ben gurion university of the negev week 3. electric potential

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

http://www.bgu.ac.il/atomchip

http://www.bgu.ac.il/nanocenter

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


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


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


Is the electric force a conservative force?


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



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.



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



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

qq

drrr

qqdW

r


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



dr

r2

r1



rF(r)

dr

r2

r1



Since F(r) · dr is hard to compute…

rF(r)

dr

r2

r1



Let’s resolve dr into dr║ and dr┴:

rF(r)

dr

r2

r1

r2

r1

rF(r)



Resolved into dr║ and dr┴, the path looks like this…

r2

r1

r2

r1

rF(r)



…which we arrange to look like this:

r2

r1

r2

r1

rF(r)



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)



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


Conclusion 1: The electrostatic force is conservative.

, 4

120

21

rr

qq

U


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

r2

r1


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


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


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



Answer:

Fq

θ

F-q

dr

rrFrrF ddU qq )(2)(2)(



Answer:

θ

F-q

drNot the same d!


Fq



Answer:

θ

F-q

dr


Fq



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


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


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.





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).





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)?



(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



(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.



(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.

ben gurion university of the negev week 3. electric potential

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