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Question 1
Which one of these statements is true?
When near the center of the object, the electric field
hardly changes with increasing distance from
A. a uniformly charged long rod.
B. a uniformly charged ring.
C. a uniformly charged disk.
D. a point charge.
Question
A solid metal ball carrying negative excess charge is
placed near a uniformly charged plastic ball.
-Q
+Q
plastic
metal
Which one of the following statements is true?
A. The electric field inside both balls is zero.
B. The electric field inside the metal ball is zero, but it is
nonzero inside the plastic ball.
C. The electric fields inside both objects are nonzero and
are pointing toward each other.
D. The electric field inside the plastic ball is zero, but it is
nonzero inside the metal ball.
Question
Which one of these statements is false?
A. The electric field of a very long uniformly charged rod
has a 1/r distance dependence.
B. The electric field of a capacitor at a location outside
the capacitor is very small compared to the field
inside the capacitor.
C. The fringe field of a capacitor at a location far away
from the capacitor looks like an electric field of a point
charge.
D. The electric field of a uniformly charged thin ring at
the center of the ring is zero.
Chapter 17
Electric Potential
Potential Energy
To understand the dynamics of moving objects we used:
forces, momenta, work, energy
The concept of electric field E deals with forces
Introduce electric potential – for work and energy
Electric potential: electric potential energy per unit charge
Practical importance:
• Reason about energy without having to worry about the details of
some particular distribution of charges
• Batteries: provide fixed potential difference
• Predict possible pattern of E field
Energy of a Single Particle
The energy of a single particle with charge q1 consists solely of its
particle energy.
q1
Particle energy
Kinetic energy
E=
mv 2
For v<<c: K »
2
mc 2
v2
1- 2
c
= mc 2 + K
Rest energy
Kinetic energy is associated with motion
A single particle has no (electric) potential energy
The kinetic energy of a single particle can be changed if positive or
negative work is done on the particle by its surroundings.
Electric Potential Energy of Two Particles
r12
Potential energy is associated with pairs of interacting objects
Energy of the system:
1. Energy of particle q1
q2
2. Energy of particle q2
3. Interaction energy Uel
q1
Esystem = E1+E2+Uel
To change the energy of particles we have to perform work.
DE1 + DE2 = Wext + Wint + Q
Wext – work done by forces exerted by other objects
Wint – work done by electric forces between q1 and q2
Q – thermal transfer of energy into the system
Electric Potential Energy of Two Particles
D(E1 + E2 ) = Wext + Wint
q2
r12
D(E1 + E2 ) - Wint = Wext
Uel  -Wint
q1
if D(mc2 ) = 0
DEsystem = DK system + DUel = Wext + Q
Total energy of the system can be changed (only) by external forces
or by adding (thermal) energy.
Work done by internal forces:
f
DU el = -Wint = - ò Fint · dr
i
Electric Potential Energy of Two Particles
Fint
f
DU el = -Wint = - ò Fint · dr
q2
r12
i
f
1 q1q2
DU el = - ò
r̂12 · r̂ dr12
2
4pe 0 r12
i
f
1 q1q2
DU el = - ò
dr12
2
4pe 0 r12
i
q1
1 q1q2
F=
rˆ12
2
4pe 0 r12
DU el = -
1
4pe 0
f
q1q2 ò
i
1
dr12
2
r12
f
é 1ù
DU el = q1q2 ê- ú
4pe 0
ë r12 ûi
1
Electric Potential Energy of Two Particles
Fint
r12
q1
é 1ù
DU el = q1q2 ê- ú
4pe 0
ë r12 ûi
æ 1 q1q2 ö
÷÷
DU el = Dçç
è 4pe 0 r12 ø
1
q2
f
1 q1q2
F=
rˆ12
2
4pe 0 r12
The potential energy of a pair of particles is:
1 q1q2
U el =
(joules)
4pe 0 r12
Electric Potential Energy of Two Particles
q2
1 q1q2
U el =
(joules)
4pe 0 r12
q1
Uel > 0 for two like-sign charges
(repulsion)
q2
q1
Uel < 0 for two unlike-sign
Charges (attraction)
Electric Potential Energy of Two Particles
æ 1 q1q2 ö
DU el = D ç
è 4pe 0 r12 ÷ø
Meaning of U0:
r12
1 q1q2
U el =
+ U0
4pe 0 r12
U el ® U 0
Choose U0=0 – no potential energy if r12 (no interaction)
q2
q1
Potential energy = amount of work
the two charges can do on each other
when moved away from each other
to 
q2
q1
Electric and Gravitational Potential Energy
q2
1 q1q2
F=
r̂
2
4pe 0 r
m1m2
F = -G 2 r̂
r
q1
m2
m1
1 q1q2
U el =
4pe 0 r
U grav
m1m2
= -G
r
Three Electric Charges
Interaction between q1 and q2 is
independent of q3
There are three interacting pairs:
q1  q2
U12
q2  q3
U23
q3  q1
U31
U= U12+ U23+ U31
1 q1q2
1 q2 q3
1 q1q3
Uel =
+
+
4pe 0 r12
4pe 0 r23
4pe 0 r13
Multiple Electric Charges
q1
q3
q6
q2
q4
Each (i,j) pair interacts:
potential energy Uij
q5
1 qi q j
U el = åU ij = å
i< j
i < j 4pe 0 rij
Notation: i<j avoids double counting: ij, ji
Electric Potential
Electric potential  electric potential energy per unit charge
U el
V=
q
Units: J/C = V (Volt)
Volts per meter = Newtons per Coulomb
Alessandro Volta (1745 - 1827)
Electric potential – often called potential
Electric potential difference – often called voltage
V due to One Particle
U el
V=
q
q2
Single charge has no electric
potential energy
Single charge has potential to
interact with other charge –
it creates electric potential
1 q1q2
U el =
4pe 0 r
1 q1
VB =
4pe 0 r
probe charge
J/C, or Volts
Electric potential at B due to
charge q1.
