# Chapter 25 Electric Potential ∫ ∫ ∫ ⋅

advertisement ```Chapter 25 Electric Potential
conservative forces -> potential energy - What is a conservative force?
Electric potential (V = U / q): the potential energy (U) per unit charge (q) is a function
of the position in space
Goal:
1.
establish the relationship between the electric field and electric potential
2.
3.
calculate the electric potential of various continuous charge distribution
use the electric potential to determine the electric field
25.1 Electric Potential and Potential Difference
r r
dU = − F ⋅ dl
B r
B r
r
r
∆U = − ∫ F ⋅ dl = −q0 ∫ E ⋅ dl (potential energy difference)
A
A
B r
r
U
∆U
V ≡
, ∆V = VB − VA =
= − ∫ E ⋅ dl (electric potential difference)
q0
q0
A
Continuity of V
Electric Field
Electric Potential
V
+
x
+
r
r
+
dV = − E ⋅ dl
The potential function is continuous everywhere.
x
Units
1 _ Volt = 1 _ V = 1 _ J / C , 1 _ N / C = 1 _ V / m
1 _ eV = 1.602 ⋅10 −19 J
300 K = ? eV, visible light: ? eV
25.2 Potential difference in a
uniform electric field
1
B r
r
V B − V A = − ∫ E ⋅ dl = − Ed , ∆U = q∆V = − qEd
A
What is equipotential surface??
Example: The electric field between two parallel
plates of opposite charge
Example: Motion of a proton in a uniform electric
field E = 8.0 x 104 V/m, d = 0.5 m (a) Find the
change in electric potential between the points A
and B. (b) Find the change in potential energy.
25.3 Electric Potential and Potential Energy Due to
Point Charges
r
E=
1
1 q r
q
rˆ =
r
2
4πε 0 r
4πε 0 r 3
rB
1
q
q
q
rˆ ⋅ rˆdr =
−
2
4πε 0 r
4πε 0 rB 4πε 0 rA
rA
VB − V A = − ∫
let rB = r and rA = ∞ --> V =
q
4πε 0 r
and U = q0V =
q0 q
4πε 0 r
Example: Potential Energy of a Hydrogen Atom
(a) What is the electric potential at a distance r = 0.529 X 10-10 m from a proton?
(b) What is the electric potential energy of the electron and the proton at this
separation?
(
)(
)
ke 9 × 109 1.602 × 10 −19
=
= 27.2 _ V
r
0.529 × 10 −10
(b) U = −eV = −27.2 _ eV
(a) V =
Example: Potential Energy of Nuclear-Fission Products
In nuclear fission, a uranium-235 nucleus captures a neutron and splits apart into two
lighter nuclei. Sometimes the two fission products are a barium nucleus (charge 56e)
and a krypton nucleus (charge 36e). Assume that immediately after the split these
2
nuclei are positive point charges separated by r = 14.6 × 10 −15 _ m . Calculate the
potential energy of this two-charge system in electron volts.
U=
