Potential Energy and
the Electric Potential
Potential Energy and
the Electric Potential
From last lecture:
Positive and Negative charges
Conductors, Insulators and Solutions
Charging processes (triboelectric, conduction,
induction)
Electric force (Coulomb’s law)
This lecture:
Work
Conservative forces and potential energy
Electric Potential

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Recall from P207
For an infinitesimal displacement ds,
the work done by the force F is:
dW = F·ds = F ds cosθ
Gravitational Force and work
F
θ
ds
WG=-mgh work done by the gravitational force
The work done by an external force to lift the block of mass m
is: W = -WG = Fh=mgh >0!
Someone has to make an effort to lift the block!!
Positive: Force is in direction of motion
Negative: Force is opposite to direction of motion
Zero: Force is perpendicular to direction of motion
(eg gravitational force for a block moving on a plane)
Along any path, a path integral is needed:
W =
B
∫ A F ⋅ ds = K B − K A
The work done by an external agent is equal
and opposite to the work done by the
gravitational force
F
with K = kinetic energy
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Conservative forces
Potential Energy of 2 charges
Fg
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Conservative Forces: the work done by the force
is independent on the path and depends only on
the starting and ending locations.
It is possible to define the potential energy U
Wconservative = −Δ U = Uinitial - Ufinal =
= -(Kfinal - Kinitial) = -ΔK

Consider 2 positive charged particles. The electric force
between them is
The work that an external agent should do to bring q 2 at a
distance rf from q1 starting from a very far away distance is
equal and opposite to the work done by the electric force.
Charges repel ⇒W>0!
F
W =
5
∫
r12
∞
F ⋅ dr = −W E
r12
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Potential Energy of 2 charges
Potential Energy of 2 charges

Since the 2 charges repel, the force on q2 due to q1
F12 is opposite to the direction of motion
 The external agent F = -F12 must do positive work!
W > 0 and W=-WE

r12
W = −W E = − ∫ ke
∞
q1q2
dr
r2



F
The external agent changes the potential energy of
the system W = ΔU = Ufinal - Uinitial
If W>0⇒U increaseas
We set Uinitial = U(∞) = 0 since at infinite distance the
force becomes null
The potential energy of the system is
dr
r12
r
 1 12
qq
W = −k e q1q2 −  = k e 1 2
 r ∞
r12
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More About U of 2 Charges
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U with Multiple Charges
Like charges ⇒ U > 0 and work must be done
to bring the charges together since they repel
(W>0)
Unlike charges ⇒ U < 0 and work is done to
keep the charges apart since the attract one the
other (W<0)
U and W (external agent against electric
force) have the same sign that is determined
by the product of the 2 charges!


If there are more than two
charges, then find U for
each pair of charges and
add them
For three charges:
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Quick Quiz 1
Work done to assemble 3 charges
Similarly if they are all positive:
Question: How much work would it take YOU to
assemble 3 negative charges?
1. W = +19.8 mJ
2. W = 0 mJ
3. W = -19.8 mJ
=(9×10 9)(1×10 -6 )(2×10 -6 )/5
=3.6 mJ
• W3 = k q1 q3/r + k q2 q3/r =
(9×10 9)(1×10-6)(3×10 -6)/5 + (9×109)(2×10-6)(3×10-6)/5 =16.2
−3µC
5m
Likes repel, so YOU will
still do positive work!
W1 = 0
• W2 = k q1 q2 /r

−1µC
• W = +19.8 mJ
• WE = -19.8 mJ
• UE = +19.8 mJ
5m
5m
−2µC
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5m
3µC
mJ
q3
5m
2µC
1µC
5m
q2
q1
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Work and Δ Potential Energy
Quick Quiz 2
Gravity
• Mass moved ∞ → rf
The total work required for YOU to assemble the set of
charges as shown below is:
1. positive
2. zero
3. negative
5m
W1 = 0
+Q
Q(−Q)
d
QQ
Q(−Q)
W3 = k
+k
d
d
Q2
Total work = −k
d
W2 = k
• Charge moved ∞ → rf
• FE = kq1q2/r2
r
qq
W = - ∫∞ k r1 2 2 dr =
kq1q2/rf
ΔUE= +kq1q2/rf
f
+Q
5m
• FG =Gm1m2/r2
r
• W= ∫∞ G m1m2 2 dr =
r
= -Gm1m2/rf
ΔUG= -Gm1m2/rf
Electric
5m
−Q
=
f
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rf
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Electric Potential

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Potential and Point Charges

The potential energy per unit charge, U/qo, is the
electric potential
 The potential is independent of the value of q
o
The electric potential is
The potential difference
between points A and B is
VB − VA =
1 1
UB UA
−
= ke q − 
q0 q0
 rB rA 
Electric force conservative ⇒
electric potential independent on
path
between A and B
Units: 1 V = 1 J/1C
It takes 1 joule of work to move a 1C charge through a
potential difference of 1 volt




Since U is a scalar quantity also V is a scalar
Just as with potential energy only differences are
meaningful
It is customary to assume to choose a reference
potential of V = 0 at r = ∞
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Quick Quiz 3
Potential of Point Charges
1 1
VB − VA = ke q − 
 rB rA 

Customary choise of reference potential:
V = 0 at rA = ∞.
In the Figure, take q 1 to be a negative source charge
and q2 to be a test charge. If q2 is initially positive and
is changed to a charge of the same magnitude but
negative, the potential at the position of q2 due to q1
(a) increases
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(b) decreases
+→-
(c) remains the same
The electric potential due to several point
charges is the sum of the potentials due to each
individual charge (superposition principle)
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Answer: (c). The potential is established only by the source
charge and is independent of the test charge.
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Electrostatic machines to create potential
differences: Van de Graaff
Generator
Quick Quiz 4
Consider the situation in QQ3 again: q 1 is a negative source
charge and q2 a test charge. When q2 is changed from
positive to negative, the potential energy of the two-charge
system
+→(a) increases
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(b) decreases
_
(c) remains the same

Answer: (a). The potential energy of the two-charge system is
initially negative, due to the products of charges of opposite sign.
When the sign of q 2 is changed, both charges are negative, and
the potential energy of the system is positive.
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The high-voltage electrode is a hollow metal dome
mounted on an insulated column
Charge is delivered continuously to dome by a
moving belt of insulating material
Large potentials can be developed by repeated trips
of the belt
Protons can be accelerated through such large
potentials
ΔV
ΔU = qΔV = −ΔK = K f
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Quick Quiz 5: Electric Dipole
Toepler-Voss Electrostatic Machine
(1860-80)
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2 small metal brushes rub against the buttons; brushes are in contact with
the disks at opposite points
2 brass collection combs are at the extremes of the horizontal rod in front
of the buttons. They collect opposite charges
spherules of the spark-gap
are charged by opposite sign
with respect to the sign of the
comb with which they are in
contact.
The combs are in contact with
the inner shields of two Leiden
jars
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
For a point P on the right of +q on the x axis
V = ke ∑
http://www.coe.ufrj.br/~acmq/electrostatic.html
 q
qi
q  2k e qa
= ke 
−
=
 x − a x + a  x 2 − a2
ri
If P is far from the dipole x>>a
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V≈
What if point P happens to be located
to the left of -q?
2k e qa
x2
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Answer: it has the same value but it’s negative
ΔV
 q
q
q 
2k qa
V = ke ∑ i = k e 
−
=− 2 e 2
 x + a x − a
ri
x −a
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(*)
p=qa
Electric dipole moment
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Verify that:
The general formula that applies to all cases
along the x axis is

 q
q 
V (x) = k e 
−

 x−a x+a

What happens for x=0?
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