The first midterm!!!
The first common hour midterm exam will be held on
Thursday October 1, 9:50 to 11:10 PM (at night) on the
Busch campus. You should go to the room corresponding to
the first 3 letters of your last name. If you have a conflict
with the exam time, please contact Prof. Cizewski
Cizewski@physics.rutgers.edu with your entire schedule for
the week of September 28 at your earliest convenience but
not later than 5:00 pm on Wednesday, September 23.
Aaa – Hoz
Hua – Moz
Mua – Shz
Sia – Zzz
ARC 103
Hill 114
PHY LH
SEC 111
Iclicker Question
A point charge +Q is placed
inside a neutral, hollow,
spherical conductor. As the
charge is moved around inside,
the electric field outside
A.
B.
C.
D.
+Q
is zero and does not change
is non-zero but does not change
is zero when centered but changes
is non-zero and changes
2
Iclicker Question
A point charge +Q is placed
inside a neutral, hollow,
spherical conductor. As the
charge is moved around inside,
the electric field outside
A.
B.
C.
D.
+Q
is zero and does not change
is non-zero but does not change
is zero when centered but changes
is non-zero and changes
3
Lecture 5. Electrostatic Potential Energy
This lecture is about forces and potential energy U(r), unit = joule, J
Next lecture is about electrical potential V(r), unit: volt, V
Don’t mix the two up.
𝑈 𝑟
𝑉 𝑟 =
𝑞
4
Lecture 5. Electrostatic Potential Energy
Coulomb’s Law and Superposition Principle – sufficient to
solve any electrostatic problem. Gauss’ Law simplifies
calculation of electric fields for symmetric charge
distributions. Why do we need anything else?
5
Lecture 5. Electrostatic Potential Energy
Coulomb’s Law and Superposition Principle – sufficient to
solve any electrostatic problem. Gauss’ Law simplifies
calculation of electric fields for symmetric charge
distributions. Why do we need anything else?
Example
Two charges q=+1C are 1 m apart at rest. We release the charges. What
would be the net kinetic energy (in SI units) of the system when the
charges are very far from one another?
+1𝐶
+1𝐶
1𝑚
A. k1
B. -k1
C. k2
D. -k2
E. 0
6
Iclicker question
Consider two concentric spherical metallic
shells, one with radius R and one with radius
2R. Both have the same charge Q. At a point
just inside the outer shell, the magnitude of
the electric field is
Q
Q
R
2R
A.
𝟐𝒌𝑸/𝑹𝟐
B.
𝒌𝑸/𝑹2
C.
𝒌𝑸/𝟐𝑹𝟐
D.
𝟎
E.
𝒌𝑸/𝟒𝑹𝟐
7
Iclicker question
Consider two concentric spherical metallic
shells, one with radius R and one with radius
2R. Both have the same charge Q. At a point
just inside the outer shell, the magnitude of
the electric field is
Q
Q
R
2R
A.
𝟐𝒌𝑸/𝑹𝟐
B.
𝒌𝑸/𝑹2
C.
𝒌𝑸/𝟐𝑹𝟐
D.
𝟎
E.
𝒌𝑸/𝟒𝑹𝟐
8
Iclicker Question
An insulating spherical shell of inner radius a and outer radius b is uniformly
charged with a positive charge density. The radial component of the electric
field, 𝑬𝒓 𝒓 , is best depicted by which figure?
A
B
C
D
9
Iclicker Question
An insulating spherical shell of inner radius a and outer radius b is uniformly
charged with a positive charge density. The radial component of the electric
field, 𝑬𝒓 𝒓 , is best depicted by which figure?
A
B
C
D
10
Lecture 5. Electrostatic Potential Energy
Coulomb’s Law and Superposition Principle – sufficient to
solve any electrostatic problem. Gauss’ Law simplifies
calculation of electric fields for symmetric charge
distributions. Why do we need anything else?
-
Electrostatic forces are conservative, and energy
conservation is a very powerful tool for solving a wide
range of problems.
-
The concept of the potential energy of the electrostatic
field is the first step towards the appreciation of the
energy of electromagnetic field.
11
Recall: Potential Energy in Gravitational Field
𝑎
𝐹𝑔
𝐹𝑢𝑠
𝑏
Work done by us while lifting the
body from a to b at constant speed:
𝑟
𝑏
𝑊𝑎→𝑏 "𝑢𝑠" =
𝑈 𝑟
𝐹𝑢𝑠 ∙ 𝑑𝑟
𝑎
𝑏
Work done by the grav. force Fg:
𝑊𝑎→𝑏 "𝑓𝑖𝑒𝑙𝑑" =
𝐹𝑔 ∙ 𝑑𝑟
𝑎
𝑏
∆𝑈 𝑟 = 𝑊𝑎→𝑏 "𝑢𝑠" =
𝐹𝑢𝑠 ∙ 𝑑𝑙
Units: Joules, J
𝑎
Change in potential energy is equal to the work done by external forces
(“us”) to move an object in gravitational field (assuming that kinetic energy
remains constant).
