Ben Gurion University of the Negev
www.bgu.ac.il/atomchip, www.bgu.ac.il/nanocenter
Physics 2 for Electrical Engineering
Lecturers: Daniel Rohrlich , Ron Folman
Teaching Assistants: Ben Yellin, Yoav Etzioni
Week 3. Electric potential energy and potential – Conservative
forces • conservation of energy • electric potential energy • potential
energy of several charges • continuous charge distributions • electric
potential
Sources: Halliday, Resnick and Krane, 5th Edition, Chap. 28.
Conservative forces
Suppose a force F acting on a particle depends only on the
position r of the particle, i.e. F = F(r). The work W done by
F in moving the particle from r1 to r2 is defined to be
r2

W  F (r )  dr .
r2
r1
If W does not depend on the choice of a path
for r, then the force F is called conservative.
Where does this name come from?
r1
Conservative forces
If F is conservative, then
r2

r2

W  F(r )  dr  F(r )  dr ,
r1
r1
r2
where the color framing the integral indicates
the path taken from r1 to r2.
r1
Conservative forces
If F is conservative, then
r2

r2

0  F(r )  dr  F(r )  dr ,
r1
r1
r2
where the color framing the integral indicates
the path taken from r1 to r2.
r1
Conservative forces
If F is conservative, then
r2

r1

0  F(r )  dr  F(r )  dr
r1
r2
r2

 F(r )  dr .
r1
Conservative forces
Conclusion: if F is conservative, then
r2

0  F(r)  dr
r1
Conservative forces
If F is conservative, then we can define a potential energy
U(r) such that
r2

U(r2) – U(r1) =  F(r )  dr ,
r1
r2
and U(r) is a single-valued function of r
(i.e. U(r) is a scalar field). With U(r) we
can define a conserved total energy, the
sum of the kinetic energy and the potential
energy.
r1
Conservation of energy
If F is conservative, then we can define a potential energy
U(r) such that
r2

U(r2) – U(r1) =  F(r )  dr ,
r1
r2
and U(r) is a single-valued function of r
(i.e. U(r) is a scalar field). With U(r) we
can define a conserved total energy, the
sum of the kinetic energy and the potential
energy.
Conclusion: a conservative force implies
a conserved total energy.
r1
Conservation of energy
Is the electric force a conservative force?
Conservation of energy
Is the electric force a conservative force? If not, then we can
use an electric field to get energy for free. Here’s how:
Conservation of energy
Is the electric force a conservative force? If not, then we can
use an electric field to get energy for free. Here’s how:
If the electric force Fq on a particle of charge q is not
conservative, then for some loop in r we can write

0  Fq (r)  dr .
Fq
Fq
Fq
Conservation of energy
Is the electric force a conservative force? If not, then we can
use an electric field to get energy for free. Here’s how:
If the electric force Fq on a particle of charge q is not
conservative, then for some loop in r we can write

0  Fq (r)  dr ,
Fq
Fq
which means that every time the charge
makes a complete loop in r, it gets more
kinetic energy!
Fq
Electric potential energy
We will prove that the electrostatic force is conservative!
First, we will calculate the work W done by the electric force
if we move a charge q1 in a straight line towards a charge q2.
Then we will prove that W does not depend on the path.
Electric potential energy
We will prove that the electrostatic force is conservative!
First, we will calculate the work W done by the electric force
if we move a charge q1 in a straight line towards a charge q2.
Then we will prove that W does not depend on the path.
The charge q2 is always at r2 while the charge q1 moves in a
straight line to r1 from a point infinitely far away.
r2
r1
Electric potential energy
We will prove that the electrostatic force is conservative!
First, we will calculate the work W done by the electric force
if we move a charge q1 in a straight line towards a charge q2.
Then we will prove that W does not depend on the path.
The charge q2 is always at r2 while the charge q1 moves in a
straight line to r1 from a point infinitely far away.
r1
W
r1
q1q2
 F(r)  dr   4 0  r2  r 2 dr
1

