mit class: particle interactions: coulomb’s law

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Workshop: Using Visualization in Teaching Introductory E&M

AAPT National Summer Meeting, Edmonton, Alberta, Canada.

Organizers: John Belcher, Peter Dourmashkin, Carolann Koleci, Sahana Murthy

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MIT Class: Particle Interactions:

Coulomb’s Law

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Gravitational Vector Field

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2ˆg

MmGr

F r

Example Of Vector Field: Gravitation

Gravitational Force:

Gravitational Field:

2

2

/ ˆ ˆg GMm r MGm m r

F

g r r

M : Mass of Earth

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Example Of Vector Field: Gravitation

Gravitational Field:

2ˆMG

rg r

M : Mass of Earth

ˆ :r unit vector from M to m

rrr

ˆ3

MGr

g r

g mF g

Created by M Felt by m

USE THIS FORM!

:r vector from M to m

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The Superposition Principle

3 13 23 F F F

1

N

j iji

F F

Net force/field is vector sum of forces/fields

In general:

Example:

1

2

13 13/F g

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In Class ProblemFind the gravitational field at point P

Bonus: Where would you put another mass m to make the field become 0 at P?

Use

g

g

NOTE: Solutions will be posted within two days of class

3

MGr

g r

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From Gravitational toElectric Fields

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Electric Charge (~Mass)Two types of electric charge: positive and negative Unit of charge is the coulomb [C]

Charge of electron (negative) or proton (positive) is

Charge is quantized

Charge is conserved

19, 1.602 10e e C

Q Ne

n p e e e

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Electric Force (~Gravity) The electric force between charges q1 and q2 is

(a) repulsive if charges have same signs (b) attractive if charges have opposite signs

Like charges repel and opposites attract !!

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Coulomb's LawCoulomb’s Law: Force on q2 due to interaction between q1 and q2

9 2 2

0

1 8.9875 10 N m /C4ek

1 212 2

ˆeq qkr

F r

ˆ :r unit vector from q1 to q2

rrr

ˆ 1 212 3e

q qkr

F r

:r vector from q1 to q2

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Coulomb's Law: Example

3132 2 2

ˆ ˆ m

1mr

r i ja = 1 m

q1 = 6 C

q3 = 3 C

q2 = 3 C

?32 F

32 3 2 3ek q qr

rF

32r

981 10 ˆ ˆ3 N

2

i j

129 2 2

3

ˆ ˆ3 m9 10 N m C 3C 3C

1m

i j

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The Superposition Principle

3 13 23 F F F

1

N

j iji

F F

Many Charges Present:Net force on any charge is vector sum of forces from other individual charges

Example:

In general:

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Electric Field (~g)The electric field at a point P due to a charge q is the force acting on a test charge q0 at that point P, divided by the charge q0 :

0

0

( ) qqq P

q

FE

For a point charge q:2

ˆ( )q eqP kr

E r

Units: N/C, also Volts/meter

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Superposition Principle

The electric field due to a collection of N point charges is the vector sum of the individual electric fields due to each charge

1 21

. . . . .N

total ii

E E E E

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Gravitational & Electric Fields

2ˆsMG

rg r

Mass Ms Charge qs (±)

2ˆs

eqkr

E r

g mF g

E qF E

This is easiest way to picture field

CREATE:

FEEL:

SOURCE:

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PRS Question:Electric Field

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PRS: Electric FieldTwo opposite charges are placed on a line as shown below. The charge on the right is three times larger than the charge on the left. Other than at infinity, where is the electric field zero?

0%0%0%0%0%0% 1. Between the two charges

2. To the right of the charge on the right3. To the left of the charge on the left4. The electric field is nowhere zero5. Not enough info – need to know which is positive6. I don’t know

:20

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PRS Answer: Electric FieldAnswer: 3. To the left of the charge on the left

Between: field goes from source to sink. On right: field dominated by qR (bigger & closer).

On left: because qL is weaker, its “push” left will somewhere be balanced by qR’s “pull” right

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Electric Field Lines1. Join end-to-end infinitesimal vectors representing E…the curve

that results is an electric field line (also known as line of force).2. By construction then, the direction of the E field at any given

point is tangent to the field line crossing that point.3. Field lines point away from positive charges and terminate on

negative charges.4. Field lines never cross each other.5. The strength of the field is encoded in the density of the field

lines.