(9 ×10 )(56)(1.602 ×10 )(36)(1.602 ×10 ) J = (9 ×10 )(56)(1.602 ×10 )(36) eV
= 199 _ MeV
V =∑
−19
9
qi
4πε 0 ri
−19
−19
9
14.6 × 10 −15
14.6 × 10 −15
-> easier for calculation without consideration of vector addition
Example: Potential Due to Two Point Charges
P1: V =
k (5nC ) k (5nC )
+
0.04
0.04
P2: V =
k (5nC ) k (5nC )
+
0.10
0.06
Example: A point charge q1 is at the origin, and a second point charge q2 is on the
x-axis at x = a. Find the potential everywhere on the x-axis.
kq
kq2
V= 1+
x
x−a
25.4 Obtaining the Value of the Electronic Field from
the Electric Potential
r r
r r
dV
dU = − F ⋅ dl --> dV = − E ⋅ dl --> E = −
dl
The electric field points in the direction in which the potential decrease most rapidly.
(1D? 2D? 3D?)
r r
dV = − E ⋅ dl = − Edl cos θ = − Et dl
E
dl
Et = − dV / dl
If the potential V depends only on x, there will be no change in V for displacements in
the y and z direction.
r r
r
dV ( x ) = − E ⋅ dl = − E ⋅ iˆdx = − E x iˆ + E y ˆj + E z kˆ ⋅ iˆdx = − E x dx
( ) (
)( )
3
--> E x = −
dV
dx
If displacements perpendicular to the radial direction give no change in V,
r r
r
dV = − E ⋅ dl = − E ⋅ (rˆdr ) = − Er dr
--> Er = −
dV
dr
Example: Find the electric field for the electric potential function V given by
V = 100 − 25 x _(V ) .
Example: Potential Due to An Electric Dipole
An electric dipole consists of a positive charge +q on the x-axis at x = +a and a
negative charge –q on the x-axis x = -a. Find the potential on the x-axis for x >> a in
terms of the electric dipole moment p = 2qa.
x > a --> V =
kq
k (− q )
2kqa
+
= 2
x−a x+a
x − a2
x >> a --> V ≈
2kqa kp
= 2
x2
x
r
General Relation Between E and V
Ex = −
dV
dV
dV
& Ey = −
& Ez = −
dx
dz
dy
r
r
 d
dV ˆ dV ˆ dV ˆ
d
d 
E = E xiˆ + E y ˆj + Ez kˆ = −
i−
j−
k = − iˆ + ˆj + kˆ V ≡ −∇V
dx
dy
dz
dy
dz 
 dx
Obtaining electric field from electric potential
r
r
∂
∂
∂
E = −∇V = −iˆ V ( x, y, z ) − ˆj V ( x, y, z ) − kˆ V ( x, y, z )
∂x
∂y
∂z
r
r
∂
1 ∂
1
∂
E = −∇V = −rˆ V (r ,θ , φ ) − θˆ
V (r ,θ ,φ ) − φˆ
V ( r ,θ , φ )
∂r
r ∂θ
r sin θ ∂φ
r
r
∂
1 ∂
∂
E = −∇V = − rˆ V (r ,θ , z ) − θˆ
V (r ,θ , z ) − zˆ V (r ,θ , z )
∂r
r ∂θ
∂z
Potential due to an electric dipole
V=
q
1
−q
r r +
r r
4πε 0 r − r1 4πε 0 r + r1
1
4
V=
=
q
4πε 0
1
(
r 2 + r1 − 2rr1 cos θ
2
1
−
r 2 + r1 + 2rr1 cos θ
2
)
r
q 1
1
1 r
q 2r1 cos θ qd cos θ
p cos θ
(1 − (−2 1 cos θ ) − (1 − 2 1 cos θ )) =
=
=
2
4πε 0 r
2
r
2 r
4πε 0 r
r
4πε 0 r
4πε 0 r 2
r
 ∂
 ∂
E = rˆ − V + θˆ −
 ∂r 
 r∂θ

