1 / 51

# Math Review -20 minutes - PowerPoint PPT Presentation

Math Review -20 minutes. Lecture 2 Electric Fields Ch. 22 Ed. 7. Cartoon - Analogous to gravitational field Topics Electric field = Force per unit Charge Electric Field Lines Electric field from more than 1 charge Electric Dipoles Motion of point charges in an electric field

Related searches for Math Review -20 minutes

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about 'Math Review -20 minutes' - tomas

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

• Cartoon - Analogous to gravitational field

• Topics

• Electric field = Force per unit Charge

• Electric Field Lines

• Electric field from more than 1 charge

• Electric Dipoles

• Motion of point charges in an electric field

• Examples of finding electric fields from continuous charges

• Demos

• Van de Graaff Generator, workings,lightning rod, electroscope,

• Field lines using felt,oil, and 10 KV supply

• Elmo

• Electron projected into an electric field

• Electric field from a line of charge

• Dipole torque

• Clickers

q1

q0

Concept of the Electric Field

• Definition of the electric field. Whenever a chrage is present and if I bring up another charge, it will feel a net Coulomb force from it. It is convenient to say that there is a field there equal to the force per unit positive charge.

• The question is how does charge q0know about chargeq1if it does not “touch it”?Even in a vacuum! We say there is a field produced by q1 that extends out in space everywhere.

• The direction of the electric field is along r and points in the direction a positive test charge would move. This idea was proposed by Michael Faraday in the 1830’s. The idea of the field replaces the charges as defining the situation. Consider two point charges:

+ q0

r

r

q1

q1

If q1 were positive

The Coulomb force is

The force per unit charge is

Then the electric field at r is due to the point charge q1 .

The units are Newton/Coulomb. The electric field has direction and is a vector.

How do we find the direction.? The direction is the direction a unit positive test

charge would move. This is called a field line.

Example of field lines for a point negative charge. Place a unit positive test charge at every point r and draw the direction that it would move

q1

r

The blue lines are the field lines.

The magnitude of the electric field is

The direction of the field is given by the line itself

q1

Important F= Eq0, thenma=q0E,

and then a = q0E/m

F unit positive test charge at every point r and draw the direction that it would move1

E

F2

Repeat with Positive Point Charge

y unit positive test charge at every point r and draw the direction that it would move

x

Moving positive charge in a field of a positive charge

y unit positive test charge at every point r and draw the direction that it would move

x

Moving negative charge in a field of a positive charge

Electric Field Lines from more unit positive test charge at every point r and draw the direction that it would movetwo positive charges

Electric Field Lines from two unit positive test charge at every point r and draw the direction that it would moveopposite charges (+ -)

This is called an electric dipole.

Electric Field Lines: a graphic concept used to draw pictures as an aid to develop intuition about its behavior.

The text shows a few examples. Here are the drawing rules.

• E-field lines either begin on + charges or begin at infinity.

• E-field lines either end on - charges or end at infinity.

• They enter or leave charge symmetrically.

• The number of lines entering or leaving a charge is proportional to the charge

• The density of lines indicates the strength of E at that point.

• At large distances from a system of charges, the lines become isotropic and radial as from a single point charge equal to the net charge of the system.

• No two field lines can cross.

Example of field lines for a uniform distribution of positive charge on one side of a very large nonconducting sheet.

This is called a uniform electric field.

• http://webphysics.davidson.edu/Applets/Applets.html

• Click on problems

• Click on Ch 9: E/M

• Play with Physlet 9.1.4, 9.1.7

Demo: Show field lines using felt, oil, and 10 KV supply

• One point charge

• Two point charges of same sign

• Two point charges opposite sign

• Wall of charge

• Direct evaluation from Coulombs Law for a single point charge

• For a group of point charges, perform the vector sum

• This is a vector equation and can be complex and messy to evaluate and we may have to resort to a computer. The principle of superposition guarantees the result.

r Physlet.

.

- q

E

1 cm away from 1 nC of negative charge

### Typical Electric Fields (SI Units)

P

downward 100 km high in the ionosphere

+ + + + + + + + + +

+ + + + + + + + +

Typical Electric Fields

E

Earth

+ Physlet.

-

r

Typical Electric Fields

• Field due to a proton at the location of the electron in the H atom. The radius of the electron orbit is

Hydrogen atom

We have Physlet.at the origin, at x=4 m and y=0.

