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Physics 2102
Jonathan Dowling
Physics 2102
Lecture 06: THU 04 FEB
Electric Potential I
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Danger!
Volume
[m3]
R
R
4 3
R
3
Area
[m2]
d
dR
Circle
L
4 R
2
 R2
Sphere
d
dR
2 R
L
R
L
Circumference
[m]
R L
2
2 RL
Cylinder
Electric Potential Energy
Electric Potential Energy U is Negative of the Work W to
Bring Charges in From Infinity:
U = –W∞
The Change in Potential Energy U Between an Initial
and Final Configuration Is Negative the Work W Done
by the Electrostatic Forces:
U = Uf - Ui = -W
+Q
• What is the potential energy of a single
–Q
+Q
a
charge?
• What is the potential energy of a dipole?
• A proton moves from point i to point f in a
uniform electric field, as shown.
- Does the electric field do positive or
negative work on the proton?
- Does the electric potential energy of the
proton increase or decrease?
Electric Potential
Electric potential difference between two points = work
per unit charge needed to move a charge between the
two points:
V = Vf – Vi = –W/q = U/q
dW  F  ds
dW  q0 E  ds
f
f
i
i
W   dW   q0 E  ds
f
W
V  V f  Vi      E  ds
q0
i
Electric Potential Energy,
Electric Potential
Units :
Potential Energy = U = [J] = Joules
Electric Potential = V = U/q = [J/C] = [Nm/C] = [V] = Volts
Electric Field = E = [N/C] = [V/m] = Volts per meter
Electron Volt = 1eV = Work Needed to Move an Electron
Through a Potential Difference of 1V:
W = qV = e x 1V = 1.60 10–19 C x 1J/C = 1.60 10–19 J
Equipotential Surfaces
f
W
V  V f  Vi      E  ds
q0
i
• The Electric Field is Tangent to the Field Lines
• Equipotential Surfaces are Perpendicular to Field Lines
• Work Is Needed to Move a Charge
Along a Field Line.
• No Work Is Needed to Move a Charge
Along an Equipotential Surface.
• Electric Field Lines Always Point
Towards Equipotential Surfaces With
Lower Potential.
Electric Field Lines and Equipotential
Surfaces
Why am I smiling?
I’m About to Be
Struck by
Lightning!
http://www.cco.caltech.edu/~phys1/java/phys1/EField/EField.html
Electric Potential and Electric
Potential Energy
The change in potential energy of a charge q moving from
point i to point f is equal to the work done by the applied
force, which is equal to minus the work done by the electric
field, which is related to the difference in electric potential:
U  U f  U i  Wapp  W  qV
We move a proton from point i to point f in
a uniform electric field, as shown.
• Does the electric field do positive or negative work on
the proton?
• Does the electric potential energy of the proton
increase or decrease?
• Does our force do positive or negative work ?
• Does the proton move to a higher or lower potential?
Example
Consider a positive and a negative charge, freely moving in a
uniform electric field. True or false?
(a) Positive charge moves to points with lower potential.
(b) Negative charge moves to points with lower potential.
(c) Positive charge moves to a lower potential energy.
(d) Negative charge moves to a lower potential energy.
(a) True
(b) False
(c) True
(d) True
+++++++++
–Q
––––––––
+Q
+V
0
–V
Conservative Forces
The potential difference between two points is independent
of the path taken to calculate it: electric forces are
“conservative”.
W U
V  V f  Vi   
   E  ds
q0
q0
i
f
Electric Potential of a Point
Charge
f
P
i

V    E  ds    E ds 
R
R
kQ
kQ
kQ
   2 dr  

r
r 
R

Note: if Q were a
negative charge,
V would be negative
Electric Potential of Many Point Charges
• Electric potential is a
SCALAR not a vector.
q4
• Just calculate the potential
due to each individual point
charge, and add together!
(Make sure you get the
SIGNS correct!)
qi
V  k
ri
i
r3
r4
q5
r5
Pr2
q2
r1
q1
q3
Electric Potential and Electric
Potential Energy
U  Wapp  qV
+Q
a
What is the potential energy of a dipole?
–Q
+Q
a
–Q
• First: Bring charge +Q: no work involved, no potential energy.
• The charge +Q has created an electric potential everywhere, V(r) = kQ/r
• Second: The work needed to bring the charge –Q to a distance a from the
charge +Q is Wapp = U = (-Q)V = (–Q)(+kQ/a) = -kQ2/a
• The dipole has a negative potential energy equal to -kQ2/a: we had to do
negative work to build the dipole (electric field did positive work).
Positive Work
+Q
a
+Q
Negative Work
+Q
a
–Q
Positive Work
Charge Moves Uphill
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+Q
a
+Q
Negative Work
+Q
a
–Q
Charge Moves Downhill
Potential Energy of A System of Charges
• 4 point charges (each +Q and equal
mass) are connected by strings,
forming a square of side L
• If all four strings suddenly snap, what
is the kinetic energy of each charge
when they are very far apart?
+Q
+Q
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• Use conservation of energy:
– Final kinetic energy of all four charges
= initial potential energy stored =
energy required to assemble the
system of charges
+Q
+Q
Let’s do this from
scratch!
Potential Energy of A System of
Charges: Solution
•
No energy needed to bring in
first charge: U1=0
+Q
• Energy needed to bring in
3rd charge =
kQ2 kQ 2
U 3  QV  Q(V1  V2 ) 

L
2L
• Energy needed to bring in
4th charge =
2kQ2 kQ 2
U 4  QV  Q(V1  V2  V3 ) 

L
2L
+Q
2L
• Energy needed to bring in
2nd charge:
kQ 2
U 2  QV1 
L
L
+Q
+Q
Total
potential energy is sum of all
the individual terms shown on left
hand side =
2

kQ
4 2
L

So, final kinetic energy of each charge
=
2


kQ
4 2
4L



Example
qi
V  k
ri
i
Positive and negative charges of equal magnitude Q are held
–Q
+Q
in a circle of radius R.
1. What is the electric potential at the center of each circle?
• VA =
k 3Q  2Q /r  kQ/r
• VC = k 2Q  4Q /r  2kQ/r
2. Draw an
arrow representing
k 2Q
 2Q  / r the
0 approximate direction of
• VB =
A
the electric field at the center of each circle.
3. Which system has the highest electric potential energy?
B
C
UB =Q•Vb has largest un-canceled charge
Electric Potential of a Dipole (on axis)
What is V at a point at an axial distance r away from the
midpoint of a dipole (on side of positive charge)?
V k
Q
k
Q
a
a
(r  )
(r  )
2
2
a
a 

 (r  2 )  (r  2 ) 
 kQ 

a
a
 (r  )(r  ) 
2
2 

Qa

2
a
2
40 (r  )
4
p
a
–Q +Q
r
Far away, when r >> a:
V 
p
40 r 2
Electric Potential on Perpendicular
Bisector of Dipole
You bring a charge of Qo = –3C
from infinity to a point P on the
perpendicular bisector of a
dipole as shown. Is the work
that you do:
a) Positive?
b) Negative?
c) Zero?
U = QoV = Qo(–Q/d+Q/d) = 0
a
-Q +Q d
P
–3C
Summary:
• Electric potential: work needed to bring +1C from infinity; units
V = Volt
• Electric potential uniquely defined for every point in space -independent of path!
• Electric potential is a scalar — add contributions from individual
point charges
• We calculated the electric potential produced by a single
charge: V=kq/r, and by continuous charge distributions:
V=kdq/r
• Electric potential energy: work used to build the system,
charge by charge. Use W=qV for each charge.