1 1 1.6 10 ev e v j =⋅ = × f qocw.nctu.edu.tw/course/physics/physics2_lecturenotes/970317.pdf ·...
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Electric potential energy:
.( : : ) :
.f i
Electrostatic force does work on a particleW F s qE sPotential energy i initial state f final state
U U U W
•
= ⋅ = ⋅•Δ = − = −
v vv v
Electric potential : : arg .
( ) : ( ) , .( )
UV potential energy per unit ch eq
J Joule N N m J VUnit V volt EC Coul C C m C m m
Δ• Δ ≡
⋅= = = = =
⋅ ⋅
E
f
i
q
191 1 1.6 10eV e V J−= ⋅ = ×
Work done by applied force .appW -W• =
sourceq testqE♁ ♁
000app
WU
W
>Δ <
<
000app
WU
W
<Δ >
>
♁
sourceq testqE
000app
WU
W
<Δ >
>
000app
WU
W
>Δ <
<
In chapter 8 we defined theassociated with a conservative force as the negative value of the work that the force must do on a particle to take it from an initial posi
U
W
Δchange in potential energy
tion to a final position .
( )f
i
i f
x
f ix
x x
U U U W F x dxΔ = − = − = − ∫
O
x. . .xi x xf
F(x)
( )f
i
x
x
U F x dxΔ = − ∫
(24 - 2)
Electric Potential Energy :
A
B
Consider an electric charge moving from an initial position at point A to a final position at point B under the influence of a
known electric field . The force exerted on the charge is:
o
o
q
E F q E=r r r
f f
oi i
U F ds q E dsΔ = − ⋅ = − ⋅∫ ∫r rr r f
oi
U q E dsΔ = − ⋅∫r r
(24 - 2)
O
Definition of voltage :
Units of : J/C known as the "Volt"
Consider a pont charge placed at the origin.We will use the defini
o
Wq
q
V
V
Δ = −
SI Units of
Potential due to a point charge
V :
P
tion given in the previous page to determine the potential at point P a distnce fro m O.
VR
14P
o
qVRπε
=
(24 - 4)
O
2
22
cos0
The electric field generated by is: 4
4
1 14
1
4
R
PR R
o
Po R
PRo o
V E ds Edr Edr
qq Er
q d drVr
q qVr R
rx x
πε
πε
πε πε
∞ ∞
∞
∞
∞
= − ⋅ = =
=
=
⎡ ⎤→ = − =⎢ ⎥⎣ ⎦
= −
∫ ∫ ∫
∫∫
r r
14P
o
qVRπε
=
(24 - 4)
V=V0
V=V0
Ev
Ev
tan
tan tan
Equi-potential surface
0
( , 0.
00 0)
Calculating V from E:
f f
i i
U WVq q
E Equipotential surfaceIf not E
q E FF S W V
dW F ds qE ds
W dW qE ds q E ds
V
•Δ
Δ = = − =
⇒ ⊥∃ ≠
⇒ ⋅ = ≠
⇒ ⋅ = ≠ ⇒ Δ ≠
•
= ⋅ = ⋅
= = ⋅ = ⋅
Δ = −
∫ ∫ ∫
v
v vv w
v vv v
cosf f
f ii i
W E ds V V E dsq
θ= − ⋅ = − = −∫ ∫v r
1 21
21
2
1 2
0. (V , 0)
V V V
E ds
E ds
EdV W
Δ = −
= − ⋅
= −
= − << >
∫∫
v vEX.
2
1
+
_d
q⊕E ds
2
1
+
_
dq⊕
E ds
2 12
12
12 2
1 1
2 1
cos180
0
( , 0)
V V V
E ds
E ds
Eds E ds Ed
V V W
Δ = −
= − ⋅
= − °
= = = >
> <
∫∫∫ ∫
v v
e-
+V~0
+V~20kV
v=?
2
19 2
7
12
1( 1.6 10 ) 20218.4 10 /4
e
W U K q V mv
k m v
v m s c
−
= −Δ = Δ ⇒ − ⋅Δ =
− − × ⋅ =
⇒ = × ≈
EX.
Potential of group of point charges•
2R 3R1R
2q 3q1q
?V =
0 i
i i
k qVR⋅
=∑
Potential of continuous charge distribution•
rQΔ
Q
Ο
?=V0
0
0 0
0
0 limQ Q
k QVrk QV
rk Q k dQV
r rΔ→
ΔΔ =
Δ=
ΔΔ→ ⇒ = =
∑
∑ ∫
r1r2
r3
P
q1
q3
q2
Consider the group of three point charges shown in the figure. The potential generated by this group at any point P is calculated using the principle of superposition
We determine th
V
1. 1 2 3
1 2 31 2 3
1 2 3
e potentials , and generated by each charge at point P.
1 1 1 , , 4 4 4o o o
V , V V
q q qV V Vr r rπε πε πε
= = =
(24 - 5)1 2 3V V V V= + +
Potential due to a group of point charges
r1r2
r3
P
q1
q3
q2
1 2 3
1 2 3
1 2
11 2
1 2 3
We add the three terms:
The previous equation can be generalized for charge
1 1 1 4 4 4
1 1 1 1
s as follo
...4 4 4
w
4
s:
o o o
nn i
o o o n o i
q qV V V V
n
qVr r r
q q q qVr r r r
πε πε πε
πε πε πε πε
= + +
= + + + =
= + +
∑
2.
