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8/11/2019 Chap 3 Gauss Law Slide
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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
+ -
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 toelectric dipol
+ -
8/11/2019 Chap 3 Gauss Law Slide
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Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
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
==
==
8/11/2019 Chap 3 Gauss Law Slide
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Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
Electric Flux for arbritary surface For arbritary surface dA
AdEdrr
=dA
=
=
S
S
AdE
d
rr
Total electric flux
For surface S isE
S
8/11/2019 Chap 3 Gauss Law Slide
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
8/11/2019 Chap 3 Gauss Law Slide
7/44Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
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
8/11/2019 Chap 3 Gauss Law Slide
8/44Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
Electric Flux, charge Q, open surface S
Electric Flux
emanating from
surface S is
=
S
ndSE 1r
1n
S
dSE
8/11/2019 Chap 3 Gauss Law Slide
9/44Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
Closing surface S, and charge Q outside S
+
1n dA
1n
2n
2
n
3n
3n
8/11/2019 Chap 3 Gauss Law Slide
10/44Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
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
8/11/2019 Chap 3 Gauss Law Slide
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8/11/2019 Chap 3 Gauss Law Slide
12/44Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
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
8/11/2019 Chap 3 Gauss Law Slide
13/44Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
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
8/11/2019 Chap 3 Gauss Law Slide
14/44Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
Gauss Surface in form spheris
This surface is applied if the source charge is pointcharge or spheris charge
dAE
8/11/2019 Chap 3 Gauss Law Slide
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8/11/2019 Chap 3 Gauss Law Slide
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8/11/2019 Chap 3 Gauss Law Slide
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Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
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
8/11/2019 Chap 3 Gauss Law Slide
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Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
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
8/11/2019 Chap 3 Gauss Law Slide
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Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
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
8/11/2019 Chap 3 Gauss Law Slide
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Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
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
8/11/2019 Chap 3 Gauss Law Slide
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Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
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
8/11/2019 Chap 3 Gauss Law Slide
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8/11/2019 Chap 3 Gauss Law Slide
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Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
Electric Field outside sperical insulator(r>R)
2
0
0
2
0
0
4
4
r
QE
QrE
Q
dSE
QSdE
=
=
=
=
rr
Rr
8/11/2019 Chap 3 Gauss Law Slide
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Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
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
8/11/2019 Chap 3 Gauss Law Slide
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Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
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
8/11/2019 Chap 3 Gauss Law Slide
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Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
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
8/11/2019 Chap 3 Gauss Law Slide
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Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
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
8/11/2019 Chap 3 Gauss Law Slide
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Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
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
8/11/2019 Chap 3 Gauss Law Slide
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Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
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 ==
8/11/2019 Chap 3 Gauss Law Slide
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Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
So the electric field is
r
rL
QE
lL
Q
rlE
qSdE
0
0
0
0
2
2
2
=
=
=
= rr
8/11/2019 Chap 3 Gauss Law Slide
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Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
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
8/11/2019 Chap 3 Gauss Law Slide
32/44
Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
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
8/11/2019 Chap 3 Gauss Law Slide
33/44
Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
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
+=
=
8/11/2019 Chap 3 Gauss Law Slide
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Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
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
8/11/2019 Chap 3 Gauss Law Slide
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Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
Gauss surface at silindrical charge
If the charge is positive
EdA
=
=
=
0
0
0
0cos
qdAE
qEdA
qAdE
rr
8/11/2019 Chap 3 Gauss Law Slide
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Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
If the charge is negative
EdA
=
=
=
0
0
0
0cos
qdAE
qEdA
qAdErr
Gauss surface at silindrical charge
8/11/2019 Chap 3 Gauss Law Slide
37/44
Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
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
8/11/2019 Chap 3 Gauss Law Slide
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Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
Outside silindrical conductor
Enclosing charge is
Qq=
8/11/2019 Chap 3 Gauss Law Slide
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Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
So the electric field is
Lr
QE
QrLE
QdAE
q
AdE
0
0
0
0
2
2
=
=
=
=
rr
8/11/2019 Chap 3 Gauss Law Slide
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Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
Electric field due to positive plane charge
Suppose there is a plane with charge density is
E
E
SSA
Q
q ==
A
S
8/11/2019 Chap 3 Gauss Law Slide
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Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
0
0
0
2
2
=
=
=
E
SSE
qAdErr
ES
ESES
SdESdESdESdE tutupungsetutup
2
0
lub
=++=
++= rrrrrrrr
8/11/2019 Chap 3 Gauss Law Slide
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Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
Electric field due to negative plane charge
Suppose there is a plane with charge density is -
E
E
SSA
Q
q =
=
A
S
8/11/2019 Chap 3 Gauss Law Slide
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Applied Physics (PU 1413) Faculty of Science Telkom Institute of Technology
0
0
0
2
)2(
=
=
=
E
SSE
qAdErr
ES
ESES
SdESdESdESdE tutupungsetutup
2
0
lub
=+=
++= rrrrrrrr
8/11/2019 Chap 3 Gauss Law Slide
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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