Medan_Listrik-fisika2

8/2/2019 Medan_Listrik-fisika2

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Summer July 2004 1

Electric Fields Cartoon - Analogous to gravitational field

Opening Demo - Bending of water stream with charged rod Warm-up problem

Physlet

Topics

Electric field = Force per unit Charge

Electric Field Lines and Electric Flux

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

List of Demos

Van de Graaff Generator, workings,lightning rod, electroscope, electric wind

Smoke remover or electrostatic precipitator Kelvin water drop generator

Transparent CRT with visible electron gun

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


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Summer July 2004 2

The Electric Field

Definition of the electric field. Whenever charges are present and if Ibring up another charge, it will feel a net Coulomb force from all the

others. It is convenient to say that there is field there equal to the forceper unit positive charge. E=F/q0. The direction of the electric field isalong r and points in the direction a positive test charge would move.This idea was proposed by Michael Faraday in the 1830s. The idea ofthe field replaces the charges as defining the situation. Consider twopoint charges:

q0q1

r


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Summer July 2004 3

The force per unit charge is E = F/q0

and then the electric field at r is E = kq1/r2 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.

q1

r E

+ q0q1

r

The Coulomb force is F= kq1q0/r2

If q1 were positive


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Summer July 2004 4

Point negative charge

E= kq1/r2

q1

q1

r


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Summer July 2004 5

Electric Field Lines

Like charges (++) Opposite charges (+ -)

This is called an electric dipole.


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Summer July 2004 6

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 begin on + charges and end on - charges. (or infinity).

They enter or leave charge symmetrically.

The number of lines entering or leaving a charge is proportional tothe charge

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

At large distances from a system of charges, the lines becomeisotropic and radial as from a single point charge equal to the netcharge of the system.

No two field lines can cross.

Show a physlet 9.1.4, 9.1.7

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


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Summer July 2004 7

Typical Electric Fields (SI Units)

1 cm away from 1 nC of negative charge

E = kq/r2=1010 *10-9/ 10-4 =105 N /C

N.m2/C2 C / m2 = N/C

Fair weather atmospheric electricity = 100 N/C downward100 km high in the ionosphere

Field due to a proton at the location of the electron in the H

atom. The radius of the electron orbit is 0.5*10-10 m.

E = kq/r2=1010 *1.6*10-19/ (0.5 *10-10 )2 = 4*1011 N /C

q

r

E

+

-r

Hydrogen atom

.

Earth

- - - - - - - - -

E+++++

1N / C = Volt/meter


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Summer July 2004 8

Example of field lines for a

uniform distribution of positive

charge on one side of a very large

nonconducting sheet.

How would the electric field change

if both sides were charged?

How would things change if the sheet

were conducting?

This is called a uniform electric field.


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Summer July 2004 9

Methods of evaluating electric fields

Direct evaluation from Coulombs Law or brute force method

If we know where the charges are, we can find E from E = kqi/ri2.This is a vector equation and can be complex and messy to

evaluate and we may have to resort to a computer. The principleof superposition guarantees the result.

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

E = dE = kdqi/r2 r where ris a unit vector directed from charge

dq to the field point.

dq = dV , or dq = dA, or dq = dl


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Summer July 2004 10

Example of finding electric field from two charges

Find x and y components of electric field due to both charges

and add them up

We have q1=+10 nC at the origin, q

2= +15 nC at x=4 m.

What is E at y=3 m and x=0? point P

x

y

q1=10 nc q2 =15 nc4

3

P

Use principle of superposition


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Summer July 2004 11

Example continued

Field due to q1

E = 1010 N.m2/C2 10 X10-9 C/(3m)2 = 11 N/C

in the y direction.

Recall E =kq/r2

and k=8.99 x 109 N.m2/C2

x

y

q1=10 nc q2 =15 nc4

3

Ey= 11 N/C

Ex= 0

Field due to q2

5

E = 1010 N.m2/C2 15 X10-9 C/(5m)2 = 6 N/C

at some angle

Resolve into x and y components

E

Ey=E sin = 6 3/5 = 18/5 = 3.6 /C

Ex=E cos = 6 (4)/5 = 24/5 = 4.8 /C

Now add all components

Ey= 11 + 3.6 = 14.6 N/CEx= -4.8 N/C

E= Ex2+Ey

2

Magnitude


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Summer July 2004 12

Example continued

xq1=10 nc q2 =15 nc

4

3

Ey= 11 + 3.6 = 14.6 N/C

Ex= -4.8 N/C

E

E=14.6( )2+4.8( )2=15.4N/C

Magnitude of electric field

E= Ex2+Ey

2

1

1 = atan Ey/Ex= atan (14.6/-4.8)= 72.8 deg

Using unit vector notation we can

also write the electric field vector as:

r

E=4.8i

+14.6j


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Summer July 2004 13

Example of two identical

charges on the x axis. What

is the filed on the y axis?

