Electrostatic Electrostatic fields fields
Sandra Cruz-Pol, Ph. Sandra Cruz-Pol, Ph. D.D.INEL 4151INEL 4151ECE UPRMECE UPRMMayagüez, PRMayagüez, PR
Some applicationsSome applications
Power transmission, X rays, lightning protectionPower transmission, X rays, lightning protection Solid-state Electronics: resistors, capacitors, FETSolid-state Electronics: resistors, capacitors, FET Computer peripherals: touch pads, LCD, CRTComputer peripherals: touch pads, LCD, CRT Medicine: electrocardiograms, Medicine: electrocardiograms,
electroencephalograms, monitoring eye activityelectroencephalograms, monitoring eye activity Agriculture: seed sorting, moisture content Agriculture: seed sorting, moisture content
monitoring, spinning cotton, …monitoring, spinning cotton, … Art: spray paintingArt: spray painting ……
We will study Electric We will study Electric charges:charges:
Coulomb's LawCoulomb's Law Gauss’s LawGauss’s Law
Coulomb’s Law (1785)Coulomb’s Law (1785) Force one charge exerts on another Force one charge exerts on another
where where kk= 9 = 9 xx 10 1099
or or kk = 1/4 = 1/4oo
221
R
QkQF
+ +R
Point charges
*Superposition applies
Force with directionForce with direction
12221
12 ˆ4
aR
QQF
o
ExampleExample
3998.1,285.1,004.12.156
4,6,8
81.82
15,15,59
4
13
2
223
1
11
F
rr
rrQQ
rr
rrQQF
x
xx
x
xx
o
Example: Point charges 5nC and -2nC are located at r1=(2,0,4) and r2=(-3,0,5), respectively.
a) Find the force on a 1nC point charge, Qx, located at (1,-3,7)
Apply superposition:
Electric field intensityElectric field intensity
Is the force per unit Is the force per unit charge when placed in charge when placed in the E fieldthe E field
Ro
aR
QE
Q
FE
ˆ4 2
Example: Point charges 5nC and -2nC are located at (2,0,4) and (-3,0,5), respectively.
b) Find the E field at rx=(1,-3,7).
3
2
223
1
11
4
1
rr
rrQ
rr
rrQE
x
x
x
x
o
If we have many chargesIf we have many charges
dvQv
v
Line charge density, Line charge density,
LL
C/mC/m
Surface charge densitySurface charge density
SS
C/mC/m22
Volume charge densityVolume charge density
vv
C/mC/m33
dSQS
S dlQL
L
The total E-field intensity The total E-field intensity isis
Ro
v
Ro
S
Ro
L
aR
dvE
aR
dSE
aR
dlE
ˆ4
ˆ4
ˆ4
2
2
2
Find E from LINE chargeFind E from LINE charge
Line charge Line charge w/uniform charge w/uniform charge density, density, LL
dlQB
A L
x
z
B
A
R
dl
dE
Ro
L aR
dzE ˆ
4
'2
0
T
tan' OTz
zRRR
R
asinacos
sec
'dzdl
(0,0,z’)
(x,y,z)
zR R
Rasinacosa
LINE chargeLINE charge Substituting in: Substituting in:
Ro
L aR
dzE ˆ
4
'2
x
z
B
A
R
dl
dE
0
T
ddz
OTz
]sec0['
tan'2
secR
(0,0,z’)
(x,y,z)zR asinacosa
]asina[cossec4
]sec[22
22
zo
L dE
]a)cos(cosa)sin(sin[4
:Charge Line finite
1212 zo
LE
a2
)90( Charge Line infinite o1,2
o
LE
More Charge More Charge distributionsdistributions
Point chargePoint charge Line chargeLine charge Surface chargeSurface charge Volume chargeVolume charge
Find E from Surface Find E from Surface chargecharge
Sheet of charge Sheet of charge w/uniform density w/uniform density SS
dSdQ S
yzhR a)a(
R
RR
a
dddS
2
322
zρ
4
aa
h
hdddE
o
S
Ro
S aR
dSdE ˆ
4 2
z
SURFACE chargeSURFACE charge Due to Due to SYMMETRYSYMMETRY
the the component component cancels out.cancels out.