r 
q1q2

4 0 r2  r1
.
F(r)
dr r1
r
r2
Electric potential energy
But now suppose the charge q1 reached the point r1 by the path
indicated. What is the work done by the electric force?
r2
r
r1
Electric potential energy
But now suppose the charge q1 reached the point r1 by the path
indicated. What is the work done by the electric force?
r2
F(r)
dr
r
r1
Electric potential energy
But now suppose the charge q1 reached the point r1 by the path
indicated. What is the work done by the electric force?
r2
F(r)
dr
r
r1
Electric potential energy
But now suppose the charge q1 reached the point r1 by the path
indicated. What is the work done by the electric force?
Since F(r) · dr is hard to compute…
r2
F(r)
dr
r
r1
Electric potential energy
But now suppose the charge q1 reached the point r1 by the path
indicated. What is the work done by the electric force?
Let’s resolve dr into dr║ and dr┴:
r2
F(r)
dr
r
r1
Electric potential energy
But now suppose the charge q1 reached the point r1 by the path
indicated. What is the work done by the electric force?
Let’s resolve dr into dr║ and dr┴:
r2
F(r)
dr
r
r1
Electric potential energy
But now suppose the charge q1 reached the point r1 by the path
indicated. What is the work done by the electric force?
Resolved into dr║ and dr┴, the path looks like this…
r2
F(r)
r
r1
Electric potential energy
But now suppose the charge q1 reached the point r1 by the path
indicated. What is the work done by the electric force?
…which we arrange to look like this:
r2
F(r)
r
r1
Electric potential energy
But now suppose the charge q1 reached the point r1 by the path
indicated. What is the work done by the electric force?
And now we are back to the same integral we had before, since
integration dr┴ does not contribute anything:
r2
F(r)
r
r1
Electric potential energy
But now suppose the charge q1 reached the point r1 by the path
indicated. What is the work done by the electric force?
And now we are back to the same integral we had before, since
integration dr┴ does not contribute anything:
r1
q1q2
W 
dr  
2
4 0
4 0 r2  r1


r

r
2

1

q1q2
F(r)
r
r1
.
r2
Electric potential energy
Conclusion 1: The electrostatic force is conservative.
Electric potential energy
Conclusion 1: The electrostatic force is conservative.
Conclusion 2: The electric potential energy of two charges q1
and q2 at points r1 and r2, respectively, is
q1q2
U
4 0 r2  r1
,
regardless of how the charges were assembled.
Remember Faraday?
We had some basic rules and observations
about field lines:
•They never start or stop in empty space –
they stop or start on a charge or extend to
infinity.
•They never cross – if they did, a small
charge placed at the crossing would show
the true direction of the field there.
• The density of field lines in one
direction is proportional to the strength of
the field in the perpendicular direction.
Remember Faraday?
Now we can add a new basic rule about
field lines:
•They never start or stop in empty space –
they stop or start on a charge or extend to
infinity.
•They never cross – if they did, a small
charge placed at the crossing would show
the true direction of the field there.
• The density of field lines in one
direction is proportional to the strength of
the field in the perpendicular direction.
• There are no closed loops!
E
E
E
E
Remember Faraday?
Now we can add a new basic rule about
field lines:
•They never start or stop in empty space –
they stop or start on a charge or extend to
infinity.
•They never cross – if they did, a small
charge placed at the crossing would show
the true direction of the field there.
• The density of field lines in one
direction is proportional to the strength of
the field in the perpendicular direction.
• There are no closed loops!
E
E
E
E
Electric potential energy
What is the electric potential energy of three charges q1, q2 and
q3 at points r1 r2 and r3, respectively?
r3
r2
r1
Electric potential energy
What is the electric potential energy of three charges q1, q2 and
q3 at points r1 r2 and r3, respectively? It is
 q1q2
q1q3
q2 q3
U



4 0  r2  r1 r3  r1 r3  r2
1
r3



.
r2
r1
Electric potential energy
What is the electric potential U(θ) energy of two charges q and
–q separated by a rod of length d, when it tilts at an angle θ in a
constant electric field E? (Assume U(90°) = 0.)
Fq
θ
F-q
Electric potential energy
What is the electric potential U(θ) energy of two charges q and
–q separated by a rod of length d, when it tilts at an angle θ in a
constant electric field E? (Assume U(90°) = 0.)
dr
θ
Fq
F-q