Ey=2*E sin = 26 3/5 = 7.2 N/C

x= 0

y

4 xq2 =15 nc

3

q2 =15 nc4

5

E = 1010 N.m2/C2 15 X10-9 C/(5m)2 = 6 N/C

3

y

4 xq2 = -15 nc q2 =15 nc

4

5

y

= 0

Ex=2*E cos = 26 4/5 = - 9.6 N/C

x

Ey

Example of two opposite

charges on the x axis. What

is the filed on the y axis?


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Summer July 2004 14

4 equal charges symmetrically spaced along a line. What is the field at

point P? (y and x = 0)

3

y

x

q2 q3q1 q4

r1r2

r3r4

21

4

Ey=k qii=1

4

cosi/ri2

P

Ey=k qii=1

cosi/ri2


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Summer July 2004 15

Find electric field due to a line of uniform + charge of length L

with linear charge density equal to

y

+xdq

r

dEy= dE cos

-x

dE = k dq /r2

dEy

dq = dx

dEy= k dx cos /r2

Ey= k q cos /r2 for a point charge

y

Ey=k cosdx/r2

L/2

L/2

x-L/2 L/20

x= y tan dx = y sec2

d

r =y sec r2 = y2sec2dx/r2=d/y

Ey=k/y cosd0

0 Ey=2k

y

sin0sin0=

L/2

y2+L2/4

Ex= dExL/2

L/2

=0

dEx

dE


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Summer July 2004 16

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?

Ey=2k

ysin0

y =10 cm.

Ey=2*10

10Nm2*100*109C/m

0.1msin90=2*104

N/C

Typical field for the electrostatic smoke cleaner

Ey

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


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Summer July 2004 17

Electric field gradient

When a dipole is 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 woodis attracted to the teflon rod and how a smoke remover or

microwave oven works.

Show smoke remover demo.


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Summer July 2004 18

Electrostatic smoke precipitator model

Negatively charged central wire has electric field that varies as1/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 tobe 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 isrepelled.

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


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Summer July 2004 19

Electric Dipoles:A pair of equal and opposite charges qseparated by a displacement d is called an electric dipole. It

has an electric dipole moment p=qd.

P

r

IEI ~ 2kqd/r3 when r is large compared to d

p=qd = the electric dipole moment

IEI ~2kp/r3 Note inverse cube law

IEI = kq/r2 Note inverse square lawfor a single charge.

+

-

d p

+q

-q


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Summer July 2004 20

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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Summer July 2004 21

H2O in a Uniform Electric Field

There exist a torque on the water molecule

To rotate it so that p lines up with E.

x

Torque about the com =

F x sin + F(d-x)sin = Fdsin =qEdsin = pEsin = p x E

Potential Energy = U = -W = -pEcos = p.

Is a minimum when p aligns with E

= p xE


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Summer July 2004 22

Motion of point charges in electric fields

When a point charge such as an electron is placed in an electricfield E, it is accelerated according to Newtons Law:

a = F/m = qE/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, In y =1/2at2 we get y =1/2(qE/m)t2


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Summer July 2004 23

Example: An electron is projected perpendicularly to a

downward electric field ofE= 2000 N/C with a horizontalvelocity v=106 m/s. How much is the electron deflectedafter traveling 1 cm.

Since velocity in x direction does not change, t=d/v =10-2/106 = 10-6 sec,so the distance the electron falls upward is

y =1/2at2 = 0.5*eE/m*t2 = 0.5*1.6*10-19*2*103/10 - 30*(10-8)2 = 0.016m

V

E

d

e

E

Demo Transparent CRT with electron gun


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Summer July 2004 24

Back to computing Electric Fields

Electric field due to a line of uniform charge

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

Next we will go on to another simpler method to calculate

electric fields that works for highly symmetric situations using

Gausss Law.


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Summer July 2004 25

dEx= k dq cos /r2

Ey= dEy

L/2

L/2

=0

s=rds=r d

Ex=k rdcos/r2

L/2

L/2

=k/r dcos0

0

dEx= k ds cos /r2

Ex=2k

r

sin0

Field due to arc of charge

What is the field at the center

of a circle of charge?

Ans. 0

Find the electric field on the axis of a uniformly charged ring


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Summer July 2004 26

Find the electric field on the axis of a uniformly charged ring

with linear charge density = Q/2R.

Ez= dEcos

dE=k

dq

r2=k

ds

r2

Ez=kcos

r2

ds

Ez=

kcos

r2

2R

dq = ds

r2 =z2+R2

cos = z/r

Ez= kQz(z2+R2)3/2

=0 at z=0

=0 at z=infinity

=max at z=0.7R

ds= Rd0

2

=R d0

2

=2

s=R


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Summer July 2004 27

Warm-up set 2

1. [153709] Can there be an electric field at a point where there is no charge?Can a charge experience a force due to its own field? Please write a one or two sentence answer

for each question.

2. [153707] An insulator is a material that...three are correctis not penetrated by electric fieldsnone of these

cannot carry an electric chargecannot feel an electrical force

3. [153708] Which of the following is true of a perfect conductor ?There can be no electric charge on the surface.There cannot be an electric field inside.There cannot be any excess electric charge inside.There cannot be any electric charges inside.

Two of the choices are correct


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Summer July 2004 28

Kelvin Water Drop Generator

Am. J. Phys. 68,1084(2000)

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