Download Electrostatic Potential Energy Example

1/30/12
QUIZ 6 January 26, 2012
Chapter 24
•  Electrostatic Potential Energy of a system of fixed
point charges is equal to the work that must be
done by an external agent to assemble the
system, bringing each charge in from an infinite
distance.
An electron moves a distance of 1.5 m through a
region where the electric field E is constant and
parallel to the displacement. The electron’s potential
energy increases by 3.2×10-19 J. What is the
magnitude of E ?
a)
1.3
N/C
b)
5.0×
10-19
N/C
c)
2.0
N/C
d)
1.0×10-19
N/C
e)
7.5×10-1
N/C
1/30/12
Point 1
Point 2
q1
e = 1.6×10-19 C
k = 8.99×109 Nm2/C2
ε0 = 8.85×10-12 C2/(Nm2)
1
Example: Electrostatic Potential Energy
Point 1
Point 2
q1
q2
1/30/12
2
Example: Three Point Charges
q3
Point 3
r1,3
Point 1
q1
r2,3
q2
r1,2
Point 2
If q1 & q2 have the same sign, U is positive because positive work by an
external agent must be done to push against their mutual repulsion.
If q1 & q2 have opposite signs, U is negative because negative work by an
external agent must be done to work against their mutual attraction.
1/30/12
3
1/30/12
4
1
1/30/12
Electrostatic Potential Energy
Calculate Electric Field from the Potential
Electric field always points in the direction of
steepest descent of V (steepest slope) and tis
magnitude is the slope.
•  We can conclude that the total work
required to assemble the three charges is
the electrostatic potential energy U of
the system of three point charges:
U = Wtotal = W2 + W3 =
Potential from a Negative
Point Charge
kq2 q1 kq3q1 kq3q2
+
+
r1, 2
r1,3
r2,3
V(r )
x
The electrostatic potential energy of a
system of point charges is the work needed
to bring the charges from an infinite
separation to their final position
1/30/12
y
y
5
Calculating the Electric Field from the Potential
Field


⎛ ∂V ˆ ∂V
∂V
E = −∇V = − ⎜
i+
ĵ +
⎝ ∂x
∂y
∂z
Ex = −
1/30/12
Potential from a Positve
Point Charge
1/30/12
-V(r )
x
6
Example: Calculating the Electric Field from the
Potential Field
What is the electric field at any point on the central
axis of a uniformly charged disk given the potential?
⎞
k̂ ⎟
⎠
∂V
∂V
∂V
, Ey = −
, and Ez = −
∂x
∂y
∂z
7
1/30/12
8
2
1/30/12
Potential due to a Group of Point Charges
r1
q1
X
r2
q2
N
V (r) = ∑ Vn (r) =
n =1
Potential from a Continuous Charge Distribution
r3
q3
r4
q4
1 N qn
∑
4πε o n =1 rn
1/30/12
9
Charge Densities
total charge
Q
1/30/12
10
Calculate Potential on the central axis of a
charged ring
small pieces
of charge
dq
Line
of
charge:
= charge per unit length [C/m]
dq
=
 dx
Surface of charge:   = charge per unit area [C/m2]
dq =
dA
Cylinder: dq = σ rdθ dz
Sphere:
dq = σ r 2 sin θ dθ dφ
Volume of Charge:  = charge per unit volume [C/m3]
dq = dV
Cylinder: dq = ρrdrdθ dz
Sphere:
1/30/12
dq = ρr 2 dr sin θ dθ dφ
11
1/30/12
12
3
1/30/12
Calculate Potential on the central axis of a
charged disk
Calculate Potential on the central axis of a
charged disk (another way)