V due to Two Particles
Electric potential is scalar:
VC = VC ,1 + VC , 2
q1
1 q2
=
+
4pe 0 r13 4pe 0 r23
1
Electric potential energy of the system:
q3
U sys
1 q1q2
= U12 =
4pe 0 r12
If we add one more charge at position C:
U sys
1 q1q2 æ 1 q1
1 q2 ö
= U12 + VC q3 =
+ç
+
q3
÷
4pe 0 r12 è 4pe 0 r13 4pe 0 r23 ø
U sys
1 q1q2
1 q1q3
1 q2 q3
=
+
+
= U12 + U13 + U 23
4pe 0 r12 4pe 0 r13 4pe 0 r23
V at Infinity
1 q1
V=
4pe 0 r
r, V=0
Positive charge
Negative charge
Exercise
What is the electrical potential at a location 1Å from a proton?
1Å
(
)
-19
2
1.6
´
10
C
æ
ö
1 q1
9 Nm
V=
= ç 9 ´ 10
= 14.4 J/C = 14.4 V
2 ÷
-10
4pe 0 r è
C ø
10 m
(
)
What is the potential energy of an electron at a location
1Å from a proton?
(
)
U el = Vq= (14.4 J/C) -1.6 ´10-19 C = -2.3´10-18 J = -14.4 eV
Exercise
2Å
1Å
What is the change in potential in going from 1Å to 2Å from the proton?
∆𝑉 = 𝑉 2Å − 𝑉 1Å = 7.2V − 14.4V = −7.2V
What is the change in electric potential energy associated with
moving an electron from 1Å to 2Å from the proton?
∆𝑈𝑒𝑙 = 𝑈𝑒𝑙 2Å − 𝑈𝑒𝑙 1Å = 𝑞𝑉 2Å − 𝑞𝑉 1Å = 𝑞Δ𝑉
(
)
DUel = -1.6 ´ 10 -19 C ( -7.2 J/C) = +1.15 ´ 10 -18 J
Does the sign make sense?
Electric Potential Energy of Two Particles
Energy of the system:
1. Energy of particle q1
2. Energy of particle q2
3. Interaction energy Uel
q2
r12
To change the energy of particles we have to perform work.
DE1 + DE2 = Wext + Wint + Q
q1
DE1 + DE2 - Wint = Wext Energy Principle
Uel  -Wint
for a multiparticle
system.
if D(mc2 ) = 0 : DK system + DUel = Wext + Q
Total energy of the system can be changed (only) by external forces.
Work done by internal forces:
f
DU el = -Wint = - ò Fint · dl
i
Example
The electron enters the
capacitor through a small hole
at A with an initial velocity to
the right.
What is the change in the
electron’s potential energy in
traveling from A to B? Change
in kinetic energy?
