.
• If there were a movement of charge, the conductor
would not be in equilibrium.
Section 15.6
Property 2
• Any excess charge on an isolated conductor resides
entirely on its surface.
– A direct result of the 1/r2 repulsion between like charges in
Coulomb’s Law
– If some excess of charge could be placed inside the
conductor, the repulsive forces would push them as far
apart as possible, causing them to migrate to the surface.
Section 15.6
Property 3
• The electric field just
outside a charged
conductor is perpendicular
to the conductor’s surface.
– Consider what would happen
if this was not true.
– The component along the
surface would cause the
charge to move.
– It would not be in
equilibrium.
Section 15.6
Property 4
• On an irregularly shaped conductor, the charge accumulates
at locations where the radius of curvature of the surface is
smallest (that is, at sharp points).
Section 15.6
Property 4, Cont.
•
•
•
•
Any excess charge moves to its surface.
The charges move apart until an equilibrium is achieved.
The amount of charge per unit area is greater at the flat end.
The forces from the charges at the sharp end produce a larger resultant
force away from the surface.
Section 15.6
An Experiment to Verify Properties of
Charges
• Faraday’s Ice-Pail Experiment
– Concluded a charged object suspended inside a metal
container causes a rearrangement of charge on the
container in such a manner that the sign of the charge on
the inside surface of the container is opposite the sign of
the charge on the suspended object
Section 15.6
Millikan Oil-Drop Experiment
• Measured the elementary charge, e
• Found every charge had an integral multiple of
e
–q=ne
Section 15.7
Van de Graaff Generator
• An electrostatic generator
designed and built by
Robert J. Van de Graaff in
1929
• Charge is transferred to the
dome by means of a
rotating belt.
• Eventually an electrostatic
discharge takes place.
Section 15.8
Link to Webassign Discussion Questions
Hand write answers to be handed in and
checked
Link to Webassign
Unit 3 A Electric Forces and Fields
Complete #7-12
Grading Rubric for Unit 3A Electric Fields and Coulomb’s Law
Name: ______________________
Conceptual notes Lesson 1 Conductors and Insulators -----------------------------------------------------____
Blue Conceptual Physics Text 500-514 --------------------------------------------------____
Lesson 3 Electrical Forces --------------------------------------------------------------------____
Advanced notes from text book:
Pgs 497-503 --------------------------------------------------------------------------------------------_____
Pgs 505-507 --------------------------------------------------------------------------------------------_____
Pgs 508-509 --------------------------------------------------------------------------------------------_____
Pgs 510-516 --- ----------------------------------------------------------------------------------------_____
Example Problems:
15.1 ------------------------------------------------------------------------------------------------------_____
15.3 ------------------------------------------------------------------------------------------------------_____
15.4 ------------------------------------------------------------------------------------------------------_____
15.5 ------------------------------------------------------------------------------------------------------_____
Web Assign 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12--------------------------------------------------------------------____