1 Math Review -20 minutes. 2 3 4 Lecture 2 Electric Fields Ch. 22 Ed. 7 Cartoon - Analogous to gravitational field Topics –Electric field = Force per.

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Math Review -20 minutes

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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– 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

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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 q0 know about charge q1 if 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:

r

q1 q0

rE =

rF / q0

r

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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.

r

q1

If q1 were positiver

rF =k

q1q0

r2 r

r

q1 + q0

r

rE =k

q1

r2 r

rE

rE =

rF / q0

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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

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

Important F= Eq0 , then ma=q0E, and then a = q0E/m

r

rE=

kq1

r2 r

r

r

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Repeat with Positive Point Charge

E

F

2

F

1

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Moving positive charge in a field of a positive charge

x

y

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Moving negative charge in a field of a positive charge

x

y

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Electric Field Lines from more two positive charges

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Electric Field Lines from twoopposite charges (+ -)

This is called an electric dipole.

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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.

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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.

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In order to get a better idea of field lines try this Physlet.

• 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

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Methods of evaluating electric fields

• 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.

rE =

kqi

ri2

i=1

N

∑ ri

rE =

kq1

r12

r

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1 cm away from 1 nC of negative charge

rE =

kq1

r2r =

8.99 ×109 Nm2

C2

⎛⎝⎜

⎞⎠⎟ ⋅(−10-9C)

10-4m2 =−8.99 ×104 N C

r

- q

r

E

.

Typical Electric Fields (SI Units)

Pr

19

E

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

Fair weather atmospheric electricity =

downward 100 km high in the ionosphereC

N 100

Earth

Typical Electric Fields

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rE =

kq1

r2r =

8.99 ×109 Nm2

C2

⎛⎝⎜

⎞⎠⎟ ⋅(+1.6 ×10-19C)

(0.5 ×10-10m)2 =5.75 ×1011

N C

r

Hydrogen atom

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

.m105.0 10−×

+

-r meter

VoltCN :Note =

Typical Electric Fields

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

We have 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?

nCq 101 += nCq 152 +=

0=xP

nCq 152 +=q1 =+10nC

Find the x and y components of the electric field for each charge and add them up.

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Recall E =

kqr2

and k =8.99 ×109 Nm2

C2

E1y =10NC

E1x =0

q1=10 nc q2 =15 nc

P

E1 =8.99 ×109 Nm2

C2 ×10 ×10−9C

(3m)2=10

NC in the y direction

rE1

Now find the magnitude of the field due to q2 ?

Find the magnitude of the

field for q1

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Now add all components

rE2 =

8.99 ×109 Nm2

C2 ×15 ×10−9C

(5m)2=5.4

NC

E2y =E2 sinφ=5.4NC

×35=1.08 ×3=3.24

NC

Ey =(10 + 3.24)NC

=13.24NC

Ex =−4.32NC

q1=10 nc q2 =15 nc

φ

φ

P

Find the magnitude of the field for q2

rE2

Ey =E1y + E2y

Ex =E1x + E2x

rE1

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Find the x and y components

E2x =E2 cosφ=5.4NC

×−45

=−1.08 ×4 =−4.32NC

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rE = −4.32( )2 + 13.24( )2 =13.93N /C

Magnitude of net electric field is

rE = Ex

2 + Ey2

Using unit vector notation we canalso write the electric field vector as:

rE =−4.32 i

∧+13.24 j

∧ N /C

Ey =13.24NC

Ex =−4.32NC

φ1 = tan−1 EyEx

⎛

⎝⎜⎞

⎠⎟= tan−1 13.24

−4.32⎛⎝⎜

⎞⎠⎟

= −71.9°

Add up the components

Direction of the total electric field is q1=10 nc q2 =15 nc

φ

φ1

rE

-71.9

y

j∧

i∧ x

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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.

.

y

4 x+15 nc

3

+15 nc4

P

ϕ

.

y

4 x+15 nc

3

+15 nc4

P

Ex =E1x + E2x =0

Because E1x =−E2xE1x E2 xBy symmetry

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rE1 = 8.99 ×109

Nm2

C2 ×15 ×10-9 C(5m)2

= 5.4 NC

Ey =2E1y =E1 sin φ = 2 ×5.4 × 35 = 6.5

NC

φ

.

y

4 x+15 nc

3

+15 nc4

Now find y component

Ey =E1y + E2y =2E1y

Because E2y =E1yφE1E1y

y

j

In unit vector notationrE = 6.5 j N/C

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y0

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.