Answer: U ( )  2 Fq (r)  dr  2Fq (r)  dr
Electric potential energy
What is the electric potential U(θ) energy of two charges q and
–q separated by a rod of length d, when it tilts at an angle θ in a
constant electric field E? (Assume U(90°) = 0.)
Not the same d!
dr
θ
Fq
F-q


Answer: U ( )  2 Fq (r)  dr  2Fq (r)  dr
Electric potential energy
What is the electric potential U(θ) energy of two charges q and
–q separated by a rod of length d, when it tilts at an angle θ in a
constant electric field E? (Assume U(90°) = 0.)
dr
θ
Fq
F-q


Answer: U ( )  2 Fq (r)  dr  2Fq (r)  dr
Electric potential energy
What is the electric potential U(θ) energy of two charges q and
–q separated by a rod of length d, when it tilts at an angle θ in a
constant electric field E? (Assume U(90°) = 0.)
Δr
θ
F-q
Fq
Δr


Answer: U ( )  2 Fq (r )  dr  2Fq (r )  dr
d

 2( Eq) cos    Eqd cos .
2

Electric potential
We have seen that “electric potential energy” can include all
the charge in a system – for example, we calculated the total
potential energy of a system of three point charges. Or it can
mean the potential energy of only some of the charges in an
external electric field – for example, we calculated the potential
energy of a dipole in a constant electric field.
Electric potential
We have seen that “electric potential energy” can include all
the charge in a system – for example, we calculated the total
potential energy of a system of three point charges. Or it can
mean the potential energy of only some of the charges in an
external electric field – for example, we calculated the potential
energy of a dipole in a constant electric field.
In the case of one point charge in an external electric field,
what we most often calculate is not the electric potential energy
but rather the electric potential.
Electric potential
We have seen that “electric potential energy” can include all
the charge in a system – for example, we calculated the total
potential energy of a system of three point charges. Or it can
mean the potential energy of only some of the charges in an
external electric field – for example, we calculated the potential
energy of a dipole in a constant electric field.
In the case of one point charge in an external electric field,
what we most often calculate is not the electric potential energy
but rather the electric potential.
The relation between the electric potential energy of a point
charge vs. the electric potential at the position of the point
charge is exactly analogous to the relation between the electric
force and the electric field.
Electric potential
We have seen that “electric potential energy” can include all
charge in a system – for example,
weiscalculated
thefield
totalat
E(r)
the electric
Fthe
q(r) is the electric force on
energy
of a system
point
charges.
Or it can
point
r.
apotential
point charge
q located
at r. of threethe
mean the potential energy of only one point charge in an
external electric field – for example, we calculated the potential
energy
of each
charge
in a dipole inV(r)
a constant
electric potential
field.
is the electric
U(r)
is the
potential
energy
at the point r.
of a point charge q at r.
In the latter case – the case of a point charge in an external
electric field – what we most often calculate is not the electric
The relation between the electric potential energy of a point
charge vs. the electric potential at the position of the point
charge is exactly analogous to the relation between the electric
force and the electric field.
Electric potential
Fq(r) is the electric force on
a point charge q located at r.
E(r) is the electric field at
the point r.
U(r) is the potential energy
of a point charge q at r.
V(r) is the electric potential
at the point r.
The units of electric potential must be energy/charge. Our unit
is the volt; 1 V = J/C = joules/coulombs.
Electric potential
Fq(r) is the electric force on
a point charge q located at r.
E(r) is the electric field at
the point r.
U(r) is the potential energy
of a point charge q at r.
V(r) is the electric potential
at the point r.
The units of electric potential must be energy/charge. Our unit
is the volt; 1 V = J/C = joules/coulombs.
Conversely, eV (the electron volt) is a unit of energy because it
is charge (e) × potential (V).
Electric potential
Example: An electron accelerates from rest through an electric
potential of 2 V. What is the final speed of the electron?
Electric potential
Example: An electron accelerates from rest through an electric
potential of 2 V. What is the final speed of the electron?
Answer: The electron acquires 2 eV kinetic energy. We have
1

