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PHY 2049 Lecture Notes
Chapter 25: Page 1 of 10
Reminder:
Kinetic Energy, Potential Energy, Energy Conservation
Kinetic energy (depends on velocity):
KE =
1
mv 2
2
Potential energy (depends on position in space):
U = PE = mgh
(gravitational potential energy in vicinity of Earth surface)
For a closed system (no external forces),
its Total Energy is conserved, i.e. does not change in time:
ETOT = KE + PE = const
vi = 0
Example:
A ball is dropped from a height h. What is the speed of the
ball when it hits the ground?
h
vf = ?
Solution:
Initial Total Energy: Ei = KEi +Ui = 0 + mgh
Final Total Energy: Ef = KEf + Uf = mvf2/2 + 0
Ei = E f
1
mgh = mv2f
2
v f = 2 gh
A. Korytov
PHY 2049 Lecture Notes
Chapter 25: Page 2 of 10
Reminder: Work, Change in Energy
In presence of an External Force,
the change in Total Energy equals to the Work done by the Force:
∆ETOT = WFexternal
f
F
Work done by the external force:
r
r
WFexternal (i → f ) = ∫ Fexternal ⋅ ds
f
i
i
Note: the result of integration does not depend on
the actual path. It depends on points i and f only!
ds
f
Fexternal
i
ds
F
Consider Special Case: Moving an object very
slowly from point i to point f with the external
force equal in magnitude, but opposite in
direction to the internal force (e.g. a person
raising a ball against gravitational force).
In this case Kinetic Energy is zero at the
beginning and at the end.
f
r
r r
r
= ∫ Fexternal ⋅ ds = − ∫ F ⋅ ds
f
∆ETOT = U f − U i = WFexternal
i
i
Thus, if one knows the internal forces (in all points of space), one
can calculate how potential energy depends on object’s position:
r r
U f − U i = − ∫ F ⋅ ds
f
i
A. Korytov
PHY 2049 Lecture Notes
Chapter 25: Page 3 of 10
Potential Energy of a charge in Electric Field.
Electric Field Potential.
f
q
i
ds
Charge q in electric field E(x,y,z)
If one knows the internal forces (we do: F=qE),
one can calculate how potential energy
depends on the charge position:
r r
 f r r
U f − U i = − ∫ F ⋅ ds = q − ∫ E ⋅ ds 
i
 i

f
E
One can define amount of energy per unit of charge q placed in some
electric field E:
f
r r
Ui
−
= − ∫ E ⋅ ds
q
q
i
Uf
Note that this quantity (amount of energy per unit of charge) does not
depend on charge. It depends only on the electric field.
This quantity is denoted V and called Electric Field Potential:
V =
U
q
The difference of Electric Field Potentials between two points (pathindependent!) is
r r
V f − Vi = − ∫ E ⋅ ds
f
i
Electric Field Potential Units: 1 Volt = 1 Joul/Coulomb
A. Korytov
PHY 2049 Lecture Notes
Chapter 25: Page 4 of 10
Electric Field from Electric Field Potential.
f
q
ds
i
Say, we knew E-field in every single point, i.e.
three numbers (Ex, Ey, Ez) per each point in
space.
Then, we converted this in one number per
each point in space:
r r
V f = − ∫ E ⋅ ds
f
E
i (V =0 )
Did we lose any information?
Consider a small displacement from point i (x,y,z) to point f (x+dx,y,z):
r r
r r
V f − Vi = V ( x + dx, y, z ) − V ( x, y, z ) = − ∫ E ⋅ ds ≈ − E ⋅ dx
f
i
r r
dV ≈ − E ⋅ dx = − E x dx
∂V ( x, y , z )
Ex = −
∂x
Similarly:
Ey = −
∂V ( x, y, z )
∂y
∂V ( x, y , z )
Ez = −
∂z
Little Miracle: No information is lost!
Knowledge of V(x,y,z) allows for recovering all three components of
the electric field vectors in all points.
A. Korytov
PHY 2049 Lecture Notes
Chapter 25: Page 5 of 10
Electric Field Potential of a point charge.
Find V for the field of a point charge Q:
f
i
ds
E
f
f
r r
Qk
Qk
V f − Vi = − ∫ E ⋅ ds = − ∫ Edx = − ∫ 2 dx =
x
x
i
i
i
f
f
i
Qk Qk
=
−
xf
xi
Therefore, one can write:
V=
Qk
+ Const
x
It is convenient to set Const=0:
Qk
V=
x
Or more generally:
V=
Q
k
r
A. Korytov
PHY 2049 Lecture Notes
Chapter 25: Page 6 of 10
Electric Field Potential (examples)
E
f
c
b
Find ∆V=Vf-Vi :
a
i
r r
a
a
a
V f − Vi = − ∫ E ⋅ ds = − ∫ E ⋅ ds ⋅ cosθ = − ∫ E ⋅ ds ⋅ = − E ∫ ds = − E c = − Ea
c
c (c )
c
i
(c)
(c)
f
Another way:
r r
r r
r r
V f − Vi = − ∫ E ⋅ ds = − ∫ E ⋅ ds − ∫ E ⋅ ds = − ∫ Edx − ∫ Edy cos 90 o = − Ea
f
i
(a )
(b )
( a)
(b )
Find ∆V for the field of a uniformly charged wire:
f
i
ds
E
f
f
r r
V f − Vi = − ∫ E ⋅ ds = − ∫ Edx = − ∫
f
i
i
i
r
λ
λ
dx = −
ln  f
2πε 0 x
2πε 0  ri



