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Practice Exam 1 has been posted on course web site
1 hand graded question and 5 multiple choose questions
Question 1 (Chap. 16)
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 2 (Chap. 16)
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.
Integration
Divide shell into rings of charge, each delimited by the angle  and
the angle +
Use polar coordinates (r, ,).

R
Distance from center: d=(r-Rcos)
Rsin
Surface area of ring:

R
Rcos
2 (R sin  )
d
r
1
(Q)d
E 
4 0 (d 2  (Rsin  )2 )3/2

 2 Rsin  R 
Q  Q 


4 R 2
1 Q
(r  Rcos )
E
sin d
3/2

4 0 2 0 (r  Rcos )2  (Rsin )2 


A Solid Sphere of Uniformly Distributed Charge
What if charges are distributed throughout an object?
Step 1: Cut up the charge into shells

1 dQ
rˆ
For each spherical outside: dE 
2
40 r
shell:
inside: dE = 0
Outside a solid sphere of charge:

E
1
Q
rˆ
2
40 r
for r>R
R
r E
A Solid Sphere of Uniformly Distributed Charge
Inside a solid sphere of charge:

E
1 Q
rˆ
2
40 r
(volume of inner shells)
Q  Q
(volume of sphere)
4 / 3  r 3
r3
Q  Q
Q 3
3
4 / 3  R
R
E
1 Qr
r̂
3
4 0 R
for r<R
Why is E~r?
 On surface:
1 QR
1 Q
E
r̂ 
r̂
3
2
4 0 R
4 0 R
R
r
E
Chapter 17
Electric Potential
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?
(
)
-19
-18
=
14.4
J/C
-1.6
´
10
C
=
-2.3
´
10
J
(
)
U el = Vq
Exercise
2Å
1Å
What is the change in potential in going from 1Å to 2Å from the proton?
DV = V æ 2 Aö - V æ 1Aö = -7.2 V
è ø
è ø
What is the change in electric potential energy associated with
moving an electron from 1Å to 2Å from the proton?
DUel = U el æ 2 Aö - U el æ 1Aö = qV æ 2 Aö - qV æ 1Aö = qDV
è ø
è ø
è ø
è ø
(
)
DUel = -1.6 ´ 10 -19 C ( -7.2 J/C) = +1.15 ´ 10 -18 J
Does the sign make sense?
Potential Energy
To understand the dynamics of moving objects consider:
forces, momenta, work, energy
Introduced the concept of electric field E to deal with forces
Introduce electric potential – to consider 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
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.
Work done by internal forces:
f
DU el = -Wint = - ò Fint · dr
i
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 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
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
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
Question 1
What is the potential in the center of uniformly charged hollow
sphere? ([k=1/(40)])
Q
R
A)
B)
C)
D)
E)
k*Q/R2
k*Q/R
k*Q/(4R2)
0
Not enough information
Potential Inside a Uniformly Charged Hollow
Sphere

A 
A 

 

V A   E  dl    E  dl    E dr
A


1 Q
VA =
4pe 0 R
=0
B 
 A 
 
VB    E  dl  (  E  dl   E  dl )
B

1 Q
VB =
4pe 0 R
A
