Chap 3 Gauss Law Slide

8/11/2019 Chap 3 Gauss Law Slide

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Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology

Line of electric field due to a point

charge

+ -


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Line of electric field due toelectric dipol

+ -


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Electric Flux Definition: number of electric field line passing through asurface

For surface dA that perpendicular with direction of electricfield, amount of electric field line across that surface is

Total electric flux is

EdAd =

dA

EAEAdAE

EdAd

A

AA

==

==


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Electric Flux for arbritary surface For arbritary surface dA

AdEdrr

=dA

=

=

S

S

AdE

d

rr

Total electric flux

For surface S isE

S


6/44Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology

An electric field represent by

Find electric flux if the surface are

a. b.c. d.

e. f. Solution

Because electric field is homogen in every surface in

question so electric flux can be rewrite in form

iS 10=r

jS 10=r

kS 10=

r

kS 10=

r

jS 10=r

iS 10=r

SEAdES

rrrr

jiE 42 +=r



Then :a.

b.

c.

d.

e.f.

010)42( =+== kjiAErr

010)42( =+== kjiAErr

4010)42( =+== jjiAErr

4010)42( =+== jjiAErr

2010)42( =+== ijiAE

rr

2010)42( =+== ijiAErr



Electric Flux, charge Q, open surface S

Electric Flux

emanating from

surface S is

=

S

ndSE 1r

1n

S

dSE



Closing surface S, and charge Q outside S

+

1n dA

1n

2n

2

n

3n

3n



Calculation of Electric Flux if charge

Q is outside the surface Look at the direction of normal of the surface and

direction of electric field Total Electric Field passing through the cubic is

0

0000

)(

)(

)(

11

33

22

11

=

++=

+

++

++=

=

SS

SS

SS

S

ndAEndAE

ndAEndAE

ndAEndAE

AdE

rr

rr

rr

rr


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Calculation of Electric Flux if charge

Q is inside the surface Look at the direction of normal of the surface and

direction of electric field Total Electric Field passing through the cubic is

0

)(

)(

)(

332211

33

22

11

+++++=

+

++

++=

=

SS

SS

SS

S

ndAEndAE

ndAEndAE

ndAEndAE

AdE

rr

rr

rr

rr



Gauss Law Number of electric field line passing through

the closing surface is proportional to number of

enclosing charge by closing surface

Some step to apply Gauss Law:

Choose the surface that electric field at the surface is

homogen/constant

Hint enclosing charge

Apply to Gauss Law

=0

qSdErr



Gauss Surface in form spheris

This surface is applied if the source charge is pointcharge or spheris charge

dAE


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Apply of Gauss Law to hint electric field due

to a point chargedA

E

0

2

0

2

0

0

0

4

4

r

qE

q

rE

qdAE

qEdA

qAdE

=

=

=

=

=

rr


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Conductor dan Insulator

Inside conductor, the charge is free moving

If given electric filed to conductor the chargeis free moving so there is a current there ischarge redistribution until the electostatic

equibirilium happen electric field insideconductor become zero Gauss law says that

charge inside conductor become zerro so thecharge must be uniformly spread out at the

surface of conductor


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Needed very quict times till electrostatic

equibrilium at the conductor is happen

So electric field inside conductor always zero andif the conductor is charging then the charge is

always at the surface of the conductor Inside insulator, the charge is not free moving

Charge will be uniformly spread inside theinsulator


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Positive conductor spheris

Suppose there is a spherical conductor. This

conductor have radius R and charge Q

dAE

The charge only spead

at the surface of sphe-rical conductor

So electric chargeinside conductor (r


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Electric field outside spherical conductor

For r>R (outside conductor), Total enclosing

charge is Q so:

where r>R

200

2

00

44

r

QE

QrE

QdSE

qSdE

==

==

rr


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Electric Field outside sperical insulator(r>R)

2

0

0

2

0

0

4

4

r

QE

QrE

Q

dSE

QSdE

=

=

=

=

rr

Rr


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Electric field inside hollow spherical insulator

Q

RR

Rrq

3

13

43

23

4

3

1343

34

=

R1

R2

r

2

0

3

1

3

2

3

1

3

0

3

1343

234

3

1343

34

0

4

1

r

Q

RR

RrE

Q

RR

RrdSE

qSdE

=

=

=

rr


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Negative charge spheris For negatif charge spheris, the principle is the same as

potitive charge spheris, the difference is in the direction of

electric field. The direction of electric field due to negative

charge spheris is radially inward

E

dA

2

0

0

2

0

0

4

4

180cos

r

QE

Q

rE

QEdS

QSdE

=

=

=

=

rr


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Example A thin spherical charge have 2Q charge and radius a.