Find ∆V in conductor:
f
r r
r
V f − Vi = − ∫ E ⋅ ds = − ∫ 0ds = 0
f
i
i
A. Korytov
PHY 2049 Lecture Notes
Chapter 25: Page 7 of 10
Electric Field Potential of many charges
Due to electric field superposition law, Potential of N point charges:
N
Qi
V =∑ k
i =1 ri
Electric Potential due to a Distribution of Charge
dV = K dQ/r
dQ
r
The electric potential from a continuous distribution of charge is the
superposition (i.e. integral) of all the (infinite) contributions from
each infinitesimal dQ as follows:
V =
∫
K
dQ
r
and
Q = ∫ dQ
A. Korytov
PHY 2049 Lecture Notes
Chapter 25: Page 8 of 10
Electric Field Potential (examples)
Example:
A total amount of charge Q is uniformily distributed along a
thin circle of radius R. What is the electric potential at a
point P at the center of the circle?
Answer:
V=
R
P
x-axis
KQ
R
Example:
A total amount of charge Q is uniformily distributed along a
thin semicircle of radius R. What is the electric potential at a
point P at the center of the circle?
R
P
x-axis
KQ
=
V
Answer:
R
Example:
A total amount of charge Q is uniformily distributed
along a thin ring of radius R. What is the electric
potential at a point P on the z-axis a distance z from
the center of the ring?
Answer: V ( z ) =
R
P
z-axis
z
KQ
z2 + R2
Example:
A total amount of charge Q is uniformily distributed
on the surface of a disk of radius R. What is the
electric potential at a point P on the z-axis a distance z
from the center of the disk?
2 KQ
V
(
z
)
=
Answer:
R2
(
z2 + R2 − z
R
P
z-axis
z
)
A. Korytov
PHY 2049 Lecture Notes
Chapter 25: Page 9 of 10
Potential Energy of two charges.
Q
q
V
Charge Q creates potential V at some
point at a distance r away from the
charge:
V=
Q
k
r
Another charge q, being placed at this
point, will give rise to potential energy of the system U=qV:
U = qV =
qQ
k
r
Example:
Two balls of mass m=1 kg each and carrying charges Q=1 C each are fixed at a
distance r=1 m from each other. Find the final velocity of each of the balls after they
have been let go.
Solution:
The energy is conserved: ETOT at the beginning must be equal to ETOT at the end.
ETOT at the beginning = KE1 + KE2 + Usystem = 0 + 0 + Q2k/r.
ETOT at the end
= KE1 + KE2 + Usystem = mv2/2 + mv2/2 + 0.
mv2= Q2k/r
v≈100 km/s
Note that velocity needed to leave the Earth is ~11 km/s
A. Korytov
PHY 2049 Lecture Notes
Chapter 25: Page 10 of 10
Potential Energy of many charges.
Q1
A
r12
r13
B Q2
r23
CQ
3
Potential Energy of a system of three charges can
be evaluated by building the system from scratch
(no charges):
- No energy is needed to bring Q1 at point A;
- Brining the second charge at point B will require
U = Q2VB = Q2
Q1
QQ
k= 1 2k
r12
r12
- Brining the third charge at point C will require
 Q1
Q
U = Q3VC = Q3  k + 2
r23
 r13
 Q1Q3
QQ
k  =
k+ 2 3k
r13
r23

Total Energy Needed:
U=
QQ
QQ
Q1Q2
k+ 1 3k+ 2 3k
r12
r13
r23
Can be easily generalized to N charges:
U=
∑
all _ different _ ij _ pairs
QiQ j
rij
k
Note: number of different pairs for N charges is N*(N-1)/2
A. Korytov