0 2322
2
04
h
dhdE
o
Sz
no
SE a2
:Charge Surface infinite
22
24
0h
ho
SzE
More Charge More Charge distributionsdistributions
Point chargePoint charge Line chargeLine charge Surface chargeSurface charge Volume chargeVolume charge
Find E from Volume Find E from Volume chargecharge
sphere of charge sphere of charge w/uniform density, w/uniform density, vv
dvdQ v
x
cos2'
'cos'2'
:cosines of Law
222
222
zRRzr
zrrzR
'''sin
zr
RdRd
.
cos
symmetry toDue
survives
dEdE
only
z
Ro
v aR
dvdE ˆ
4 2
’
’
(r’,’’
P(0,0,z)
v
dE
(Eq. *)
Differentiating (Eq. *)
''''sin'2 drddrdv
Find E from Volume Find E from Volume chargecharge
Substituting… Substituting… zo
vz a
R
dvdE ˆcos
4 2
x
'''sin
zr
RdRd
''''sin'2 drddrdv
’
’
(r’,’’
P(0,0,z)
v
dE
2
222'
'
2
0'
2
0
1
2
''
'''
4 RzR
rRzdr
zr
RdRrdE
rz
rzR
a
ro
vz
ro
zo
v ar
Qa
r
aE ˆ
4ˆ
3 22
3
De donde salen los limites de R?
P.E. 4.5P.E. 4.5 A square plate at plane z=0 and A square plate at plane z=0 and
carries a charge mC/m2 . carries a charge mC/m2 . Find the total charge on the plate and Find the total charge on the plate and the electric field intensity at (0,0,10).the electric field intensity at (0,0,10).
2,2 yx
dyydxdyydxQyxyx
2
0
2
2
2
2
2
2
)2(1212
y12
mC1922
)2(1242
0
2y
ro
s adSr
E ˆ4 2
3'
'
4 rr
rrdS
o
s
)10,','()0,','()10,0,0(' yxyxrr
Cont…Cont…
2/322
2
2
2
2 100
)10,,(
4
12
yx
yxdxdyyE
y o
2
2
2
2
2
22/322
2
2
2
2
2
22/3222/322
6
100
ˆ10
100
ˆ
100
ˆ10108
x
z
x x
yx
yx
adxdyyy
yx
adxdyyy
yx
axdxdyy
sheet of charge
x=2
y=2
z
Due to symmetry only Ez survives:
mMVa
yx
aydxdyE
z
z
/ˆ5.16
100
ˆ10210108
2
2
2
02/322
6
Electric Flux DensityElectric Flux Density
][
:isflux electric Then the
]/[ˆ4
22
CSdD
mCaR
dvED R
vo
D is independent of the medium in which the charge is placed.
Gauss’s LawGauss’s Law
vS
S
venc
enc
S
dvDSdD
SdDdvQ
QSdD
Dv
Gauss’s LawGauss’s Law
The The total electric total electric flux flux ,, through any through any closed surface is closed surface is equal to the equal to the total total charge enclosedcharge enclosed by by that surface.that surface.
v
v
S
enc
Rv
o
dvSdDQ
aR
dvED
ˆ
4 2
Some examples:Some examples: Finding Finding D D at point P from the charges:at point P from the charges:
Point Charge is at the Point Charge is at the origin.origin.
Choose a spherical dSChoose a spherical dS Note where D is Note where D is
perpendicular to this perpendicular to this surface.surface.
