ben gurion university of the negev week 3. electric potential
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Ben Gurion University of the Negevwww.bgu.ac.il/atomchip, www.bgu.ac.il/nanocenter
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 Source: Halliday, Resnick and Krane, 5th Edition, Chap. 28.
Physics 2B for Materials and Structural EngineeringPhysics 2B for Materials and Structural Engineering
Lecturer: Daniel Rohrlich Teaching Assistants: Oren Rosenblatt, Irina Segel
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
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?
. )(2
1
r
r
rrF dW
r1
r2
, )()(2
1
2
1
r
r
r
r
rrFrrF ddW
Conservative forces
If F is conservative, then
where the color framing the integral indicates the path taken from r1 to r2.
r1
r2
r1
r2
Conservative forces
If F is conservative, then
where the color framing the integral indicates the path taken from r1 to r2.
, )()(02
1
2
1
r
r
r
r
rrFrrF dd
Conservative forces
If F is conservative, then
r1
r2. )(
)()(01
2
2
1
rrF
rrFrrF
r
r
r
r
d
dd
Conservative forces
Conclusion: if F is conservative, then
)(0 rrF d
r1
r2
Conservative forces
If F is conservative, then we can define a potential energy U(r) such that
U(r2) – U(r1) =
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
r2
, )(2
1
r
r
rrF d
Conservation of energy
If F is conservative, then we can define a potential energy U(r) such that
U(r2) – U(r1) =
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 impliesa conserved total energy. r1
r2
, )(2
1
r
r
rrF d
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 rrF dqFq 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
which means that every time the charge makes a complete loop in r, it gets more kinetic energy!
, )(0 rrF dqFq Fq
Fq
Electric potential energy
We will prove that the electrostatic force is conservative!First, we will calculate the work W done by the electric forceif 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 forceif 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 forceif 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
rF(r)dr
. 4
4
1 )(
120
21
22
21
0
11
rr
rrF
r
r
drrr
qqdW
r
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
r1
r
r2
r1
F(r)r
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?
dr
r2
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?
rF(r)
dr
r2
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…
rF(r)
dr
r2
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┴:
rF(r)
dr
r2
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┴:
rF(r)
dr
r2
r1
r2
r1
rF(r)
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
r1
r2
r1
rF(r)
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
r1
r2
r1
rF(r)
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
r1
r2
r1
rF(r)
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:
.
44
1
120
212
2
21
0
1
rr
qqdr
rr
qqW
r
Electric potential energy
Conclusion 1: The electrostatic force is conservative.
, 4
120
21
rr
U
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
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
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?
r2
r1
r3
r2
r1
. 4
1
23
32
13
31
12
21
0
rrrrrr
qqqqqqU
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
r2
r1
r3
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.)
Answer:
Fq
θ
F-q
dr
rrFrrF ddU qq )(2)(2)(
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.)
Answer:
θ
F-q
drNot the same d!
rrFrrF ddU qq )(2)(2)(
Fq
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.)
Answer:
θ
F-q
dr
rrFrrF ddU qq )(2)(2)(
Fq
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.)
Answer:
Fq
θ
F-q
Δr
Δr
. coscos2
)(2
)(2)(2)(
Eqdd
Eq
ddU qq
rrFrrF
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 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 one point charge in an external electric field – for example, we calculated the potential energy of each charge in a dipole in a constant electric field.
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.
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.
Electric potential
The units of electric potential must be energy/charge. Our unit is the volt; 1 V = J/C = joules/coulombs.
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.
Electric potential
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).
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.
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
and since the mass of the electron is me = 9.1 × 10–31 kg, the speed is
, J 10 3.2eV
J 10 1.6 eV] 2[
2
1 19192
vme
. m/s 10 4.8kg 10 9.1
J 10 3.22 531–
19
v
Electric potential
•The electric potential V(r) at a point r, due to a system of n point charges qi located at ri, is
•The electric potential V(r) at a point r, due to a continuous charge density ρ(r), is
. 4
)(1 0
n
i i
iqV
rrr
. '4
)'()(
0
3
rr
rr
r'dV
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
d/2(x,y,z)
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(x,y,z)
. )2
( re whe
, 1
1
4
),,(
222
0
dzyxr
rr
qzyxV
r+
r–
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
y
–L/2
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.
. 4/2/
4/2/ln
4
ln4
4)(
22
22
0
/2
/2
22
0
2/
2/22
0
yLL
yLL
yzz
yz
dzyV
L
L
L
L
z
y
–L/2
L/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
and C = 0 yields V(0) = 0.
, 8
4
)()(3
0
2
30
CR
qr
R
qrdrrdrErV
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)/8πε0R
3 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.
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