V = ?

25.5 Electric Potential Due to Continuous Charge
Distributions
dV =
kdq
r
V =∫
kdq
1
=
4πε 0
r
∫
dq
r
V Due to an Infinite Line Charge:
METHOD 1:
1
λdx
V=
∫
4πε 0
d 2 + x2
let x = d tan θ
V=
λ
4πε 0
tan −1
L
d
tan −1
λ
1
1
1
(
+
)d sin θ
2 1 − sin θ 1 + sin θ
−1 0
∫ secθdθ = 4πε ∫
tan −1
0
d
0
tan −1
λ 1  1 + sin θ 
V=
ln
4πε 0 2  1 − sin θ  0
tan
L
d
L
d
=
d
 L + L2 + d 2
λ
ln
4πε 0 
d




METHOD 2:
Obtain E by applying Gauss’s law:
r 4πkλL
2kλ
E=
rˆ =
rˆ
2πrL
r
r_P
VP − Vref = −
∫
r _ ref
 R
2kλ
dr = 2kλ ln P
R
r
 ref




a
V on The Axis of a Charged Ring
r
V =∫
2π
kdq
λds
λadθ
k 2πaλ
= k∫ 2
=
k
=
=
∫
r
x + a2
x2 + a 2
x2 + a2
0
x
P
kQ
x2 + a2
5
E x = −dV / dx = ?
V on The Axis of a Uniformly Charged Disk:
kdq
kσ 2πrdr
kσ 2πrdr
V =∫
=∫
=∫
= 2πkσ
2
2
r
z +r
z2 + r2
0
R
V=
V =
1
4πε 0
σ
2ε 0
(
R
2π
0
0
∫ dr ∫ rdθ
(z
+ r2
2
)
R
0
σ
z2 + r2
z2 + R2 − z
)
for z >> R -->
  R 2 1 / 2

  1 R 2   kπR 2σ kQ
− z =
V ( z ) = 2πkσ  z 1 + 2  − z  = 2πkσ  z 1 +
=
 

z 
2 z 2  
z
z




Example: Find the electric field along z direction.
Ez = −
∂
σ
V =
∂z
2ε 0

1 −




2
2 
z +R 
z
25.6 Electric potential Due to a Charged Conductor
r
E = 0 inside a conductor -> V = const
The conductor is a three dimensional equipotential surface.
The potential V has the same value everywhere on an equipotential surface.
V Due to an Infinite Plane of Charge
+
Q
METHOD 2:
Obtain E by applying Gauss’s law:
E = 2πkσ
x
Vp
Vp
r r
x > 0 --> V = − ∫ E ⋅ dl = − ∫ 2πkσdx = −2πkσx + V0
reff
reff
Vp
Vp
reff
reff
r r
x < 0 --> V = − ∫ E ⋅ dl =
∫ 2πkσdx = 2πkσx + V
0
6
V Inside and Outside a Spherical Shell of Charge
METHOD 2:
Obtain E by applying Gauss’s law:
r >= R , V =
kQ
r
kQ
R
r < R, V =
V for a Uniformly Charged Sphere
(a) r > R
E=
r r
Q
1 Q
ˆ
r
,
E
=
−
E ⋅ rˆdr =
2
∫
4πε 0 r
4πε 0 r
∞
1
(b) r < R
E=
r
Q
4πε 0 R
V (r ) =
rrˆ , V (r ) − V ( R ) = − ∫ E ⋅ rˆdr =
3
R
1 2 1 2
 R − r 
2 
4πε 0 R  2
Q
3
3 2 1 2
 R − r 
2 
4πε 0 R  2
Q
3
Example: A hollow uncharged spherical conducting
shell has an inner radius a and an outer radius b. A
positive charge q is in the cavity, at the center of the
sphere. (a) Find the charge on each surface of the
conductor. (b) Find the potential.
Qa = − q , Qb = + q
r ≥ b, V =
kq
r
b ≥ r ≥ a, V =
a≥r, V =
kq
b
kq kq kq
− +
r
a
b
Example: The two spheres are separated by a
distance much greater than R1 and R2. Find
the charges Q1 and Q2 on the two spheres if
7
the total charge is Q. Find the ratio of the magnitudes of the electric fields at the
surfaces of the spheres.
(1) Q1 + Q2 = Q
kQ1
kQ
R1
= V1 = V2 = 2 --> Q1 =
Q
R1
R2
R1 + R2
kQ1
E
R
R2
(2) 1 = 1 = 2
E2 kQ2
R1
2
R2
A charge is placed on a conductor of nonspherical shape.
V =
kQ k 4πR 2σ
=
= 4πkσR
R
R
σ=
V
4πkR
small R --> Large V
25.7 The Millikan Oil-Drop Experiment
8
25.8 Applications of Electrostatics
The Van de Graaff Generator
The Electrostatic Precipitator
Xerography and Laser Printers
9
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