p361 lecture3a coulombs law

8/13/2019 P361 Lecture3a Coulombs Law

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P361: Electromagnetic

Theory

Electrostatic Fields 1

Patrick Sibanda


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Outline1. Coulombs Law2. The Electric FieldExamples


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Coulombs Law

Where k=9x109N!m2/C2. and we have used the value of!

0= 8.854187282x10-12 farad/meter

Coulomb

s law quantifies the interaction between charged particles i.e. gives the force (in Newtons) between charges Q1and Q2,

where r12is the distance in meters between the charges

F12

=

Q1Q

2

4!"0r12

2r = k

Q1Q

2

r12

2 r newtons

The force is attractive for unlike &repulsive for like

charges

+ -

Q1 Q2

r12


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Coulombs law is valid for point charges. If the charged objects are spherical and the charge is

uniformly distributed, r12is the distance between the centersof the spheres.

If the charges are at rest For a charge in motion if the other charge is at rest-------------------------------------------------------------------------------------


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Solving Problems Involving Coulombs Law andVectors

x

y

Q2=+50C

Q3=+65C

Q1=-86C

52 cm

30c

m

!=30

Example: Calculate the net electrostatic force on charge Q3due to the charges Q1 and Q2.


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Step 0: Remember to Think Think Think!

This is a Coulombs Law problem (all we have to work with, sofar).

We only want the forces on Q3.

Forces are additive, so we can calculate F32and F31and addthe two.

If we do our vector addition using components, we must resolve

our forces into their x- and y-components.


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Draw and label forces (only those on Q3).

Draw components of forces which are not along axes.

x

y

Q2=+50C

Q3=+65C

Q1=-86C

52 cm

30cm

!=30

F31

F32Draw a representative

sketch.

Draw and label relevant

quantities.

Draw axes, showing

origin and directions.

Step 1: Diagram


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F 032,x = (from diagram)

F32,y= 330 N and F32,x= 0 N.

x

y

Q2=+50C

Q3=+65C

Q1=-86C

52 cm

r

32=30cm

!=30

F31

F32

Step 3: Replace Generic Quantities by Specifics

F32 = k Q3Q2r32

2

repulsive

F32,y = k

Q3Q2

r32

2


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F3x= F31,x+ F32,x= 120 N + 0 N = 120 N

F3y= F31,y+ F32,y= -70 N + 330 N = 260 N

You know how to calculate the magnitude F3and the angle

between F3and the x-axis.

F3The net force is thevector sum of all the

forces on Q3.

x

y

Q2=+50C

Q3=+65C

Q1=-86C

52 cm

30cm

!=30

F31

F32

Step 3: Complete the Math


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That was a sample Coulombs law calculation using three pointcharges.

How do you apply Coulombs law to objects that containdistributions of charges?

Well use another tool to do that"


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The Electric FieldCoulomb's Law (demonstrated in 1785) shows that chargedparticles exert forces on each other over great distances.

How does a charged particle "know" another one is there?


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Faraday, beginning in the 1830's, was the leader in developingthe idea of the electric field. Here's the idea:

A charged particle emanates a "field"into all space.

Another charged particle senses the field,and knowsthat the first one is there.

+

+-

like

chargesrepel

unlikechargesattract

F12

F21

F31

F13


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We define the electric field by the force it exerts on a testcharge Q0:

This is your second starting equation. By convention the direction of the electric fieldis the direction of the force exerted on a POSITIVEtest charge. The absence of

absolute value signs around Q0means you must include the sign of Q0in your work.

If the test charge is "too big" it perturbs the electric field, so thecorrectdefinition is

Any time you know the electric field, you can use this equation to calculate the force

on a charged particle in that electric field.

You wont be required to usethis version of the equation.

E =

F0Q

0

E = limQ0!0

F0

Q0

F =Q

E


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The units of electric field are Newtons/Coulomb.

Later you will learn that the units of electric field can also be

expressed as volts/meter:

[ ]N V

E = =C m

The electric field exists independent of whether there is acharged particle around to feelit.

E!" #$=

F0

!

"

#

$Q

0[ ] = NC


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+Remember: the electric field direction is thedirection a + charge would feel a force.

A + charge would be repelled by another + charge.

Therefore the direction of the electric field is away from positive(and towards negative).


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The equation for the electric field of a point charge then

becomes:

We define as a unit vector from the source point to the fieldpoint:

r

+

source point

field point

r

You may start with either equationfor the electric field (this one or the

one on the previous slide). Butdont use this one unless youREALLY know what you are doing!

E = kQ

r2

r


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1. A DipoleA combination of two electric charges with equal magnitude andopposite sign is called a dipole.

+ -+q -q

d

The charge on this dipole is q (not zero, not +q, not q, not2q). The distance between the charges is d. Dipoles areeverywherein nature.

This is an electricdipole. Later in the course well study magnetic dipoles.


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Dipole

i

2i i

qE=k rr!

!

Example: calculate the electric field at point P, which lies on theperpendicular bisector a distance L from a dipole of charge q.

+ -+q -q

d

P

Lto be worked at the blackboard

3

o

qdE

4 r

=

!"


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+ -+q -q

d

P

L

3

o

qdE

4 r=

!"

Caution! The aboveequation for E appliesonly to points alongthe perpendicularbisector of the dipole.

It is not a startingequation.

.companion notes


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Motion of a Charged Particlein a Uniform Electric Field

A charged particle in an electric field experiences a force, and ifit is free to move, an acceleration.

