Electrostatics

Electric Field

Electric Field Lines

Shielding and Charging by Induction

Electric flux and Gauss’s law

Electrical energy

potential difference and electric potential

Assignments: For next class: Read Ch. 20 HW4 Set due next Wed, 9/18 Exam 1 (Wed 9/25): (Ch 19-20)

Question/Observation Mondays

Research Q/O with HW (Wednesday)

Assignments: For next class: Read Ch. 20 HW4 Set due next Thurs 9/19 Exam 1 (Thurs 9/26): (Ch 19-20)

Question/Observation Thurs

Research Q/O with HW

5

Problem solving steps

1. Visualize problem – labeling variables

2. Determine which basic physical principle(s) apply

3. Write down the appropriate equations using the variables

defined in step 1.

4. Check whether you have the correct amount of

information to solve the problem (same number of

knowns and unknowns).

5. Solve the equations.

6. Check whether your answer makes sense (units, order of

magnitude, etc.).

From observations: one finds that whenever multiple charges are present, the net force on a given charge is the vector sum of all forces exerted by other charges.

Electric force obeys a superposition principle.

6

How to work the problem?

Find the electrical forces between pairs of charges separately

Then add the vectors

Remember to add the forces as vectors

8

F12 F13

Superposition Principle

Example: 3 charges in a

line

q1 = 6.00 uC q2 = 1.50 uC q3 = -2.00 uC d1 = 3.00 cm d2 = 2.00 cm K = 8.99 x 109 N.m2 /C2

Consider three point charges at the corners of a triangle, as shown in the next slide. a) Find the resultant force on q3 b) Find the direction on f3 (this problem looks like the one presented in pg. 662-663 Walker)

q1 = 6.00 x 10-9 C

q2 = -2.00 x 10-9 C

q3 = 5.00 x 10-9 C

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The force exerted by q1 on q3 is

The force exerted by q2 on q3 is

The total force exerted on q3 is the vector sum of

and

13F

13F

23F

23F

11 9/12/2013

Consider three point charges at the corners of a triangle, as shown below. Find the resultant force on q3.

Solution:

The net force = 7.16 x 10-9 N Θ = 65.2˚

Spherical Charge Distribution

Read and work on pg 663/664

Field

Action/Force at a distance: Gravitational field

Point charge

The lines radiate equally in all directions

For a positive source charge, the lines will radiate outward

For a negative source charge, the lines will point inward

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

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

Unequal and unlike charges

Note that two lines leave the +2q charge for each line that terminates on -q

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

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

Electric forces act through space even in the absence of physical contact.

Suggests the notion of electrical field (first introduced by Michael Faraday (1791-1867).

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Mathematically,

SI units are N / C

o

FE

q

Given: One finds:

2

o

e

q qF k

r

2e

qE k

r

Apply Coulomb’s Law

For each charge, find the force on the charge of interest

Determine the direction of the force

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 trig

a) E will be constant (a charged plane)

b) E will decrease with distance as (a point charge)

c) E in a line charge, decrease with a distance 1/r

2

1

r

An electron moving horizontally passes between two horizontal planes, the upper plane charged negatively, and the lower positively.

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- vo

- - - - - - - - - - - - - - - - - - - - - -

+ + + + + + + + + + + + + + + + + + + + + +

A uniform, upward-directed electric field exists in this region. This field exerts a force on the electron. Describe the motion of the electron in this region.

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- vo

- - - - - - - - - - - - - - - - - - - - - -

+ + + + + + + + + + + + + + + + + + + + + +

Observations:

Horizontally:

No electric field

No force

No acceleration

Constant horizontal velocity

0

0

0

x

x

x

x o

o

E

F

a

v v

x v t

28

- vo

- - - - - - - - - - - - - - - - - - - - - -

+ + + + + + + + + + + + + + + + + + + + + +

29

- vo

- - - - - - - - - - - - - - - - - - - - - -

+ + + + + + + + + + + + + + + + + + + + + +

Observations:

Vertically:

Constant electric field

Constant force

Constant acceleration

Vertical velocity increase linearly with time.

2

/

/

1/

2

y o

y o o

y o o o

y o o o

o o o

E E

F q E

a q E m

v q E t m

y q E t m

30

-

- - - - - - - - - - - - - - - - - - - - - -

+ + + + + + + + + + + + + + + + + + + + + +

Conclusions:

The charge will follow a parabolic path downward.