Potential energy: always the energy of interaction between masses, not a
characteristic of a single mass (the potential energy of the system “Earth +
mass m”).
12
Gravitational Potential Energy of the system “Earth + astronaut”
𝑎
𝐹𝑔
𝐹𝑢𝑠
𝑏
𝑟
Potential energy with respect to “” :
𝑟
𝑈 𝑟 −𝑈 ∞ =
𝑟
𝐹𝑢𝑠 ∙ 𝑑𝑟 =
∞
𝐺
∞
𝑈 𝑟
𝑟2
𝑈+𝐾
𝐾 𝑟2
𝑚𝐸 𝑚
𝑟 ∙ 𝑑𝑟𝑟 =
𝑟2
𝑟1
𝑟
𝑚𝐸 𝑚
𝑚𝐸 𝑚
𝑑𝑟
=
−𝐺
𝑟2
𝑟
𝐺
∞
𝑟
𝐾 𝑟1 = 0
𝑈 𝑟 + 𝐾 𝑟 = 𝑐𝑜𝑛𝑠𝑡
Energy conservation
13
Reference Point(s)
𝑟
𝑈 𝑟 =
𝐹𝑢𝑠 ∙ 𝑑𝑟
𝑠𝑜𝑚𝑒
𝑟𝑒𝑓.𝑝𝑜𝑖𝑛𝑡
Reference point: matter of convenience, only U
matters (because only forces can be measured,
and the forces depend on U, not U).
𝐹 𝑟 =−
𝜕𝑈 𝑟
𝑟
𝜕𝑟
“sea level” as a reference:
𝑈 𝑟 − 𝑈 𝑟𝐸 = −𝐺𝑚𝐸 𝑚
≅ 𝐺𝑚𝐸 𝑚
𝑟 − 𝑟𝐸
𝐺𝑚𝐸
=
𝑚
∆ℎ = 𝑚𝑔∆ℎ
𝑟𝐸 2
𝑟𝐸 2
≡
𝑟 − 𝑟𝐸 ≪ 𝑟𝐸
1 1
−
𝑟 𝑟𝐸
𝑔
U  h   mg h
14
Conservative Vector Fields
Gravitational and Electrostatic Fields are conservative.
The reason: both fields are central (a central force
depends only on the distance between interacting
objects and is directed along the line joining them).
∆𝑈 depends only on the initial and final
points of the trajectory:
𝐹𝐸 ∙ 𝑑𝑙 = 0
- for any loop
No closed E field lines in electrostatics:
15
Work in Electrostatic Field
Consider two point charges, +𝑞1 and +𝑞2 . Place +𝑞1 at the origin.
𝑞2
𝑞1
r
𝐹𝑢𝑠
𝐹𝑢𝑠
𝑑𝑙
𝑞2 𝐸 𝑟
𝑞2 𝐸 𝑟
Let’s bring 𝑞2 from  to () P at a distance r from the origin. The work done by us:
𝑟
𝑊∞→𝑟
"𝑢𝑠"
=
𝑟
𝐹𝑢𝑠 ∙ 𝑑𝑙 =
∞
∞
1 𝑞1 𝑞2
𝑞1 𝑞2 1 1
𝑞1 𝑞2 1
−
𝑟 ∙ 𝑑𝑟 =
−
=
2
4𝜋𝜀0 𝑟
4𝜋𝜀0 𝑟 ∞
4𝜋𝜀0 𝑟
along the green
trajectory
𝑈 𝑟
We worked against the electrostatic
forces and increased the potential
energy of this charge distribution.
∝ 1/𝑟
r
16
Electrostatic Potential Energy
𝑟
𝑟
𝑈 𝑟 =
𝐹𝑢𝑠 ∙ 𝑑𝑙 = −
𝑠𝑜𝑚𝑒
𝑟𝑒𝑓.𝑝𝑜𝑖𝑛𝑡
- the potential energy of the
system “charge 𝑞 + external
electrostatic field 𝐸 𝑟 ”
𝑞𝐸 𝑟 ∙ 𝑑𝑙
𝑠𝑜𝑚𝑒
𝑟𝑒𝑓.𝑝𝑜𝑖𝑛𝑡
𝑞1
𝑞2
r
()P
()P
𝑈 𝑟 =−
()Q
()Q
𝑞2 𝐸1 𝑟 ∙ 𝑑𝑙 = −
∞
𝑞1 𝐸2 𝑟 ∙ 𝑑𝑙 =
∞
𝑈 𝑟
𝑈 𝑟
𝑞1 𝑞2 1
4𝜋𝜀0 𝑟
r
𝐸𝑡𝑜𝑡 = 𝐾 𝑟 + 𝑈 𝑟
()P
++ or --
()Q
r
+- or -+
17
Iclicker Question
Two charges q=+1C are 1 m apart at rest. We release the charges. What
would be the net kinetic energy of the system when the charges are very
far from one another?