(AB)= 4mm; 𝐸 = 2x103N/C.
x
∆𝑈𝑒𝑙 = −𝐹𝑒𝑙𝑒𝑐𝑡𝑟𝑖𝑐 ⋅ ∆𝑙 = − −𝑒𝐸𝑥 ∆𝑥 = 𝑒𝐸𝑥 ∆𝑥
= (1.6x10-19 C)(2x103 N/C)(0.004m) =1.3x10-18 J
K = -Uelectric = -1.3x10-18 J
Electric Potential Difference in a
Uniform Field
Electric potential  electric potential energy per unit charge
DU electric
DUelectric
= qDV, DV =
q
Units: J/C = V (Volt)
Volts per meter = Newtons per Coulomb
Electric potential, V – often called potential
Electric potential difference, V – often called voltage
∆𝑈𝑒𝑙 = −𝐹𝑒𝑙𝑒𝑐𝑡𝑟𝑖𝑐 ⋅ ∆𝑙
𝑞
𝑞
∆𝑉 = −𝐸 ⋅ ∆𝑙
Example
∆𝑙
300
∆𝑉 = −𝐸 ∙ ∆𝑙 = −𝐸∆𝑙 cos(𝜃)
𝑁
= −(100 )(2𝑚) cos 30°
𝐶
= −173 𝑉
V at Infinity
1 q1
V=
4pe 0 r
V
Positive charge
r
r, V=0
Electric Potential Energy of Two Particles
q2
1 q1q2
U el =
(joules)
4pe 0 r12
q1
Uel > 0 for two like-sign charges
(repulsion)
q2
q1
Uel < 0 for two unlike-sign
Charges (attraction)
Sign of the Potential Difference
DUel = qDV
The potential difference V can be positive or negative.
The sign determines whether a particular charged particle
will gain or lose energy in moving from one place to another.
If qV < 0 – then potential energy decreases and K increases
If qV > 0 – then potential energy increases and K decreases
Path going in the direction of E: Potential is decreasing (V < 0)
Path going opposite to E: Potential is increasing (V > 0)
Path going perpendicular to E: Potential does not change (V = 0)
Sign of the Potential Difference
DUel = qDV
If freed, a positive charge will
move to the area with a lower
potential:
Vf – Vi < 0
DU el = qDV < 0
DK = -DU el > 0 (no external forces)
V1 < V2
Moving in the direction of E means that potential is decreasing
Sign of the Potential Difference
DUel = qDV
To move a positive charge to the
area with higher potential:
Vf – Vi > 0
DU el = qDV > 0
Need external force to perform work
V2 > V1
Moving opposite to E means that potential is increasing
Question 1
A proton is free to move from right to left
in the diagram shown. There are no other
forces acting on the proton. As the proton
moves from right to left, its potential energy:
A)
B)
C)
D)
Is constant during the motion
Decreases
Increases
Not enough information
V1 < V2
Potential Difference in a
Nonuniform Field
C
x
∆𝑉 = −𝐸 ∙ ∆𝑙
𝐸1 = 𝐸1𝑥 𝑥
A to C: V1 = -|E1x|(xC-xA)
𝐸2 = − 𝐸2𝑥 𝑥
C to B: V2 = |E2x|(xB-xC);
A to B: V = V1+ V2 = -|E1x|(xC-xA) + |E2x|(xB-xC)
Potential Difference with Varying
Field
𝑥𝑓
𝑓
𝐸 ∙ 𝑑𝑙 = −
∆𝑉 = −
𝑖
𝑦𝑓
𝐸𝑥 𝑑𝑥 −
𝑥𝑖
𝑧𝑓
𝐸𝑦 𝑑𝑦 −
𝑦𝑖
𝐸𝑧 𝑑𝑧
𝑧𝑖
Example: Different Paths near Point Charge
1. Along straight radial path:
rf
ri
f
DV = - ò E · dl
rf
DV = - ò
ri
i
q
r̂ · r̂ dr
2
4pe 0 r
1
1
+q
q
E=
r̂
2
4pe 0 r
1
Origin at +q
rf
1
DV = q ò 2 dr
4pe 0 ri r
rf
é 1ù
DV = q ê- ú
4pe 0 ë r û ri
1
é 1 1ù
DV =
q ê - ú = V f - Vi
4pe 0 êë rf ri úû
1
Example: Different Paths near Point Charge
f
2. Special case
DV = - ò E · dl
i
iA: DViA = 0
AB: DVAB
é1 1ù
=
qê - ú
4pe 0 ë r3 r1 û
1
BC: DVBC = 0
𝑞
é1 1ù
qê - ú
Cf: DVCf =
4pe 0 ë r2 r3 û
1
+
DViABCf
é1 1 1 1ù
1 é1 1ù
=
qê - + - ú =
q ê - ú = DVif
4pe 0 ë r3 r1 r2 r3 û 4pe 0 ë r2 r1 û
1
Example: Different Paths near Point Charge
f
DV = - ò E · dl
3. Arbitrary path
i
E · dl = Edl cosq
E · dl = Edlradial
f
DV = - ò Edlradial
i
+
é 1 1ù
DV =
q ê - ú = V f - Vi
4pe 0 êë rf ri úû
1
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