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 one-dimensional 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 y-direction, 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.

q = mg

= (2.93 10-15 kg)(9.80 m/s2)

= -4.85 10-19 C Ey -5.92 104 N/C

(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.

q = m(ay + g)

= (2.93 10-15 kg)(-3.30 m/s2 + 9.80 m/s2)

Ey -5.92 104 N/C

= -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. The acceleration would be

upward. The magnitude of the acceleration would be 0 m/s2. The

magnitude of the acceleration would be 9.80 m/s2. The magnitude of the acceleration would be 19.60 m/s2.

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 1 + 2, where 1 is the

field due to the positive charge q1 and 2 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:

E1 = ke |q1| = (8.99 109 N · m2/C2)

(7.00 10-6 C) r1

2 (0.400 m)2 = 3.93 105 N/C

The vector 1 is vertical, making an angle of 90° with respect to the positive x-axis. 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:

E2 = ke |q2| = (8.99 109 N · m2/C2)

(5.00 10-6 C) r2

2 (0.500 m)2 = 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:

sin θ = opp

= 0.400

= 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:

= tan-1( Ey )= tan-1 (

2.49 105 N/C ) = 66.6° Ex 1.08 105 N/C

(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 remain the same.

It would first increase and then decrease. It would first decrease and then increase.

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 = n e

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--------------------------------------------------------------------____