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Quiz August 30, 2010
From Mechanics (PHYS 220)
There is a certain net flux Φο through a Gaussian sphere of
radius r enclosing an isolated charged particle. Suppose
the enclosing Gaussian surface is changed to
• Energy
Kinetic Energy: associated with the state of motion
1 A Gaussian cube with edge length r
2 A larger Gaussian sphere of radius 2r
3 A Gaussian cube with edge length 2r
Potential Energy: associated with the configuration of
the system
Rank the net flux for each Gaussian surface:
• Conservative Forces:
(A) Φο = Φ1 < Φ2 = Φ3
(B) Φ2 = Φο > Φ1 = Φ3
(C) Φο = Φ1 > Φ2 = Φ3 (D) Φο = Φ1 = Φ2 = Φ3
(Ε) Φ1 = Φ3 > Φ2 = Φο 8/31/10
Work done by a conservative force is independent of path.
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Potential Energy
m1m2
r2
PE = −G
m1m2
r
2
Potential Energy of Two Point
Gravitational Potential Energy
Fg = G
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U=0 at r=∞
Electric Potential Energy
Fe = k
q1q2
r2
PE = k
q1q2
r
U g = −G
m1m2
r
PEelec = k
q1q2
r
Potential energy: can be positive or negative
Field forces push object into area with lower potential energy:
Potential energy decreases, work is done on object –
for example, kinetic energy of the object may increase (speed)
Gravity: always attraction.
U is more negative (“lower”)
when objects are closer
q1q2 < 0 : attraction.
U is more negative (“lower”)
when objects are closer
q1q2 > 0 : repulsion.
U is lower when objects
are farther apart
Total energy is conserved
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Potential Energy of Several Point Charges:
Electric Potential
q5
An Example with 3 charges
q4
Suppose a test charge q moves
through the area with many fixed
charges around
q1
Total potential energy PE:
+q1
-q3
PE = PE12 + PE23 + PE13
q6
q
q3
q2
+q2
⎛ qq qq
⎞
qq
qq
PEelec = k ⎜ 1 + 2 + ... 1 2 + 1 3 + ...⎟
r2
r12
r13
⎝ r1
⎠
The changes in potential energy are proportional to that charge
Electric potential V: electric potential energy per unit charge:
Potential energy of many charges is a scalar sum of potential energies of
all the pairs of charges in the system.
Note: each pair should be counted only once.
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*Electric potential energy PEelec ≠ electric potential (or potential) V
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Electric Potential versus Electrical Potential
Energy
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Potential & Electric Fields
The electric field points in the direction in which the
potential decreases most rapidly.
Electric Potential is a property of an electric field and is
measured in J/C or V
Electric Potential Energy is an energy of system
consisting of the charged object and the external
electric field, and is measured in Joules.
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Potential Difference
Potential Difference
*independent of path
*independent of test charge
To find ΔV, find work done on a positive charge qo by
the electric field as it moves from i to f.
The potential difference between two points is the
negative of the work done by the electrostatic force
to move a unit charge from one point to another.
At any point, the differential work done on the
particle by electrostatic force:
Work:
Potential Difference:
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Potential due to a point charge:
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Potential due to a Group of Point Charges
V in space around a charged particle relative to the
zero potential at infinity:
Find the Potential at the center of the square.
N
V (r) = ∑ Vn (r) =
For a point charge
n =1
V(r) to ∞
V=
+
q
4πε 0 r
q1 = +12 nC
r
q2 = -24 nC
+
-
+
+
q3 =+31 nC
1 N qn
∑
4πε o n =1 rn
q3 =+17 nC
V (r = 0) = ∞
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Equipotentials
Definition: locus of points with the same potential.
• Calculate Electric Potential from the
Electric Field
• General Property: The electric field is always
perpendicular to an equipotential surface.
ΔV = −EΔx
• Calculate Electric Field from the Electric
Potential
E=−
ΔV
Δx
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Equipotentials: Examples
Point charge
infinite positive
charge sheet
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E due to an infinite line charge
electric dipole
Corona discharge around a high voltage power line,
which roughly indicates the electric field lines.
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Equipotentials
Demo
Get energy out
ΔPE = q ΔV
charge flow
+
+
V=
+
R
+
+
+
+
+
+
+
17
ΔV = V (r1 ) − V (r2 )
+
across fluorescent
light bulb
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Supplemental:
Potential Energy: Gravitational & Electric
Potential inside & outside a conducting sphere
Gravitational potential energy
Vref = 0
ΔPE = -Wgrav.field
at r = ∞.
Common mistake: thinking that potential must be zero
inside because electric field is zero inside.
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r2
r1
Also try an
elongated neon bulb.
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kQ
R
19
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W = F.d.cos(θ)
mm
Fg = G 1 2 2
r
mm
PE = −G 1 2
r
Electric potential energy
ΔPE = -Welec.field
U=0 at r=∞
W = F.d.cos(θ)
qq
Fe = k 1 2 2
r
qq
PE = k 1 2
r
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