+1𝐶
+1𝐶
𝑘𝑞1 𝑞2
𝑈 𝑟 =
𝑟
1𝑚
A. k1 J
B. -k1 J
C. k2 J
D. -k2 J
E. 0
18
Iclicker Question
Two charges q=+1C are 1 m apart at rest. We release the charges. What
would be the net kinetic energy (in SI units) of the system when the
charges are very far from one another?
+1𝐶
+1𝐶
1𝑚
A. k1
B. -k1
C. k2
D. -k2
E. 0
19
Example
𝑒𝑛𝑑
∆𝑈 𝑟 = −
𝑄𝐸 𝑟 ∙ 𝑑𝑙
𝑠𝑡𝑎𝑟𝑡
1,0
= −𝑄
1,−2
𝐸 𝑥, 𝑦 ∙ 𝑑𝑥 𝑥 +
0,0
𝐸 𝑥, 𝑦 ∙ 𝑑𝑦 𝑦
1,0
1,0
= −𝑄
1,−2
2𝑥 + 3𝑦 ∙ 𝑑𝑥 𝑥 +
0,0
2𝑥 + 3𝑦 ∙ 𝑑𝑦 𝑦
1,0
= +3 2 − 6
J = −12 J
20
From 𝑼 𝒓 to 𝑭 𝒓
𝑟
𝑈 𝑟 =−
𝐹 𝑟 ∙ 𝑑𝑙
One-dimensional case:
𝑟𝑒𝑓.𝑝𝑜𝑖𝑛𝑡
+𝑞
r
()P
𝑈 𝑟
−𝑞
()Q
r0
r
𝑑𝑈 𝑥
𝐹 𝑥 =−
𝑥
𝑑𝑥
𝑞1 𝑞2 1
𝑞2
𝑈 𝑟 =
=−
4𝜋𝜀0 𝑟
4𝜋𝜀0 𝑟
𝐹 𝑟 =−
𝑑𝑈 𝑟
𝑟
𝑑𝑟
𝑑𝑈 𝑟 /𝑑𝑟 – positive, 𝐹 𝑟 is directed along −𝑟.
(as it should be, because of attraction)
Q: is it possible to find E(r) if we know U(r) only at this particular point?
21
Iclicker Question
U(x) is shown on the plot:
U(x)
x
Which plot shows
the correct F(x) ?
F(x)
𝑑𝑈 𝑥
𝐹 𝑥 =−
𝑥
𝑑𝑥
22
Iclicker Question
U(x) is shown on the plot:
U(x)
x
Which plot shows
the correct F(x) ?
F(x)
23
Choice of Reference Point
The reference point can be any point. By shifting the reference point, we add the same
constant values to the potential energies at all other points:
𝑈 𝑟
𝑟
𝑈 𝑟 =−
r
𝑞𝐸 ∙ 𝑑𝑙
𝑟𝑟𝑒𝑓
This ambiguity shouldn’t bother us, because only the potential difference is meaningful
(𝐹𝐸 𝑟 = −
𝑑𝑈 𝑟
𝑑𝑟
𝑟), and the difference doesn’t depend on the reference point.
𝑟
Localized (“finite”)charge distribution.
The natural choice is to set U=0 at an infinitely distant point.
Again, it’s just a matter of convenience. For many problems,
it’s convenient to set the potential energy of charges on the
“ground” to zero.
O
𝑟𝑟𝑒𝑓 
24
“Infinite” Charge Distributions
- charge distribution that extends to infinity (e.g., a charged infinite plane).
The choice of an infinitely distant reference point is inconvenient – all electrostatic
energies would be infinitely large. The remedy is to choose less remote reference
point.
Example: uniformly charged plane.
y
Convenient reference point:
any point within the plane (x=0). All points
within the plane have the same 𝑈 (we can move
the charge over the plane without doing any
work).