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PRS Questions:Electric Field

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PRS: ForceThe force between the two charges is:

0%

25%

25%

50% 1. Attractive2. Repulsive3. Can’t tell without more information4. I don’t know

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PRS Answer: Force

One way to tell is to notice that they both must be sources (or sinks). Hence, as like particles repel, the force is repulsive.

You can also see this as tension in the field lines

The force between the two charges is:2) Repulsive

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PRS: Field LinesElectric field lines show:

Remember: Don’t pick up until you are ready to answer

0%

33%

0%0%

67%

1. Directions of forces that exist in space at all times.

2. Directions in which charges on those lines will accelerate.

3. Paths that charges will follow.4. More than one of the above.

5. I don’t know..

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PRS Answer: Field Lines

NOTE: This is different than flow lines (3). Particles do NOT move along field lines.

Answer: 2. Directions charges accelerate.

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In-Class Problem

d

s

q q

P

Consider two point charges of equal magnitude but opposite signs, separated by a distance d. Point P lies along the perpendicular bisector of the line joining the charges, a distance s above that line. What is the E field at P?

ij

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Two PRS Questions:E Field of Finite # of

Point Charges

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PRS: Equal Charges

Electric field at P is:

3/ 2 3/ 22 22 2

3/ 2 3/ 22 22 2

2 2ˆ ˆ1. 2.

4 4

2 2ˆ ˆ3. 4.

4 4

5. I Don't Know

e e

e e

k qs k qd

d ds s

k qd k qs

d ds s

E j E i

E j E i

1 2 3 4 5

25%

75%

0%0%0%

1. 1

2. 23. 34. 45. 5

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PRS Answer: Equal ChargesElectric field at P is:

3/ 222

2 ˆ1.

4

ek qs

ds

E j

There are a several ways to see this. For example, consider d0. Then,

2

2 ˆe

qks

E j

which is what we want (sitting above a point charge with charge 2q)

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PRS: 5 Equal Charges

Six equal positive charges q sit at the vertices of a regular hexagon with sides of length R. We remove the bottom charge. The electric field at the center of the hexagon (point P) is:

2 2

2 2

2 2ˆ ˆ1. 2.

ˆ ˆ3. 4.

5. 0 6. I Don't Know

kq kqR Rkq kqR R

E j E j

E j E j

E

1. 1

2. 2

3. 3

4. 4

5. 5

6. 6

1 2 3 4 5 6

100%

0% 0%0%0%0%

1. 12. 23. 34. 45. 56. 6

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PRS Answer: 5 Equal Charges

• E fields of the side pairs cancel (symmetry)• E at center due only to top charge (R away)• Field points downward

Alternatively:• “Added negative charge” at bottom• R away, pulls field down

2ˆ4. kq

RAnswer : E j

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Charging

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How Do You Get Charged?

• Friction• Transfer (touching)• Induction

+q Neutral----

++++

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Demonstrations:Instruments for

Charging

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Electric Dipoles

A Special Charge Distribution

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Electric Dipole Two equal but opposite charges +q and –q, separated

by a distance 2a

charge×displacementˆ ˆ×2 2q a qa

p

j j

points from negative to positive chargep

q

-q

2a

Dipole Moment

p

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Why Dipoles?

Nature Likes To Make Dipoles!

Animation

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Dipoles make Fields

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Electric Field Created by Dipole

3 3x ex xE k q

r r

3 3y ey yE k q

r r

Thou shalt use components!

3/ 2 3/ 22 2 2 2( ) ( )e

x xk qx y a x y a

3/ 2 3/ 22 2 2 2( ) ( )e

y a y ak qx y a x y a

2 3 3 3

ˆ ˆ ˆx yr r r r

r r i j

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PRS Question:Dipole Fall-Off

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PRS: Dipole Field

As you move to large distances r away from a dipole, the electric field will fall-off as:

0%

0%

0%

100% 1. 1/r2, just like a point charge2. More rapidly than 1/r2

3. More slowly than 1/r2

4. I Don’t Know

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PRS Answer: Dipole Field

We know this must be a case by thinking about what a dipole looks like from a large distance. To first order, it isn’t there (net charge is 0), so the E-Field must decrease faster than if there were a point charge there.