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y

4 x-15 nc +15 nc

4φ

x P

E = 8.99 ×109 Nm2

C2 ×15 ×10-9 C(5m)2

= 5.4 NC

Ex =2 ×E cos φ = 2 ×5.4 × 45 = - 8.6

NC

Find the x and y component of the electric field

rE = -8.6 i N/C

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θ−=+ cosE2EE x2x1

icosE2E ,So net θ−=r

2akq

E =

a2L

acos 2

L

θ

ia2

Lakq

2E ,So2net −=

r

rEnet =−

kqLa3 i

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2Lra ⎟

⎠⎞

⎜⎝⎛+=

( )( )23

23 2

r2L32

2L2

net )(1(1

rp

kr

pkE ⎟⎟

⎠

⎞⎜⎜⎝

⎛

+=

+=

rqLp =

rEnet ;

kp

r3

Electric Dipole

aa

E1

E1x

θ

θ

ij

y

For large r

E1x =−Ecosθ and E1x =E2x

E2

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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

r

when r is large compared to L

p=qL = the electric dipole moment

Note inverse cube law

+-

L

+q-q

rEnet ;

kp

r3

The dipole moment p is defined as a vector directedfrom - to +

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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 )p

rextE

r

1Fr

2Fr

x

Torque about the com =

=FL sinθ

= qEL sinθ = pE sinθ =rp ×

rE

Eprrr

×=So,

When the dipole rotates through the electric field does work:

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Water (H2O) is a molecule that has a permanent dipole moment.

When a dipole is an electric field, the dipole

moment wants to rotate to line up with

the electric field. It experiences a torque.

GIven p = 6.2 x 10 - 30 C m And q = -10 e and q = +10e

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.

Leads to how microwave ovens heat up food

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Why do Electric Dipoles align with Electric fields ?

Potential Energy

Ans. The energy is minimum when aligns with

EpcospEUrr⋅−=θ−==

Integrating,

€

So, U = -r p ⋅

r E

pr

€

rE

€

dW = −τdθ = −pE sinθdθWork 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

θθ+=−= dsinpEdWdU

0UcospEdsinpEdWdUU +θ−=θθ−=−==∫ ∫ ∫°=θ= 90 when0U choose We

Problem 22-50

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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.(a)parallel to the field(b) perpendicular to the field(c) antiparallel to the field

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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.

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Smoke Remover

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

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Motion of point charges in electric 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

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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.

v

E

d

•

E

y

electron

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v

E

d

•

E

y

electron

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Finding Electric Fields for Continuous Distributions of Charge

• 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

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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

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

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Continuous distribution of charges

• 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.

dq =dV volumedq=dA area

dq=λdl line

E = dE = kdq / r2∫∫

y

+x

dq-x

- L/2 L/2

dE

r. .

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Find electric field due to a line of uniform + charge of length L with linear charge density equal to λ

θr

dEy= dE cos θ

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

dq = λdxdEy= k λ dx cos θr2

0€

Ex = dEx−L / 2

L / 2

∫ = 0dE

y

+xdq

-x- L/2 L/2

dE

dE = k λdx /r2

λ

€

Ey = kλ cosθ dx /r2

−L / 2

L / 2

∫

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

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θ

y

+x

dq

r

-x

dEy

y

x-L/2 L/20

x= y tanθ dx = y sec2

θ dθ

r =y sec θ r2 y2sec2 θ

dx / r2 = dθ / y

€

Ey = kλ cosθ dθ / y−θ 0

θ 0

∫ = kλ / y cosθ dθ−θ 0

θ 0

∫ = kλ / y 2sinθ0

dE

€

Ey =2kλ

ysinθ0

€

sinθ0 =L /2

y 2 + L2 /4

€

Ey = kλ cosθ dx /r2

−L / 2

L / 2

∫

€

Ey =kλ

y

L

y 2 + L2 /4

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

45

€

Ey = dEy−θ 0

θ 0

∫ = 0

s=r θds=r dθ

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

dq=λds

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

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€

Ey = dEy−θ 0

θ 0

∫ = 0

s=r θds=r dθ

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

dq=λds

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

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dEx= k dq cos θ r2

s=r θds=r dθ

€

Ex = kλ rdθ cosθ /r2

−θ 0

θ 0

∫ = kλ /r dθ cosθ−θ 0

θ 0

∫

dEx= k λds cos θ r2

€

Ex =2kλ

rsinθ0

Now find the x component

dq=λds

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Find the electric field on the axis of a uniformly charged ring with linear charge density λ Q/2πR

€

E z = dE cosθ∫

€

dE = kdq

r2= kλds

r2

€

E z =kλ cosθ

r2ds∫

€

E z =kλ cosθ

r22πR

dq = λdsr2 =z2+R2

cos θ z/r

€

E z =kQz

(z2 + R2)3 / 2

=0 at z=0=0 at z=infinity=max at z=0.7R

€

ds = Rdθ0

2π

∫∫ = R dθ0

2π

∫ = 2πR€

s = Rθ

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Chapter 22 Problem 30 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?

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Chapter 22 Problem 44At 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?

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Chapter 22 Problem 50 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