2
19 J 
19
me v  [2 eV] 1.6  10

3.2

10
J ,

2
eV 

and since the mass of the electron is me = 9.1 × 10–31 kg, the
speed is
v 
2  3.2  10 19 J
9.1  10 – 31 kg
 8.4  10 5 m/s .
Electric potential
•The electric potential V(r) at a point r, due to a system of n
point charges qi located at ri, is
n
V (r ) 

i 1
qi
.
4 0 ri  r
•The electric potential V(r) at a point r, due to a continuous
charge density ρ(r), is
 (r ' )
V (r )  d r'
4 0 r 'r

3
.
Electric potential
Example 1: What is the electric potential at a point (x,y,z) due
to a dipole made of charges q at (0,0,d/2) and –q at (0,0, –d/2)?
z
d/2
(x,y,z)
–d/2
Electric potential
Example 1: What is the electric potential at a point (x,y,z) due
to a dipole made of charges q at (0,0,d/2) and –q at (0,0, –d/2)?
Answer:
z
d/2
d 2
whe re r  x  y  ( z  )
2
2
r–
r+
–d/2
1
1
V ( x, y , z ) 
   ,
4 0  r r 
q
(x,y,z)
2
.
Electric potential
Example 2: A rod with a uniform charge density λ lies along
the z-axis between (0,0,–L/2) and (0,0,L/2). Let’s calculate the
electric potential V(y) at a point (0,y,0) on the y-axis.
z
L/2
y
–L/2
Electric potential
Example 2: A rod with a uniform charge density λ lies along
the z-axis between (0,0,–L/2) and (0,0,L/2). Let’s calculate the
electric potential V(y) at a point (0,y,0) on the y-axis.
L/2
V ( y) 
z
L/2
y
–L/2

dz
2
2
4

z

y
L / 2
0


4 0
 
2
2
ln
z

z

y
 

L/2


  L/2
 L / 2  L2 / 4  y 2


ln 
4 0   L / 2  L2 / 4  y 2


 .


Halliday, Resnick and Krane, 5th Edition, Chap. 28, Prob. 4:
The electric field inside a nonconducting, uniform sphere of
radius R has magnitude E = qr/4πε0R3, where q is the total
charge on the sphere and r is the distance from its center.
(a) Given V(0) = 0, compute V(r).
(b) What is the difference in electric potential between the
surface of the sphere and its center?
(c) Given V(∞) = 0, show that V(r) = q(3R2 –r2)/8πε0R3 inside
the sphere. Why does this answer differ from the answer in
part (a)?
Halliday, Resnick and Krane, 5th Edition, Chap. 28, Prob. 4:
The electric field inside a nonconducting, uniform sphere of
radius R has magnitude E = qr/4πε0R3, where q is the total
charge on the sphere and r is the distance from its center.
(a) Given V(0) = 0, compute V(r).
Answer: Symmetry dictates that E is radial. So we have to
integrate E(r) = qr/4πε0R3 with respect to dr and, if necessary,
add a constant to make V(0) = 0. Thus


V (r )   drE (r )   dr
and C = 0 yields V(0) = 0.
qr
4 0 R
3

qr 2
8 0 R
3
C ,
Halliday, Resnick and Krane, 5th Edition, Chap. 28, Prob. 4:
The electric field inside a nonconducting, uniform sphere of
radius R has magnitude E = qr/4πε0R3, where q is the total
charge on the sphere and r is the distance from its center.
(b) What is the difference in electric potential between the
surface of the sphere and its center?
Answer: The question asks what is V(R) – V(0), and the
answer is – q/8πε0R.
Halliday, Resnick and Krane, 5th Edition, Chap. 28, Prob. 4:
The electric field inside a nonconducting, uniform sphere of
radius R has magnitude E = qr/4πε0R3, where q is the total
charge on the sphere and r is the distance from its center.
(c) Given V(∞) = 0, show that V(r) = q(3R2 –r2)/8πε0R3 inside
the sphere. Why does this answer differ from the answer in
part (a)?
Answer: Outside the sphere, the potential must be V(r) =
q/4πε0r, because the sphere looks just like a point charge q, and
then we also satisfy V(∞) = 0. This outside potential must
match the inside potential V(r) = q(3R2 –r2)/8πε0R3 at r = R
and, indeed, both expressions yield V(R) = q/4πε0R. So V(r) is
the same potential as in (a); they differ only by a constant so
that in (a) we have V(0) = 0 and here we have V(0) = 3q/8πε0R.