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Physics 2113
Jonathan Dowling
Physics 2113
Lecture 11: FRI 18 SEP
Electric Potential I
Danger!
R
R
Volume
[m3]
Area
[m2]
4p
R
3
4p R
d
3 dR
Circle
L´
pR
2
d
2 dR
Sphere
2p R
L´
R
L
Circumference
[m]
2p RL
p R2 L
d
dR
Cylinder
R
R
4p
V=
R
3
d
3 dR
Circle
L´
R
L
V = pR L
2
d
dR
dV =
Sphere
4p R dR
2
A = pR
2
dV =
2p RLdR
d
dR
dA =
2p RdR
L´
Cylinder
dV =
d
dL
p R dL
2
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 Voltage
Electric potential voltage difference between two points
= work per unit charge needed to move a charge
between the two points:
V = Vf – Vi = –W/q = U/q
Electric Potential Energy vs
Electric Potential Voltage
Units :
Potential Energy = U = [J] = Joules
Potential Voltage = V = U/q = [J/C] = [Nm/C] = [V] = Volts
Electric Force = F = [N] = Newtons
Electric Field = E = [N/C] = Newtons/Coulomb= [V/m] = Volts/m
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 = 1.0eV
Electric Potential Voltage 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
voltage:
DU = U f -U i = Wapp = -Wfield = qDV
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?
Positive Work
+Q
a
+Q
Negative Work
+Q
a
–Q
Applied Positive Work:
Potential Energy
Increases
+Q
a
+Q
Charge Moves Uphill:
I’m doing work against
Field.
Applied Negative Work:
Potential Energy
Decreases
+Q
a
–Q
Work done by field is negative of
Applied work done by me.
Charge Moves Downhill:
I’m Doing Work With Field
(d) Potential Voltage:
higher? ✔
lower?
+ work?
(a) E-field does:
– work? ✔
+Vhigh = + + + + + + +
increase? ✔
(c) Potential Energy:
decrease?
+ work? ✔
(b) Work done by me:
– work? -V
low
= -------
ICPP:
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 voltage.
(b) Negative charge moves to points with lower potential voltage.
(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
electron
(d) Potential Voltage:
(-)
higher? ✔
lower?
+ work? ✔
(a) E-field does:
– work?
-Vlow = - - - - - - -
-
increase?
(c) Potential Energy:
decrease? ✔
+ work?
(b) Work done by me:
– work? ✔
+Vhigh = + + + + + + +
Units :
Electric Potential Energy,
Electric Potential
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
F = qE (Force is charge times Field)
U = qV (Potential Energy is charge times Potential)
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
Electric Potential Energy = Joules
Electric potential energy difference U between two
points = work needed to move a charge between the
two points:
U = Uf – Ui = –W
Electric Potential Voltage = Volts =
Joules/Coulomb!
Electric potential — voltage! — difference V between
two points = work per unit charge needed to move a
charge between the two points:
V = Vf – Vi = –W/q = U/q
Equal-Potential = Equipotential Surfaces
• 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 (Or Back to the Surface
Where it Started).
• 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
Equipotential Surfaces: IPCC
For which paths is work done W = 0?
I ? II ? III ? IV ?
For which paths is work done, W ¹ 0?
I ? II ? III ? IV ?
For which paths is work done, W , the same?
I ? II ? III ? IV ?
++++++
------
DV = V f - Vi = -W / q0
DVI = V1 - V1 = 0
DVII = V3 - V3 = 0
DVIII = V1 - V2 ¹ 0
DVIV = V1 - V2 = DVIII ¹ 0
Conservative Forces
The potential difference between two points is independent
of the path taken to calculate it: electric forces are
“conservative”.
2
+
+
+
+
downhill
(a) ® ?
¬?
­?
¯?
(b) 1 = + 2 = + 3 = + 4 = - 5 = +
W1 = +10eV
(c) 3 >1 = 2 = 5 > 4
W3 = +20eV
W2 = +10eV
W4 = -10eV
W5 = +10eV
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:
dV=kdq/r
• Electric potential energy: work used to build the system,
charge by charge. Use W=qV for each charge.