Electric Potential Energy - Wake Forest...
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Transcript of Electric Potential Energy - Wake Forest...
Electric PotentialFlashback to PHY113…
Work done by a conservative force is independent of the path of the object.Gravity and elastic forces are examples.This leads to the concept of potential energy and helps us avoid tackling problems using only forces.
Electrostatic forces are also conservative.
UdxFWf
i
x
xs Δ−== ∫
Ev
A
Electric Potential Energy
q0
sEsF
EFvvvv
vv
dqdU
q
pathpath∫∫ −=−=Δ
=
.. 0
0
…but F is a conservative force
∫−=ΔB
A
dqU sE wv.0
…so the path we take does not matter
B
q0
Electric PotentialWork, ΔU, is dependent on the magnitude of the test charge, q0.We’d like to have a quantity independent of the test charge and only an attribute of the electric field.
0qUV =
Electric Potential(J/C=V)
∫−=Δ
=ΔB
A
dqUV sE vv
.0
Potential Difference between A and B
∫∞
−=P
P dV sE.
Electric Potential at P
Unit Pit Stop
Potential
Energy
Electric Field
CJV =
mNJ ⋅=
mVmN
JJCV
CNCN =⎟
⎠⎞
⎜⎝⎛⎟⎠⎞
⎜⎝⎛=
..
( ) ( )( ) JCJCVeeV 1919 106.11106.111 −− ×=×==Electron-Volt
Electric Potential in a Uniform Field
∫ ∫∫ −=°−=−=ΔB
A
B
A
B
A
dsEdsEdV 0cos. sE
EdV −=Δ EdqU 0−=Δ
Electric field lines point to decreasing potential.
A positive charge will lose potential energy and gain kinetic energy when moving in the direction of the field.
Equipotential Surfaces
B
A
s
E
Cθ
sEsEsE ... −=−=−=Δ ∫∫B
A
B
A
ddV
θcos90cos0cos
)()(
EsVEsEsV
V
CBAC
CBAC
−=Δ°−°−=Δ
−−=Δ E.sE.s
θcos0cos
)()(
0
0
00
EsqUEsqU
qqU
AC
CBAC
−=Δ°−=Δ
−−=Δ E.sE.s
D
θcos)()(... EsddV CDAC
D
A
D
AAD −=−−=−=−=−=Δ ∫∫ E.sE.ssEsEsE
No work is done moving a charged particle perpendicular to a field (along equipotential surfaces)
Equipotential Surfaces
Uniform Field Point Charge Electric Dipole
Electric Potential of a Point Charge
∫−=−B
AAB dVV sE.
drrqkds
rqkd
rqkd eee 222 cos.ˆ. === θsrsE
⎥⎦
⎤⎢⎣
⎡−=−=−
−=−
∫
∫
ABe
r
reAB
rAB
rrqk
rdrqkVV
drEVVB
A
112
0=∞=AV
rqkV e=
Two Point Charges
⎟⎟⎠
⎞⎜⎜⎝
⎛+=
2
2
1
1
rq
rqkV eP
A System of Point Charges
12
21
rqqkU e=
⎟⎟⎠
⎞⎜⎜⎝
⎛++=
31
13
23
32
12
21
rqq
rqq
rqqkU e
Concept QuestionTwo test charges are brought separately into the vicinity of a charge +Q. First, test charge +q is brought to a point a distance r from +Q. Then this charge is removed and test charge –q is brought to the same point. The electrostatic potential energy of which test charge is greater:
1. +q2. –q3. It is the same for both.
Getting From V to E
∫−=B
A
dV sE.
sE ddV .−=
dxdVEx −=E
E
drdVEr −=
In general:xVEx ∂∂
−=zVEz ∂∂
−=yVEy ∂∂
−=
Electric Potential of a Dipole
22
xqakV e≈ (x >> a)
3
4xqak
dxdVE e
x =−=
222
axqakV e
P −=
222
xaqxkV e
P −=
( ) ⎟⎟⎠
⎞⎜⎜⎝
⎛
−
+−=−= 222
22
2xaxaqk
dxdVE ex
Between the Charges
Electric Potential Due to Continuous Charge Distributions
Start with an infinitesimal charge, dq.
rdqkdV e=
Then integrate over the whole distribution
∫=r
dqkV e
Electric Potential Due to a Uniformly Charged Ring
22 axQkV e
+=
( ) 2/322 axQxkE e
+=
Electric Potential Due to a Finite Line of Charge
⎟⎟⎠
⎞⎜⎜⎝
⎛ ++=
aall
lQkV e
22
ln
Electric Potential Due to a Uniformly Charged Sphere
2rQkE er =
∫∫∞∞
−=−=r
e
r
rB rdrQkdrEV 2
rQkV eB =
RQkV eC =r
RQkE e
r 3=
( )2232
rRRQkdrEVV e
r
RrCD −=−=− ∫
⎟⎟⎠
⎞⎜⎜⎝
⎛−= 2
2
32 R
rRQkV e
D
r > R
r < R
Potential Due to a Charged Conductor
Charges always reside at the outer surface of the conductor.The field lines are always perpendicular to surface.Then E.ds=0 on the surface at any point.Which means, VB-VA=0along the surface.The surface is an equipotential surface.Finally, since E=0 inside the conductor, the potential V is constant and equal to the surface value.
Connected Charged Conducting Spheres
1
2
2
1
rr
EE
=
2
1
2
1
rr
=
Cavity Within a Conductor
0. =−=− ∫B
AAB dVV sE
We can always find a path where E.ds
is
non-zero.But, since ΔV=0 for all paths, E must be zero everywhere in the cavity.
A cavity without any charges enclosed by a conducting wall is field free.
SummaryCharges can have different electric potential energy at different points in an electric field.Electric potential is the electric potential energy per unit charge.All points inside a conductor are at the same potential.
For Next ClassReading Assignment
Chapter 26 – Capacitance and Dielectrics
WebAssign: Assignment 3