Electric potential §8-5 Electric potential Electrostatic field does work for moving charge --E-field possesses energy 1.Work done by electrostatic force.
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§8-5 Electric potential Electric potential
Electrostatic field does work for moving charge--E-field possesses energy
1.Work done by electrostatic force Let test charge q0 moves a b along arbitrary path in the E-field set up by point charge q .
The work done by electrostatic force=?
c
a
b
q
ar
br
q0
ldFdW
ldEq
0
dlEq cos0
Edrq0
ld
: displacement
ld
c
a
b
q
ar
br
dr
r E
b
aoab EdrqW
b
a
r
rdr
r
qq2
0
0 1
4
)11
(4 0
0
ba rr
The work depends on only the initial position and final position of qo, and has nothing to do with the path.
c
a
b
q
ar
br
q0
b
aab ldEqW
0
ibia
n
i
i
rr
qq 11
41 0
0
b
a n ldEEEq
)( 210
---has nothing to do with path
When qo moves in the E-field set up by charges’ system q1, q2, qn ,
n
iiW
1
When qo moves in the E-field set up by charged body,
Conclusion:-- the work has nothing to do with path
2. Circular theorem of electrostatic field
When q0 moves along a closed path L ,
00 LLldEqdWW
Electrostatic force is conservative force.
E-force does work:
3. Electric potential energy
b
a
ab ldEqW
0
-- The E-potential energy when q0 at point a and b.
PbPa EE
PbEPaE 、
Electrostatic field is conservative field.
The work done by electrostatic force = the decrease of the electric potential energy
0 ldEL
Circular theorem of electrostatic field
q0 moves in E-field a b ,
Notes
(1) EP a is relative quantity. If we want to decide the
magnitude of EP when q0 at a point , we must choose zero reference point of E-potential energy.
PaE a
ldEq
0
Z
PEW
The work done by E-force for q0=- increment of E-potential energy.
Z--zero E-potential energy pointZ 0PE
The choice of zero E-potential energy point : Choose zero point at when the charge distr
ibution is finite.
Choose zero point at the finite distance point w
hen the charge distribution is infinite.
(2) EP is scalar. It can be positive, negative or zero.
0q(3) EP depends on E-field and , it belongs the system.
0q
EU Pa
a
4. Electric potential
a
ldEZ
Definition
b
a
ldE
E-potential difference:
0q
EEUUV PbPa
baab
)(0 baab UUqW
--Describe the character of E-field.
Work:
5.Calculation of E-potential(1) The E-potential of a point charge q
aa ldEU
r
drr
q2
04
1
a
Edr
r
q
04
1
Take U=0 , then the E-potential of a point a :
The distance from q to a
DiscussionU
r+
U
r
If q>0 , U>0 for any point in the space.
when r , U U() 0
If q<0 , U<0 for any point in the space.
when r , U U() 0
r
1
aa ldEU
a naa
ldEldEldE
21
nUUU 21
n
iiU
1
n
i i
i
r
q
1 04
( 2 ) The E-potential of a system
The system of point charges : q1,q2,,qn
qi
a
ri
q1
r1
--superposition principle of E-potential
dq
ar
( 3 ) The E-potential of charged body
Divide q many of dq
For any dq :
r
dqdU
04
1
For entire charged body :
charge element
r
dqdUUa
04
1
Integrating for charged body
Caution !!
This method can be used for the finite distribution charged body.
[Example 1] Four point charges q1 = q2 = q3 = q4
= q is put on the vertexes of a square with edge of a respectively.
Calculate (1)The E-potential at point 0. (2)If test charge q0 is moved from to 0, how much work does the E-force do?
2
ar
1q 2q
3q4q
0a
a
a
a
6. Examples of calculating E-potential
a
qa
q
r
qUU
ii
00
0
4
10
2
2
1
4
14
( 1 )
a
qqUqEEW PP
0
0000
2
( 2 )
Definition method
a
a ldEUZ
Integrating for path
Calculate the E-potential set up by a charged body
Two methods
Use the definition of E-potential as the distribution of is known. – definition methodE
Use the superposition principle of E-potential-- superposition method.
Integrating for charged body
r
dqU
04q
superposition method
xPxrR
0
[Example 2] Calculate the E-potential on the axis of a uniform charged ring. q 、 R are known.Solution
Method Use the E-field distribution of the ring that was calculated before.
23
220 )(4 Rx
qxE
The direction: along x axis
P
ldEU
xdx
xR
qx2/32
0 )(4
x
dxE
2204 xR
q
r
dqdU
04
1
Method -- superposition method
q dq
dUU
2204
1
xR
q
qdq
r04
1
xPxrR
0
U
x0
Discussion
(1) ox
R
qU
00 4
1
Rx(2) x
qU
04
1
If the charged body is a half circle, ?0U0
R
R
+
++++
+
++
q
[Example 3] Calculate the E-potential distribution of a charged spherical surface
E-field distribution:
E0
204
1
r
q
Rr( )
Rr( )
Solution
Definition method
Zero potential point :
R
+
++++
+
++
q
P
ldEU
drr
q
r2
04
1
(1) For any point P outside the sphere
r
q
04
1
Rr( )
Pr
(2) For any point inside the sphere surface
+
++++ q
++
+
+
++
+
R
P
ldEU
R
r
dr0
P
rdr
q
R2
04
1
R
q
04
1
The sphere is equipotential.
