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Material Type: Notes; Professor: Ucer; Class: General Physics II; Subject: Physics; University: Wake Forest University; Term: Unknown 1989;
Typology: Study notes
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dx F
xf xi
s^
q
0
s E s F E F v v v v v v d
q
d
U
q
path
path
Δ
=
.
.^
0
0
…but
is a conservative force
Δ
B A
d
q
U
s
E
w
v
.
0
…so the path we takedoes not matter
q
0
Potential Energy Electric Field
C J
V
=
m
N
J
⋅
=
m
V
m N
J
J C V
N C
C
N
= ⎞ ⎟ ⎠
⎛ ⎜ ⎝ ⎞ ⎟ ⎠
⎛ ⎜ ⎝
=
.
.
J C J C V e
eV
19
19
−
−^
Electron-Volt
Electric Potential in a Uniform Field
∫
∫
∫
B A
B A
B A
ds
E
ds
d
cos
s
E
Ed
V
Δ
Ed q
U
0
Δ
Electric field lines point to decreasingpotential. A positive charge will lose potential energyand gain kinetic energy when moving in thedirection of the field.
Point Charge
Electric Dipole
∫ − =
−
B A
A
B^
d
V
V
s E
.
dr q r k
ds q r k
d
q r k
d
e
e
e^
2
2
2
cos
ˆ.
.^
=
=
=
θ
s r
s E
⎤ ⎥ ⎦
⎡^ ⎢ ⎣
−
=
−
−
∫ ∫
A
B
e
r r e
A
B
r
A
B
r
r q k
dr r q k
V
V
dr E
V
V
B A
1
1
2
∞= A V
q^ r
k
V
e
=
12
2 q (^1) r q
k
U
e
=
⎞ ⎟ ⎟ ⎠
⎛ ⎜⎜ ⎝
=
31
1 3
23
3 2
12
2 1
r
q q
r
q q
r
q q
k
U
e
distance
charge –
potential energy of which test charge is greater:
2
e
(x >> a)
3
e
x^
2
2
e
P
2
2 2
x
a
qx k
V
e
P^
−
=
⎞ ⎟ ⎟⎠
⎛ ⎜ ⎜ ⎝
2 2
2
2
2
2
x
a
x
a
q k
dVdx
E
e
x
Between the Charges
Electric Potential Due toContinuous Charge Distributions
Start with an infinitesimalcharge,
dq
. dq^ r
k
dV
e
=
Then integrate over thewhole distribution
∫
e
Electric Potential Due toa Finite Line of Charge
e
2
2
Electric Potential Due toa Uniformly Charged Sphere
2 Q r k
E
e
r^
=
∫
∫^
∞
∞
r e
r
r
B^
dr r Q k
dr E
V
2
Q^ r k
V
e
B^
=
Q^ R k
V
e
C^
=
r R Q k
E
e
r^
3
=
(^
2 )
2 3 2
r
R Qk R
dr E
V
V
e
r R
r
C
D^
−
=
−
⎞⎟ ⎟⎠
⎛^ ⎜⎜⎝
−
=
2 2
3
2
r R
Q R k
V
e
D
r > R
r < R
Connected Charged ConductingSpheres
2 1
1 2
r^ r
E E
=
1 2
1 2
r r
q q
=
A
B
We can always find apath where E.
s
is
non-zero. But, since
paths, E must be zeroeverywhere in the cavity.
A cavity without any charges enclosed by a
conducting wall is field free.