1/30/12

13
Calculate Potential due to an infinite sheet
1/30/12
E due to an infinite line charge
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
1/30/12
14
Corona discharge around a high voltage power line,
which roughly indicates the electric field lines.
15
1/30/12
16
4
1/30/12
Equipotentials
Equipotentials: Examples
Definition: locus of points with the same potential.
• General Property: The electric field is always
perpendicular to an equipotential surface.
Point charge
1/30/12
17
Equipotential Lines on a Metal Surface
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
–
–
–
–
–
–
–
–
–
–
–
–
–
Locally E⊥ =
electric dipole
1/30/12
18
Potential inside & outside a conducting sphere
σ
ε0
Gauss: E|| = 0
at electrostatic
equilibrium
Vref = 0
in electrostatic equilibrium
all of this metal is an equipotential;
i.e., it is all at the same voltage
1/30/12
infinite positive
charge sheet
19
at r = ∞.
Common mistake: thinking that potential must be zero
inside because electric field is zero inside.
1/30/12
20
5
1/30/12
Quiz 7 Jaunary 31, 2012
Summary
The graph gives the electric potential V as a function of
x. Rank the five regions according to the magnitude of
the x-component of the electric field within them,
greatest first.
(A)  4 > 2 > 3 > 1 > 5
•  If you know the functional behavior of the potential V
at any point, you can calculate the electric field.
•  The electric potential for a continuous charge
distribution can be calculated by breaking the
distribution into tiny pieces of dq and then integrating
over the whole distribution.
(B)  3 > 1 > 5 > 2 = 4
V
(C)  1 = 2 > 4 = 3 > 5
•  Finally no work needs to be done if you move a charge
on an equipotential, since it would be moving
perpendicular to the electric field.
(D)  2 > 4 > 1 = 3 = 5
1 2 3
•  The charge concentrates on a conductor on surfaces
with smallest radius of curvature.
1/30/12
21
4
5
x
1/30/12
22
Another Method: Electrostatic Potential Energy
Leave the charges in place and add all the
electrostatic energies of each charge and divide
it by 2.
where:
qi -- is the charge at point i
Vi -- is the potential at
location i due to all of
the other charges.
1/30/12
23
1/30/12
24
6
1/30/12
Another Method: Electrostatic Potential Energy
Summary: Electrostatic Potential Energy for a
Collection of Point
q3
Point 3
r2,3
r1,3
Point 1
q1
1. Bringing charge by charge from infinity
Point 2
U=
1/30/12
⎛ kq kq ⎞
⎛ kq kq ⎞ ⎤
1 ⎡ ⎛ kq2 kq3 ⎞
+
+ q2 ⎜ 1 + 3 ⎟ + q3 ⎜ 1 + 2 ⎟ ⎥
⎢ q1 ⎜
⎟
2 ⎢⎣ ⎝ r1,2 r1, 3 ⎠
⎝ r1,2 r2, 3 ⎠
⎝ r1, 3 r2, 3 ⎠ ⎥⎦
1 ⎡ kq1q2
kq q
kq q ⎤ kq q
kq q kq q
+2 3 1 +2 2 3⎥= 1 2 + 3 1 + 2 3
⎢2
2 ⎣ r1,2
r1, 3
r2, 3 ⎦
r1,2
r1, 3
r2, 3
1 n
∑ qiVi and trying
2 i =1
2. Or using the equation:
not to forget the factor of ½ in front.
q2
r1,2
U=
U=
Note: The second way, could involve more calculations for point
charges. However, it becomes very handy whenever you try to
calculate the electrostatic potential energy of a continuous
distribution of charge, the sum becomes an integral.
25
1/30/12
26
V due to an infinite line charge (continued)
V due to an infinite line charge
Remember Gauss’s Law

 
dV = −E ⋅ dl = −E R Rˆ ⋅ dl = −E R dR
Gaussian
cylinder
VP − Vref = − ∫
RP
Rref
dR
R
RP
= −2kλ ln
Rref
Problem : you can' t choose
Rref = 0 or Rref = ∞ because
ln0 = −∞ and ln ∞ = +∞.
∴0 < Rref < ∞ and
Rref
V = 2kλln
Where we define V = 0 at R = Rref
R
R
line charge with
charge density l


s
s
n
∫ E dA =E ∫ dA
s
n
R
barrel
RP
Rref
Qinside
εo
Qinside = λ L
+ E ∫ dAEndcaps
E R ABarrel = E R (2π RL) =
1/30/12
= −2kλ ∫
L
∫ E ⋅ dA = ∫ E dA =
ER dR
λL
εo
0
λ
or
2π Rε o
2 kλ
ER =
R
ER =
27
1/30/12
28
7
1/30/12
Work and Equipotential Lines
Which of the following
statements is true about the
work done by the electric
field in moving a positve
charge along the paths?
(A)  III = IV > 0, I = II = 0
(B)  I = II = 0, III > IV
(C)  III = IV < 0, I = II = 0
1/30/12
electric field
lines
V1 = 100 V
V2 = 80 V
V3 = 60 V
V4 = 40 V
29
8