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

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

-

F

If the only force is due to the

electric field, then

If E is constant, then a is constant, and you can use theequations of kinematics.

E

F! =ma =Q

E


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Example: an electron moving with velocity v0in the positive xdirection enters a region of uniform electric field that makes aright angle with the electrons initial velocity. Express the

position and velocity of the electron as a function of time.

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

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

Ex

y

-

v0


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The Electric FieldDue to a Collection of Point Charges

(Principle of superposition) If more than one charge is involved, the net force is the

vector sum of all forces (principle of superposition). For objects with complex shapes, you must add up all the

forces acting on each separate charge (turns into calculus!).

+

++ -

--


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The electric field due to a small "chunk" "Q of charge is

The electric field due to collection of "chunks" of charge is

unit vector from "Qto wherever you

want to calculate "E

unit vector from "qito wherever you

want to calculate E

As "Q#dQ#0, the sum becomes an integral.

!

E =1

4!"0

!Q

r2 r

E = !

Eii" =

1

4!"0

!Q

i

ri

2

i" ri


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If charge is distributed along a straight line segment parallelto the x-axis, the amount of charge dQ on a segment of length dxis $dx.

$ is the linear density of charge (amount of charge per unitlength). $may be a function of position.

Think $% %length. $times the length of line segment is thetotal charge on the line segment.

!

x

dx

$ $dx


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The electric field at point P due to the charge dQ is

x

dQ

P

r

dE

Im assuming positively charged objectsin these distribution of chargesslides.

r

dE =

1

4!"0

dQ

r'2

r =1

4!"0

!dx

r'2

r


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If charge is distributed over a two-dimensional surface, theamount of charge dQ on an infinitesimal piece of the surface is&dS, where &is the surface density of charge (amount ofcharge per unit area).

x

y

area = dS

&

charge dQ = &dS


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dE

The electric field at P due to the charge dq is

x

y

P

rr

dE =

1

4!"0

dQ

r'2

r =1

4!"0

!ds

r'2

r


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The net electric field at P due to the entire surface of charge is

x

y

P

r

E

E =1

4!"0

r!(x,y)dS

r'2

S

!

r


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x

zP

r

After you have seen the above, I hope you believe that the netelectric field at P due to a three-dimensional distribution ofcharge is

y

E

r

E =

1

4!"0

r!(x,y,z)dV

r'2

V

! !is the volume chargedensity


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Summarizing:

Charge distributed along a line:

Charge distributed over a surface:

Charge distributed inside a volume:

If the charge distribution is uniform, then $, &, and 'can be taken outsidethe integrals.

E =1

4!"0

r!dx

r'2

!

E =

1

4!"0

r!dS

r'2

S

!

E =1

4!"0

r!dV

r'2

V

!


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The Electric Field

Due to a Continuous Charge Distribution(worked examples)


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Example: A rod of length L has a uniform charge per unit length$and a total charge Q. Calculate the electric field at a point Palong the axis of the rod at a distance d from one end.

P x

y

d L

Lets put the origin at P. The linear charge density and Q are

related by Q= and Q = L

L! !

Lets assume Q is positive.


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P x

y

d L

dE xdx dQ = $dx

E= d

Ed

d+L

! x= "k

!dx

x2d

d+L

! x = "k!

dx

x2d

d+L

! x = "k! "

1

x

#

$%

&

'(d

d+L

x

E= !k! ! 1

d+L

+

1

d

"

#$

%

&'x = !k!

!d+ d+L

d(d+L)

"

#$

%

&'x = !k

!L

d(d+L)

x = ! kQ

d(d+L)

x


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Example: A ring of radius a has a uniform charge per unitlength and a total positive charge Q. Calculate the electric fieldat a point P along the axis of the ring at a distance x0from its

center.

P

a

dQ

r

x0

dE

x!

!

By symmetry, the y- and z-components of E are zero,

and all points on the ringare a distance r from pointP.


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P

a

dQ

r

dE

x!

!

2

dQdE=k

r

x 2dQdE =k cosr!

2 2

0r = x a+

0x

cosr

! =

x0

( )

0 0 0 0x x 3/ 22 3 3 2 2

ring ring ring 0

x x x kx QdQE dE k k dQ k Q

r r r r x a

! "= = = = =# $

% & +

' ' '

Or, in general, on the ring axis

( )

x,ring 3/ 22 2

kxQE .

x a

=

+

For a given x0, r is a constantfor points on the ring.

No absolute valuesigns because Q ispositive.


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Example: A disc of radius R has a uniform charge per unit area&. Calculate the electric field at a point P along the central axisof the disc at a distance x0from its center.

P

r

dQ

x

x0R

The disc is made ofconcentric rings. Thearea of a ring at aradius r is 2(rdr, andthe charge on each ringis &(2(rdr).

We can use the equation on the previous slide for the electricfield due to a ring, replace a by r, and integrate from r=0 tor=R.

( )

0ring 3/ 2

2 2

0

kx 2 rdr dE .

x r

! "

=

+Caution! Ive switched

the meaningof r!


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P

r

dQ

xx0R

( ) ( )

R0

x x 03/ 2 3/ 202 2 2 2

disc disc0 0

kx 2 rdr 2r drE dE kx

x r x r

! "= = = "!

+ +

# # #

( )

( )

R1/ 2

2 2

00 0

x 0 1/ 22 2

00

0

x r x xE kx 2k

1/ 2 x x R

!" # $ %+

& '( )= *+ = *+ !& '( )!

+, - . /


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p361 lecture3a coulombs law

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