𝟎, 𝟎
x
𝟎, 𝒚
𝑥,𝑦
𝒙, 𝒚
∆𝑈 = 𝑈 𝑥, 𝑦 − 𝑈 0,0 = −
𝑞𝐸 𝑥 ∙ 𝑑𝑙
0,0
U(x)
x
=−
0,𝑦
𝑞𝐸
0,0
𝑥 ∙ 𝑑𝑦 −
𝑥,𝑦
𝑞𝐸
0,𝑦
𝜎
𝑥 ∙ 𝑑𝑥 = −𝑞 2𝜖 𝑥
0
25
Superposition: several interacting charges
𝑞2
𝑟02
𝑟01
𝑞1
𝑞3
If the test charge interacts with several charges, the potential
energy of the interaction between the test charge and all other
charges is an algebraic sum of 𝑈0𝑖 :
𝑟03
𝑞0
𝑈0 =
Example: a positive
test charge interacting
with a dipole
𝑈0 =
𝑈0
1 𝑞0 𝑞1 𝑞0 𝑞2 𝑞0 𝑞3
𝑞0
+
+
=
4𝜋𝜀0 𝑟01
𝑟02
𝑟03
4𝜋𝜀0
𝑖
𝑞𝑖
𝑟0𝑖
𝑞0 𝑞1 1
𝑞0 𝑞2 1
𝑞0 𝑞 1 1
+
=
−
4𝜋𝜀0 𝑟01
4𝜋𝜀0 𝑟02
4𝜋𝜀0 𝑟1 𝑟2
+𝑞 𝑞0
−𝑞
𝑈0 𝑥
𝑥
26
Conclusion
Conservative vector fields  Potential energy.
Electrostatic potential energy.
Calculation of U for simple charge distributions.
Next time: Lecture 6. Calculation of Electric
Potentials
§§ 23.3 - 23.5
27
Appendix I. Conservative Vector Fields
Conservative vector fields: the work by the field on a “charge” depends only on
the initial and final points of a trajectory, but not on the shape of the trajectory.
Math involved:
𝑐𝑢𝑟𝑙 𝐸 𝑟 =
𝑐𝑢𝑟𝑙 𝑎 𝑟 = 0
𝑎 ∙ 𝑑𝑙 = 0
𝜕𝐸𝑦 𝜕𝐸𝑥
𝜕𝐸𝑧 𝜕𝐸𝑦
𝜕𝐸𝑥 𝜕𝐸𝑧
−
𝑥+
−
𝑦+
−
𝑧
𝜕𝑦
𝜕𝑧
𝜕𝑧
𝜕𝑥
𝜕𝑥
𝜕𝑦
- circulation of
a vector field
- looks scary…
But we can tell at a glance if the curl is zero or not: think of a vector
field as a water flow and place a paddle wheel at the point in question.
If the wheel rotates, the curl is non-zero.
𝑦
𝑦
𝑎 𝑥, 𝑦 = 𝑦𝑦
𝑥
𝜕𝐸𝑦 𝜕𝐸𝑦
=
=0
𝜕𝑧
𝜕𝑥
𝑐𝑢𝑟𝑙 𝑎 𝑟 = 0
𝑎 𝑥, 𝑦 = 𝑥𝑦
𝑥
𝜕𝐸𝑦
=1
𝜕𝑥
𝑐𝑢𝑟𝑙 𝑎 𝑟 ≠ 0
28
Appendix II: Total energy of interactions between several charges
𝑞2
𝑟02
The total work required to assemble this charge distribution:
𝑟12
𝑞0
𝑞1
𝑈Σ =
𝑟03
𝑟13
𝑞3
𝑞0 𝑞1 1
𝑞0 𝑞2 1
𝑞1 𝑞2 1
1
+
+
+⋯=
4𝜋𝜀0 𝑟01
4𝜋𝜀0 𝑟02
4𝜋𝜀0 𝑟12
4𝜋𝜀0
𝑖<𝑗
𝑞𝑖 𝑞𝑗
𝑟𝑖𝑗
- all possible pairs, but each pair
we count just once.
Example: four equal charges in the corners of a square.
𝑞
𝑎
𝑞
𝑎
𝑞
𝑞
𝑈𝑒
𝑡𝑜𝑡𝑎𝑙
𝑞𝑖 𝑞𝑗
𝑞2
𝑞2
=
=4
+2
4𝜋𝜖0 𝑟𝑖𝑗
4𝜋𝜖0 𝑎
4𝜋𝜖0 𝑎 2
𝑎𝑙𝑙 𝑝𝑎𝑖𝑟𝑠
each pair
counts only
once
The energy is positive (the charges repel
each other), the system can do some work
if we let the charges go.
On the other hand, the potential energy of one of these charges in the field due to the
other three charges:
𝑞2
𝑞2
𝑈𝑒 = 2
+
4𝜋𝜖0 𝑎 4𝜋𝜖0 𝑎 2
29