Motion similar to motion under gravitational field only except the downward acceleration is now larger.

Question:

Charge q1=7.00 mC is at the origin, and charge q2=-10.00 mC is on the x axis, 0.300 m from the origin. Find the electric field at point P, which has coordinates (0,0.400) m.

31

x

y

0.300 m q1 q2

0.4

00 m

P

E1

E2

E

Observations:

First find the Electric field at point P due to charge q1 and q2.

Field E1 at P due to q1 is vertically upward.

Field E2 at due to q2 is directed towards q2.

The net field at point P is the vector sum of E1 and E2.

The magnitude is obtained with

2e

qE k

r

32

Set up the problem:

q1=7.00 mC

q2=-10.00 mC

K = 8.99 x 109 N. m2 /C2

r1 = 0.400m

r2= ? What do we do?

33

1) Calculate r2

2) Calculate the electric field at P

3) It looks like example in pg. 669

4) See solutions in the next pag.

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Solution:

35

Question:

Is it safe to stay inside an automobile during a lightning storm? Why?

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Question:

Is it safe to stay inside an automobile during a lightning storm? Why?

Answer:

Yes. It is. The metal body of the car carries the excess charges on its external surface. Occupants touching the inner surface are in no danger.

SAFE

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

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)

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

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

Field lines penetrating an area A perpendicular to the field

Compare the flux, Φ, to a water hose.

In general:

ΦE = E A cos θ

ΦE = E A cos θ The perpendicular to the area A is at an angle θ to

the field

SI unit: N.m2 /C

If the surface is close, the sign is The flux is positive for field lines that leave the

enclosed volume of the surface

The flux is negative for field lines that enter the enclosed volume of the surface.

Let’s calculate the total flux on the surface of a sphere (it has to be symmetrical).

Gauss’ Law states that the electric flux through any closed surface is equal to the net charge Q inside the surface divided by εo

εo is the permittivity of free space and equals 8.85 x 10-

12 C2/Nm2

The area in Φ is an imaginary surface, a Gaussian surface

insideE

o

Q

SI unit;

N.m2 /C

Consider a thin hollow shell of uniformly distributed charge Q. Find the electric field inside and outside the shell.

The calculation of the field outside the shell is identical to that of a point charge

The electric field inside the shell is zero

2e

o

2 r

Qk

r4

QE

Use a cylindrical Gaussian surface

On pg. 664, we learned that if a charge Q is distributed uniformly, then:

Q = σA

Where σ is the surface charge density

A

Q

Note, the field is uniform

o2E

The field must be perpendicular to the sheet

The field is directed either toward or away from the sheet

The device consists of plates of positive and negative charge

The total electric field between the plates is given by

The field outside the plates is zero

o

E

The Coulomb force is a conservative force

A potential energy function can be defined for any conservative force, including Coulomb force

The notions of potential and potential energy are important for practical problem solving

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The electrostatic force is conservative

As in mechanics, work is

Work done on the positive charge by moving it from A to B

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A B

E

d

cosW Fd

cosW Fd qEd

The work done by a conservative force equals the negative of the change in potential energy, DPE

This equation is valid only for the case of a uniform electric field

allows to introduce the concept of electric potential

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PE W qEdD

The potential difference between points A and B, VB-VA, is defined as the change in potential energy (final minus initial value) of a charge, q, moved from A to B, divided by the charge

Electric potential is a scalar quantity Electric potential difference is a measure of

electric energy per unit charge Potential is often referred to as “voltage”

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B A

PEV V V

q

DD

Electric potential difference is the work done to move a charge from a point A to a point B divided by the magnitude of the charge. Thus the SI units of electric potential

In other words, 1 J of work is required to move a 1 C of charge between two points that are at potential difference of 1 V

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1 1V J C

Units of electric field (N/C) can be expressed in terms of the units of potential (as volts per meter)

Because the positive tends to move in the direction of the electric field, work must be done on the charge to move it in the direction, opposite the field. Thus,

A positive charge gains electric potential energy when it is moved in a direction opposite the electric field

A negative charge looses electrical potential energy when it moves in the direction opposite the electric field

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1 1N C V m

Electric Force (Coulomb’s force): Superposition of forces

l Electric Field (point charge):

Definition of electric Flux, :

Gass’s Law electric Flux (q is enclosed by a surface):

2

21

er

qqkF

2e

qE k

r

ΦE = E A cos θ

insideE

o

Q