24 rDdSDQ r
S
r
S
SdDQ
D
P
r
charge
rar
QD ˆ
4 2
Some examples:Some examples: Finding Finding D D at at point P from the charges:point P from the charges:
Infinite Line ChargeInfinite Line Charge
Choose a cylindrical dSChoose a cylindrical dS Note that integral =0 at Note that integral =0 at
top and bottom surfaces top and bottom surfaces of cylinderof cylinder
S
l SdDQdl
D
P
Line
charge
lDdSDQS
2
aD L ˆ
2
Some examples:Some examples: Find Find DD at at point P from the charges:point P from the charges:
Infinite Sheet of chargeInfinite Sheet of charge
Choose a cylindrical Choose a cylindrical box cutting the sheetbox cutting the sheet
Note that D is parallel to Note that D is parallel to the sides of the box.the sides of the box.
S
s SdDQdS
bottomtop
sS dSdSDQA
zS aD ˆ
2
AADA sS
sheet of charge
D
DArea A
P.EP.E. .
4.74.7
A point charge of 30nC is located at the origin, A point charge of 30nC is located at the origin, while plane y=3 carries charge 10nC/mwhile plane y=3 carries charge 10nC/m22..
Find Find DD at (0, 4, 3) at (0, 4, 3)
ns
rQ aar
QDDD ˆ
2ˆ
4 2
yan
D ˆ2
10)0,0,0()3,4,0(
344
10303
22
9
2
3
9
nC/m ˆ057.0ˆ08.5
ˆ5)3,4,0(54
1030
zy
y
aa
anD
P.E. 4.8P.E. 4.8If C/m2 . If C/m2 . FindFind
:: volume charge density at (-1,0,3)volume charge density at (-1,0,3)
Flux thru the cube defined by Flux thru the cube defined by
Total charge enclosed by the cubeTotal charge enclosed by the cube
zyx axaxyazyD ˆˆ4ˆ2 2
CQ 2
10,10,10 zyx
1
0
1
0
1
0
4 dzdydxxdvQv
venc
3C/m44)3,0,1( xDv
ReviewReview
nS aD ˆ
2
rar
QD ˆ
4 2
aD L ˆ2
Point charge or volume
Charge distribution
Line charge distribution
Sheet charge distribution
We will study Electric We will study Electric charges:charges:
Coulomb's Law (general cases)Coulomb's Law (general cases) Gauss’s Law (symmetrical cases)Gauss’s Law (symmetrical cases) Electric PotentialElectric Potential (uses scalar, (uses scalar,
not vectors)not vectors)
Electric Potential, VElectric Potential, V The work done to move a charge The work done to move a charge QQ from from AA to to B B isis
The (-) means the work is done by an external force.The (-) means the work is done by an external force. The total work= potential energy required in The total work= potential energy required in
moving Q:moving Q:
The energy per unit charge= potential difference The energy per unit charge= potential difference between the 2 points:between the 2 points:
ya dlEQ
dlFdW
B
A
ldEQW
VC
J
B
A
AB ldEQ
WV
V is independent of the path taken.
The The PotentialPotential at any point is the potential at any point is the potential difference between that point and a difference between that point and a chosen reference point at which the chosen reference point at which the potential is zero. (choosing infinity):potential is zero. (choosing infinity):
For many For many Point charges at Point charges at rrkk::
(apply superposition)(apply superposition)
For For Line ChargesLine Charges::
For For Surface chargesSurface charges::
For For Volume chargesVolume charges::
v
v
o rr
dvrrV
'ˆˆ''ˆ
4
1)ˆ(
V ˆˆ
Q
4
1 )(
n
1k
k
ko rr
rV
L
L
o rr
dlrrV
'ˆˆ''ˆ
4
1)ˆ(
V 4
'
1
4 ˆ'ˆ
'4)(
2 r
Q
r