(24 - 5)1 2 3V V V V= + +
EX. Potential for uniformly charged thin ring
Qr z
R
φddQ
?)( =ZV 0
20
0
200
0
0
2 2
, 2 2
2
2
Q
k dQ Q QV dQ Rd dr R
k Q dr
k Q dr
k Qr
k QR z
π
π
ϕ ϕπ π
ϕπ
ϕπ
= = ⋅ =∫
= ⋅∫
= ∫
=
=+
A
B
C
pr Consider the electric dipole shown in the figure.We will determine the electric potential created at point P by the two charges of the dipole using superposition.Point P is at a distance from the
V
r
( ) ( )( ) ( )
( ) ( ) ( ) ( )
center O of the dipole.Line OP makes an angle with the dipole axis
14 4o o
r rq q qV V Vr r r r
θ
πε πε− +
+ −+ − − +
⎛ ⎞ −= + = − =⎜ ⎟⎜ ⎟
⎝ ⎠
(24 - 6)
Example :
Potential due to an electric dipole
A
B
C
pr( ) ( )
2( ) ( ) 2 2
We assume that where is the charge separation. From triangle ABC we have: cos
cos 1 cosAlso: 4 4
where the electric dipole moment.o o
r d d
r r d
q d pr r r Vr r
p qd
θ
θ θπε πε
− +
− +
− ≈
≈ → ≈ =
= =
?
(24 - 6)
2
1 cos4 o
pVr
θπε
=
:V from E gDeterminin•
sdρ
⊕q
Eρ
dVsdEqsdFdW ρρρρ
⋅=⋅=
dVsdEq
dW−=⋅=
ρρ
) Calculus. of principle lFundamenta
(
path toparallel field ,||
||
||
∫−=−⇔
−=
dsEVV
dsdVEdsEif
if
kEjEiEE zyxˆˆˆ
general,In
++=ρ
,xVEx ∂∂
−= ,yVEy ∂∂
−= ,zVEz ∂∂
−= .)ˆˆˆ()ˆˆˆ( Vz
ky
jx
ikzVj
yVi
xVE
∂∂
+∂∂
+∂∂
−=∂∂
+∂∂
+∂∂
−=ρ
,ˆˆˆ)(
kz
jy
ixdel ∂
∂+
∂∂
+∂∂
≡∇ .VE −∇=ρ
EX. Check field for single charge
rqkV 0=
r
q
20
|| rqk
drdV
dsdVE =−=−=
rr
qkE ˆ
20=
ρ
EX. Check filed for uniformly charged thin ringknown
Q
V 0=∂∂
−=xVEx 0=
∂∂
−=yVEy
....=−=∂∂
−=dzdV
zVEz
dqO A
2 2
Potential created by a line of charge oflength L and uniform linear charge densityλ at point P. Consider the charge element
at point A, a distance from O.From triangle OAP we have:
dq dx x
r d x
λ=
= + Here is the distance OPThe potential created by at P is:
ddV dq
(24 - 8)
Example :
dqO A
(24 - 8)
( )( )( )
2
2 2
2 20
2 2
0
2
2
2
2
2
1 14 4
4
ln
l
4
n ln4
n
l
o o
L
o
L
o
o
dx x d
dq dxdVr d x
dxVd x
V x d x
V x
d
L
x
L
x
d
λπε πε
λπε
λπελπε
= =+
=+
⎡ ⎤= + +⎢ ⎥⎣ ⎦
⎡ ⎤= +
= + +
+ −⎢ ⎥⎣ ⎦
+∫
∫
Lets assume that the electric field forms an angle with thepath . The work done by the electric field is:
cos cosWe also know that 0. Thus: cos 0
0, 0, 0
Er
W F r F r qE rW qE r
q E r
θ
θ θθ
Δ
= ⋅Δ = Δ = Δ= Δ =
≠ ≠ Δ ≠
r
r
r r
Thus: cos 0 The correct picture is shown in the figure belo
9w
0θθ = → = °
(24 - 11)
SV
Er
Examples of equipotential surfaces and the corresponding electric field lines
Uniform electric field Isolated point charge Electric dipole
(24 - 12)
A
B
V V+dV
Now we will tackle the reverse problem
i.e. determine if we know . Consider two equipotential surfaces thatcorrspond to the values and separated by a distance as shown in the figure.
E V
V V dVds
+
r
Consider an arbitrary direction represented by the vector . We will allow the electric field to movea charge from the equipotenbtial surface to the surface .
o
ds
qV V dV+
r
(24 - 13)
E VCalculating the electric field from the potential r
The work done by the electric field is given by: (eqs.1)
also cos cos (eqs.2)If we compare these two equations we have:
cos cos
From triangle PAB we see that
o
o
o o
W q dVW Fds Eq ds
dVEq ds q dV Eds
θ θ
θ θ
= −= =
= − → = −
cos is the component of along the direction s.
Thus:
s
s
EE E
VEs
θ
∂= −
∂
r
sVEs
∂= −
∂(24 - 13)
A
B
V V+dV
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