Answer: 2) More rapidly than 1/r2

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Point Dipole Approximation

Take the limit r a

30

3 sin cos4x

pEr

23

0

3cos 14y

pEr

Finite Dipole

Point Dipole

You can show…

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Shockwave for Dipole

Dipole Visualization

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Dipoles feel Fields

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Demonstration:Dipole in Field

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Dipole in Uniform FieldˆEE i

ˆ ˆ2 (cos sin )qa p i j

( ) 0net q q F F F E E

Total Net Force:

Torque on Dipole:

tends to align with the electric field p

τ r F

2 sin( )a qE

p E

sin( )rF sin( )pE

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Torque on DipoleTotal Field (dipole + background) shows torque:

• Field lines transmit tension• Connection between dipole field and

constant field “pulls” dipole into alignment

Animation

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PRS Question:Dipole in Non-Uniform Field

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PRS: Dipole in Non-Uniform FieldA dipole sits in a non-uniform electric field E

E

Due to the electric field this dipole will feel:

0%

0%

33%

67% 1. force but no torque2. no force but a torque3. both a force and a torque4. neither a force nor a torque

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PRS Answer: Non-Uniform Field

Because the field is non-uniform, the forces on the two equal but opposite point charges do not cancel.As always, the dipole wants to rotate to align with the field – there is a torque on the dipole as well

Answer: 3. both force and torqueE

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Continuous Charge Distributions

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?P E

V

Continuous Charge Distributions

ii

Q q Break distribution into parts:

2ˆe

qkr

E r

E field at P due to q

Superposition:

E E

V

dq

d E

2ˆe

dqd kr

E r

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Continuous Sources: Charge Density

LQ

QA

QV

Length L

L

w

L

Area A wL

R

L

2Volume V R L

dLdQ

dAdQ

dVdQ

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Examples of Continuous Sources: Line of charge

LQ

Length L

L

dLdQ

Link to applet

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Examples of Continuous Sources: Line of charge

LQ

Length L

L

dLdQ

Link to applet

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Examples of Continuous Sources: Ring of Charge

2Q

R

dLdQ

Link to applet

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Examples of Continuous Sources: Ring of Charge

2Q

R

dLdQ

Link to applet

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Example: Ring of Charge

P on axis of ring of charge, x from centerRadius a, charge density .

Find E at P

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Ring of Charge

Symmetry!0E 1) Think about it

Mental Picture…

2) Define Variables

dq dl

22 xar

a d

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Ring of Charge3) Write Equation dq a d

2

ˆe

rd k dqr

E

3x exdE k dqr

22 xar

3erk dqr

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Ring of Charge4) Integrate

3x x exE dE k dqr

Very special case: everything except dq is constant

2a

22 xar

3exk dqr

dq2 2

0 0a d a d

Q

dq a d

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Ring of Charge5) Clean Up

3x exE k Qr

3/ 22 2x exE k Q

a x

3/ 22 2ˆ

exk Q

a x

E i

3/ 2 22

ex e

k QxE k Qxx

6) Check Limit 0a

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r

2L

2L

s

P

j

i

In-Class: Line of Charge

Point P lies on perpendicular bisector of uniformly charged line of length L, a distance s away. The charge on the line is Q. What is E at P?

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r

2L

2L

xd

x

xddq s

22 xsr

P

j

i

Hint: Line of Charge

Typically give the integration variable (x’) a “primed” variable name. ALSO: Difficult integral (trig. sub.)

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E Field from Line of Charge

2 2 1/ 2ˆ

( / 4)eQk

s s L

E j

Limits:

2ˆlim e

s L

Qks

E j

ˆ ˆ2 2lim e es L

Qk kLs s

E j j

Point charge

Infinite charged line

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In-Class: Uniformly Charged Disk

P on axis of disk of charge, x from centerRadius R, charge density .

Find E at P

( 0)x

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Disk: Two Important Limits

1/ 22 2ˆ1

2disko

x

x R

E i

2

1 ˆlim4disk

x R o

Qx

E i

ˆlim2disk

x R oE i

Limits:

Point charge

Infinite charged plane

***

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Scaling: E for Plane is Constant

1) Dipole: E falls off like 1/r3

2) Point charge: E falls off like 1/r2

3) Line of charge: E falls off like 1/r4) Plane of charge: E constant