Qadra
r
QldErV
o
r
or
r
ro
r
S
s
o rr
dSrrV
'ˆˆ''ˆ
4
1)ˆ(
P.E. P.E. 4.104.10
A point charge of -4A point charge of -4C is located at (2,-1,3)C is located at (2,-1,3)
A point charge of 5A point charge of 5C is located at (0,4,-2)C is located at (0,4,-2)
A point charge of 3A point charge of 3C is located at the originC is located at the origin
Assume V(∞)=0 and Assume V(∞)=0 and Find the potential at (-1, Find the potential at (-1, 5, 2)5, 2)
Crr
QrV
k ko
k
3
1 4)(
30
3
18
5
46
4
109/1
10)2,5,1(
30)0,0,0()2,5,1(
18)2,4,0()2,5,1(
46)3,1,2()2,5,1(
9
6
3
2
1
V
rr
rr
rr
=10.23 kV
ExampleExampleA line charge of 5nC/m is located on line x=10, y=20A line charge of 5nC/m is located on line x=10, y=20
Assume V(0,0,0)=0 and Assume V(0,0,0)=0 and Find the potential at A(3, 0, 5)Find the potential at A(3, 0, 5)
adaldErV
o
L ˆˆ2
)ˆ(
0=|(0,0,0)-(10,20,0)|=22.36 and A=|(3,0,5)-(10,20,0)|= 21.2
8.40
lnln2
ln2
)ˆ(
A
Aoo
LAorigin
o
L
V
VV
CrV
VA=+4.8V
P.E. P.E. 4.114.11
A point charge of 5nC is located at the A point charge of 5nC is located at the originorigin
V(0,6,-8)=2V and V(0,6,-8)=2V and Find the potential at A(-3, 2, 6)Find the potential at A(-3, 2, 6)
Find the potential at B(1,5,7), the potential difference Find the potential at B(1,5,7), the potential difference VVABAB
Cr
QV
o
4 C
n
r
o
104
52
10)8,6,0()0,0,0(
5.2C
VCn
Vo
A 93.3)0,0,0()6,2,3(4
5
Vn
Vo
B 696.25.2)0,0,0()7,5,1(4
5
VVVV ABAB 233.1
Relation between E and Relation between E and VV
0
ldEVV
VV
BAAB
BAAB
A
B
Esto aplica sólo a campos estáticos.
Significa que no hay trabajo NETO en mover una carga en un paso cerrado donde haya un campo estático E.
0 SdEldES
V is independent of the path taken.
Static E satisfies:Static E satisfies:
dzEdyEdxE
ldEdV
zyx
0 E
dzz
Vdy
y
Vdx
x
VdV
A
BCondition for Conservative field = independent of path of integration
VE
P.E. P.E. 4.124.12
Given that Given that E=(3xE=(3x22+y)a+y)axx +x a +x ayy kV/m, kV/m, find the work donefind the work done in in moving a -2moving a -2C C charge from (0,5,0) to (2,-1,0)charge from (0,5,0) to (2,-1,0) by by taking the straight-line path.taking the straight-line path.
xdydxyxdlEQ
W 23
1
5
2
0
23 xdydxyxQ
W
)3(353 2 dxxdxxxdlEQ
W
2
0
2 563 dxxxQ
W
)1218)(( QW
610128 Q
W
a) (0,5,0)→(2,5,0) →(2,-1,0)
b) y = 5-3xdxdy 3
mJW 12)2(6
mJW 12
ExampleExample Given the potential Given the potential
Find D at . Find D at .
cossin10
2rV
VED oo
0,2
,2
2/ˆ1.22 mCaD r
a
V
ra
ra
rE r ˆsin
10ˆcoscos
10ˆcossin
20333
In spherical coordinates:
aaaED roo ˆ0ˆ0ˆ8
20)0,2/,2(
a
V
ra
V
ra
r
VE r ˆ
sin
1ˆ
1ˆ
Electric DipoleElectric Dipole
Is formed when 2 point charges of equal Is formed when 2 point charges of equal but opposite sign are separated by a small but opposite sign are separated by a small distance.distance.
21
12
21 4
11
4 rr
rrQ
rr
QV
oo
2
cos
4 r
dQV
o
P
y
r1
r2
r
z
d
Q+
Q-
For far away observation points (r>>d):
Energy Density in Energy Density in Electrostatic fieldsElectrostatic fields
It can be shown It can be shown that the total that the total electric work done electric work done is: is:
v
o
v
E dvEdvEDW 2
22
1
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