What is the magnitude and direction of the electric field E at y=3 m and x=0?

### Example of finding electric field from two charges lying in a plane

P

Find the x and y components of the electric field for

each charge and add them up.

Find the magnitude of the Physlet.

field for q1

P

q2 =15 nc

q1=10 nc

P

5

φ

φ

Find the x and y components

q2 =15 nc

q1=10 nc

Now add all components

y Physlet.

φ

q1=10 nc

q2 =15 nc

x

Add up the components

φ1

Magnitude of net electric field is

Direction of the total electric field is

Using unit vector notation we can

also write the electric field vector as:

-71.9

y Physlet.

y

.

.

P

P

3

3

4

4

4

4

x

x

+15 nc

+15 nc

+15 nc

+15 nc

5

ϕ

Example of two identical charges of +15 nC on the x axis. Each is 4 m from the origin. What is the electric field on the y axis at P where y= 3 m? Example using symmetry to simplify.

Find x and y components of the electric field at P due to each charge

and add them up. Find x component first.

By symmetry

y Physlet.

.

3

4

4

x

+15 nc

+15 nc

y

5

φ

Now find y component

φ

Example of two opposite charges on the x axis. Each is 4 m from the origin. What is the electric field on the y axis at P where y= 3 m? Example using symmetry to simplify.

y

Ex

P

5

3

φ

4

4

x

-15 nc

+15 nc

Find the x and y component of the electric field

Ey=0 by symmetry

Electric Dipole from the origin. What is the electric field on the y axis at P where y= 3 m? Example using symmetry to simplify.

y

E1

θ

E1x

L

L

q

=

=

cos

2

a

a

2

a

a

θ

E2

For large r

This is called an Electric Dipole: A pair of equal and opposite charges q separated by a displacement L. It has an electric dipole moment p=qL.

P

when r is large compared to L

p=qL = the electric dipole moment

r

Note inverse cube law

-q

+q

The dipole moment p is defined as a vector directed

from - to +

+

-

L

x opposite charges q separated by a displacement L. It has an electric dipole moment

Electric Dipoles in Electric fields

A uniform external electric field exerts no net force on a dipole, but it does exert torque that tends to rotate the dipole in the direction of the field (align with )

Torque about the com = τ

So,

When the dipole rotates through q, the electric field does work:

Water (H opposite charges q separated by a displacement L. It has an electric dipole moment 2O) is a molecule that has a permanent dipole moment.

And q = -10 e and q = +10e

GIven p = 6.2 x 10 - 30 C m

What is d?

d = p / 10e = 6.2 x 10 -30 C m / 10*1.6 x 10 -19 C = 3.9 x 10 -12 m

Very small distance but still is responsible

for the conductivity of water.

When a dipole is an electric field, the dipole

moment wants to rotate to line up with

the electric field. It experiences a torque.

Leads to how microwave ovens heat up food

Why do Electric Dipoles align with Electric fields ? opposite charges q separated by a displacement L. It has an electric dipole moment

Work done equals

The minus sign arises because the torque opposes any increase in θ.

Setting the negative of this work equal to the change in the potential energy, we have

Integrating,

Potential Energy

Ans. The energy is minimum when aligns with

Problem 22-50 opposite charges q separated by a displacement L. It has an electric dipole moment

• An electric dipole consists of charges +2e and -2e separated by 0.61 nm. It is in an electric field of strength 3.8 x 106 N/C. Calculate the magnitude of the torque on the dipole when the dipole moment has the following orientations.

• parallel to the field

• (b) perpendicular to the field

• (c) antiparallel to the field

In Nonuniform Electric fields with a gradient, the dipole gets attracted towards regions of stronger fields

• When a dipole is in an electric field that varies with position, then the magnitude of the electric force will be different for the two charges. The dipole can be permanent like NaCl or water or induced as seen in the hanging pith ball. Induced dipoles are always attracted to the region of higher field. Explains why wood is attracted to the teflon rod and how a smoke remover or microwave oven works.

• Show smoke remover demo.

Smoke Remover gets attracted towards regions of stronger fields

Negatively charged central wire has electric field that varies as 1/r (strong electric field gradient). Field induces a dipole moment on the smoke particles. The positive end gets attracted more to the wire.