Rr
分段积分
Conclusion
R
q
04
1
Rr( )
r
q
04
1
Rr( )
U
The distribution curve of E-potential
The distribution curve of E-field : E
R
R r
rO
O
88
r2
r
1
1
superposition method :
Integrating for charged body
r
dqU
04q
It’s very complex !!
Conclusion
When the E-field distribution is symmetry and it can be calculated by using Gauss’s Law conveniently, it is simpler to calculate potential by using definition method.
When the E-field distribution is not symmetry an
d it can’t be calculated by using Gauss’s Law conveniently, it is simpler to calculate potential
by using superposition method.
P
[Example 4] Calculate the E-potential distribution of an infinite line with uniform charge(the linear density isλ).
Solution
Use definition method
rE
02
pr
How to choose the zero potential point?
Choose any point b as zero potential point
b
PP ldEU
When rp<rb , U >0. When rp>rb , U <0.
b
P
P
b
r
rln
2 0
Finite distance to the charged line
§8-6 Equipotential surface and Potential gradient
1. Equipotential surface
--the potential has the same value at all points on the surface.
Positive point chargeElectric dipole
Dash line-- equipotential surface
Real line-- -lineE
A parallel plate capacitor
++ ++++ +++
0)(0 baab UUqW
Prove : Assume q0 moves along equipotential surface a b ,then :
The properties of equipotential surfaces
No net work is done by the E-field as a charge moves between any two points on the same equipotential surface.
ab
q0
ldEqdW
0
Prove :
0
ldE
EAssume on an equipotential surface, the field
at point P is
then :
q0 moves along equipotential surface, ld
q0Pld
E
-lines are always normal to equipotential surfacesE
b
aba ldEUU
E
a b
Prove :Assume there are two equipotential surface U a , U b
then
ld
b
adlE 0
ba UU
-line points on the direction of the increase of the potential.
E
Ua Ub
E
Uc
r2
r1
the density of equipotential surfaces shows the magnitude of E-field.Prove :
Assume there is a family of equipotential surfacesUa 、 U b 、 Uc 、 E1
E2
then :11 rEUU ba
22 rEUU cb
1
2
2
1
r
r
E
E
2. Potential gradient
Equipotential surface
P
P ldEU
-lineE
If the distribution of U is known, How can we calculate ? E
(1) Special example : uniform field
unit positive charge ab, the work done by E-force:
ba UUW nE
nEU
n
UE
E
C
l
na b
nE
Express that the ’s direction is the direction of U decrease.
E
q=1
q=1 moves ac , the work done by E-force :
lE cosca UUW
lEU l
l
UEl
lEl
’s component at the direction of l
E
’s component at any direction = the negative magnitude of the rate of U change with distance on that direction.
E
(2) Any E-field:
U
dUU 0dU
nd
ld
dl
dUEl
dn
dUE
definition :potential gradient
grad U ndn
dU
ndn
dUE
E
grad U
n
The normal direction of equipotential surface,point on U increasing
= U
U
In Cartesian coordinate system
x
UE x
y
UE y
z
UEz
E
)( kz
Uj
y
Ui
x
U
a.the magnitude of at any point = the maximum of the rate of U change with distance on that point. the direction of is perpendicular to the equipotential surface at that point and along the the direction of U decreasing.
E
E
b. In the space that U is constant, U = 0 , E = 0
c. E is not sure = 0 in the space of U= 0.
U is not sure = 0 in the space of E= 0.
conclusion
[Example 1] Calculate the E-field of an electric dipole.
l
rr
r
x
y
q q
P( )x y
r
q
r
qU
o
o
4
1
4
1
2rrr coslrr
24
cos
r
pU
o
21
22 yx
x
r
xcos
23
220 )(4 yx
pxU
25
220
22
)(4
)2(
yx
yxp
x
UE x
25
220 )(4
3
yx
pxy
y
UE y
E
jEiE yx
R
P xx0
[example 2] calculate the E-field on the axis of the
round charged plate using its potential gradient .
Solution The potential of any ring on the axis:
r
dqdU
04
1
dsdq d2
22 xr 2202 x
dd
R
P xx0
dr
Ed
xxRU 22
02
i.e., the potential on the axis: xUU
idx
dUE
220
12 xR
x
R
x
ddUU
0 2202
§8-7 E-Force Exerted on a Charge§8-7 E-Force Exerted on a Charge
1. The E-force exerted on a charged particle
The torque exerted on the dipole:
EqF
EPM e
sinlFM
F
F
q
+q
l
sin l qE sinEPe
lqPe
--electric dipole moment
(1) E-dipole in the uniform field
The E-dipole will rotate under the action of the torque until the direction of is the same with E
eP
F
F
q
+q
l
Assume the potential energy of E-dipole is zero at the position =900 Ep (=900) =0
Then the potential energy of E-dipole at any orientation is
00
90)90(sin dEPWE ep
cosEPe
EPE ep
(2) E-dipole in non-uniform field
21 FFF
2211 sin2
sin2
ee rF
rFM
)( 21 EEq
ee
ee r
EEP
r
EEqr 2121
F1
F2
q
+q
1
l
2
0 --Move to the area of larger E-field
0 --rotating
2. A moving charged particle in a uniform E-field
EqF
m
qEa x
m
qExavv 222
02
qUmvmv 20
2
2
1
2
1
Ev
//)1( E
vq
td
vdmam
Ex=U
22
2
1
2
1t
m
qEatx
20
2
2
1
v
yE
m
qx
tvy 0
Ev 0)2(
E
0v
q x
yv
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