Inside thin spherical there is a rigid conductor spheris

with radius b and charge of this spheris is -3Q.

Find electric field outside athin spherical charge (r>a).

ab


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2

00

2

00

4

4

180cos

r

QE

QrE

QEdS

qSdE

==

==

rr

Elecric field for r>aMake Gauss surface in form of spheris with

radius r>aTotal charge enclose this surface Gauss is

q=2Q+(-3Q)=-Q

So the electric field is


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Electric Field due to infinite straight wire Choose Gauss surface in silindrical form

For positive straight wire, the direction of electric field

is radially outward from the center of silindris For negative straight wire, the direction of electric field

is radially inward to the center of silindris

dAE


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rlE

EdS

EdSEdS

SdESdESdESdE

tutup

ungsetutup

tutupungsetutup

2

90cos

0cos90cos lub

lub

=

+

+=

++=

rrrrrrrr

Electric flux passing throught silindrical surface is

If lenght of the wire is L then charge enclose by

Silindrical surface is

llL

Qq ==


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r

rL

QE

lL

Q

rlE

qSdE

0

0

0

0

2

2

2

=

=

=

= rr


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Example

Find magnitude and the direction of electric

field at 20 cm from long straight wire which the

density charge of this wire is=10 mC/m.

Solution:

025,0

4

1,0

)2,0(2

10.10

2

3

====

r

E

A

B

N/C


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Example

Find magnitude and the direction of electric field

at 20 cm from long straight wire which the density

charge of this wire is

=-10 mC/m.

Solution:

025,0

4

1,0

)2,0(2

10.10

2

3

====

r

E

A

B

N/C


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Example Two infinite straight wires have density charge dan

-2. The distance between two wires is a. Find

electric field at distance b from -2

wire. 2Q.

EEEtotalrrr

+= 2

-2

ba P

E-2E

)(2

2

)(2

2

00

2

bab

EEEtotal

+=

=


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Electric field due to silindrical charge

Suppose we have silindrical charge with

radius: R , lenght: L, and charge: Q.

We must choose Gauss surface in silindricalform with radius: r and lenght: L

For positive silindrical charge, the directionof electric field is radially outward from the

center of silindris

For negative silindrical charge, the direction

of electric field is radially inward to the center

of silindris


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Gauss surface at silindrical charge

If the charge is positive

EdA

=

=

=

0

0

0

0cos

qdAE

qEdA

qAdE

rr


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If the charge is negative

EdA

=

=

=

0

0

0

0cos

qdAE

qEdA

qAdErr

Gauss surface at silindrical charge


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Electric field due to rigid silindrical charge

conductor Inside silindrical conductor

The enclosing charge is zero because the charge isonly spread at the surface, so electric field inside

silindrical conductor E=0


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Outside silindrical conductor

Enclosing charge is

Qq=


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Lr

QE

QrLE

QdAE

q

AdE

0

0

0

0

2

2

=

=

=

=

rr


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Electric field due to positive plane charge

Suppose there is a plane with charge density is

E

E

SSA

Q

q ==

A

S


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0

0

0

2

2

=

=

=

E

SSE

qAdErr

ES

ESES

SdESdESdESdE tutupungsetutup

2

0

lub

=++=

++= rrrrrrrr


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Electric field due to negative plane charge

Suppose there is a plane with charge density is -

E

E

SSA

Q

q =

=

A

S


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0

0

0

2

)2(

=

=

=

E

SSE

qAdErr

ES

ESES

SdESdESdESdE tutupungsetutup

2

0

lub

=+=

++= rrrrrrrr


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Electric field due to two plane Two plane have and - density charge.

02

== EE

-

E1 E2 E3

0)()(

)()(

0)()(

3

0

2

1

=+=

=+=

=+=

iEiEE

iEiEE

iEiEE

r

r

r

Documents

Chap 3 Gauss Law Slide