In the meantime a corona discharge is created. This just means that induced dipole moments in the air molecules cause them to be attracted towards the wire where they receive an electron and get repelled producing a cloud of ions around the wire.

When the smoke particle hits the wire it receives an electron and then is repelled to the side of the can where it sticks. However, it only has to enter the cloud of ions before it is repelled.

It would also work if the polarity of the wire is reversed

Motion of point charges in electric fields gets attracted towards regions of stronger fields

• When a point charge such as an electron is placed in an electric field E, it is accelerated according to Newton’s Law:

a = F/m = q E/m for uniform electric fields

a = F/m = mg/m = g for uniform gravitational fields

If the field is uniform, we now have a projectile motion problem- constant acceleration in one direction. So we have parabolic motion just as in hitting a baseball, etc except the magnitudes of velocities and accelerations are different.

Replace g by qE/m in all equations;

For example, y =1/2at2 we get y =1/2(qE/m)t2

Example illustrating the motion of a charged particle in an electric field: An electron is projected perpendicularly to a downward uniform electric field of E= 2000 N/C with a horizontal velocity v=106 m/s. How much is the electron deflected after traveling a distance d=1 cm. Find the deflection y.

y

v

electron

d

E

E

y electric field:

v

electron

d

E

E

• Electric Field due to a long wire

• Electric field due to an arc of a circle of uniform charge.

• Electric field due to a ring of uniform charge

• Electric field of a uniform charged disk

What is the electric field from an infinitely long wire with linear charge density of +100 nC/m at a point 10 away from it. What do the lines of flux look like?

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

Typical field for the electrostatic smoke remover

Continuous distribution of charges linear charge density of +100 nC/m at a point 10 away from it. What do the lines of flux look like?

• Instead of summing the charge we can imagine a continuous distribution and integrate it. This distribution may be over a volume, a surface or just a line. Also use symmetry.

y

dE

.

.

r

+x

-x

- L/2

L/2

dq

Find electric field due to a line of uniform + charge of length L with linear charge density equal to λ at a point on the y axis that bisects the rod.

λ= Q / L

y

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

dE

dE

r

θ

-x

+x

- L/2

0

L/2

dq

dE = k dq /r2 Apply Coulombs law to a slug of charge dq

dq = λdx

dEy= k λ dx cos θ/r2

dE = k λdx /r2

dEy= dE cos θ

y length L with linear charge density equal to

dEy

dE

r =y sec θ r2 =y2sec2θ

r

x= y tanθ dx = y sec2 θ dθ

y

θ

-x

+x

x

-L/2

0

L/2

dx / r2=dθ / y

dq

What is the field at the center due to arc of charge uniformly distributed along arc?

dq=λds

s=r θ

ds=r dθ

0 due to symmetric element of charge above and below the x axis

What is the field at the center due to arc of charge uniformly distributed along arc?

dq=λds

s=r θ

ds=r dθ

0 due to symmetric element of charge above and below the x axis

Now find the x component uniformly distributed along arc?

dEx= k dq cos θ /r2

dq=λds

s=r θ

ds=r dθ

dEx= k λds cos θ /r2

Find uniformly distributed along arc?the electric field on the axis of a uniformly charged ring with linear charge density λ = Q/2πR.

r2 =z2+R2

dq = λds

cos θ =z/r

=0 at z=0

=0 at z=infinity

=max at z=0.7R

Chapter 22 Problem 30 uniformly distributed along arc?

A disk of radius 2.7 cm has a surface charge density of 4.0 µC/m2 on its upper face. What is the magnitude of the electric field produced by the disk at a point on its central axis at distance z = 12 cm from the disk?

Chapter 22 Problem uniformly distributed along arc?44

At some instant the velocity components of an electron moving between two charged parallel plates are vx = 1.7 x 105 m/s and vy = 2.8 x 103 m/s. Suppose that the electric field between the plates is given by E = (120 N/C) j.

(a) In unit vector notation what is the electron's acceleration in the field?

(b) What is the electron's velocity when its x coordinate has changed by 2.7 cm?

Chapter 22 Problem 50 uniformly distributed along arc?

An electric dipole consists of charges +2e and -2e separated by 0.74 nm. It is in an electric field of strength 3.4 x106 N/C. Calculate the magnitude of the torque on the dipole when the dipole moment has the following orientations.

(a) parallel to the field

(b) perpendicular to the field

(c) antiparallel to the field