Lecture 4 Work, Electric Potential and Potential Energy Ch. 25people.virginia.edu/~ral5q/classes/phys632/summer08... · 2008-07-10 · 1 Lecture 4 Work, Electric Potential and Potential

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Lecture 4 Work, Electric Potentialand Potential Energy Ch. 25

• Topics• Work, electric potential energy and electric potential• Calculation of potential from field• Potential from a point charge• Potential due to a group of point charges, electric dipole• Potential due to continuous charged distributions• Calculating the electric field from a potential• Electric potential energy from a system of point charges• Equipotential Surface• Potential of a charged isolated conductor

• Demos• Teflon and silk• Charge Tester, non-spherical conductor, compare charge density at different radii.

•Elmo•Potential in the center of four charges•Potential of a electric dipole

• Polling

2

W =!F !d!s

i

f

" = q!E !d!s

i

f

"

!U = "W

3

!V =!U

q

4

Work, Potential Energy and Electric potential• The electric force is mathematically the same as gravity so it too must

be a conservative force. We want to show that the work done isindependent of the path and only depends on the endpoints. Then

the force is said to be a conservative force.

• First start with work Work done by the electric force =

• Then we will find it useful to define a potential energy as is the casefor gravity.

• And the electric potential

W =!F !d!s

i

f

" = q!E !d!s

i

f

"

!V =!U

q

!U = "W

5

Lets start with a uniform electric field and findthe work done for a positive test charge.

b

a

E

P2 P3

P1

c

W = q!E !d!s

i

f

" = qE cos# s

6

Find work done along path W12 for a positive testcharge

b

a

E

P2 P3

P1

c

F

W12= 0W

12= qE cos90 a

W = q!E !d!s

i

f

" = qE cos# s

7

Find Work along path W23

b

a

E

P2 P3

P1

c

F

W

23= qEb

W = q!E !d!s

i

f

" = qE cos# s

W23= qE cos0 b

8

W12 + W23 = 0 + qEb =qEb

Compare this work done along path W13

9

Work done along path W13

b

a

P2 P3

P1

c

Fθ

W13= qEcos! c = qE

b

cc = qEb

W = q!E !d!s

i

f

" = qE cos# s

θ

10

Conclusion• Work done along path W12 + W23 = W13.

• Work is independent of the particular path.• Although we proved it for a uniform field, it istrue for any field that is a only a function of r andis along r.• It only depends on the end points i and f.

•This means we can define a function at everypoint in space and when we take the difference ofthat function between any two points, it is equalto the negative of the work done.

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b

a

P2 P3

P1

c

Fθ

When we go from P1 to P3 we evaluate the Work function atP3 and subtract the value at P1 and then the a differenceequals the negative of the work done in going form P1 to P3.This function is called the potential energy function

12

Example of finding the Potential EnergyFunction U in a Uniform Field

What is the electric potential difference for a unit positivecharge moving in an uniform electric field from a to b?

E

Ed

a b

!U =Ub "Ua = "W = " q!E #d!s

a

b

$ = "Eq dxa

b

$ = "Eq(xb " xa)

abUUU !="

If we set the origin at xb = 0, and measure from b to a, then

!U = qEd

This is analogy with gravitation where we U =mgh.

Ub=0 and the potential energy function is U=qEd

13

Now define the Electric Potential Difference ΔVwhich does not depend on charge.

!V =!U

q

!U = "W

!V ="W

q0

• The potential difference is the negative of the work done per unit charge by an electric field on a positive unit charge when it moves from one point to another.

14

!V = Ed

!V =!U

q

!V =qEd

q

For a battery of potential difference of 9 volts you would say that the positive terminal is 9 volts above the negative terminal.

!U = qEd

Find the potential difference ΔV for a uniform electric field

15

Note relationship betweenpotential and electric field

• V is a scalar not a vector. Simplifies solving problems.

• We are free to choose V to be 0 at any location. Normally V is chosen to be 0 at the negative terminal of a battery or 0 at infinity for a point charge.

!V ="W

q0

= "

!F

q0

#d!s$ = "

!E #d!s$ dV = !Edx

E = !dV / dx

16

Example for a battery in a circuit• In a 9 volt battery, typically used in IC circuits, the

positive terminal has a potential 9 v higher than thenegative terminal. If one micro-Coulomb of positivecharge flows through an external circuit from the positiveto negative terminal, how much has its potential energybeen changed?

q

17

Generalize concept of electricpotential energy and potentialdifference for any electric field

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(independent of path, ds)

Therefore, electric force is a conservative force.

= - Work done by the electric force =

!!F "d!s

i

f

#!U =U f "Ui

q

UV

!=!

!V = Vf "Vi = "

!E #d!s$

y

x

19

Find the electric potential when moving fromone point to another in a field due to a pointcharge?

!V = "

!E #d!r$

!E =

kq

r2r

Vf !Vi = !!E "d!r

i

f

#

20

Vf !Vi = !!E "d!r

R

#

$ = !kqcos0"1

r2d

R

#

$ r = kq1

r R

#

= kq(1

#!1

R)

R

kqV =

04

1

!"=k

eqn 25-26

Replace R with r

V =1

4!" 0

q

r

Vf !Vi = 0 !Vi = !kq

R

Potential of a point charge at a distance R

Vf !Vi = !!E "d!r

i

f

#

21

Electric potential for a positive point charge

V (r) =kq

r

r = x2+ y

2

• V is a scalar• V is positive for positive charges, negative for negative charges.• r is always positive.• For many point charges, the potential at a point in space is the

simple algebraic sum (Not a vector sum)

22

Hydrogen atom.• What is the electric potential at a distance of 0.529 A fromthe proton? 1A= 10-10 m

Electric potential due to a positive point charge

r = 0.529 A

•What is the electric potential energy of the electron at that point?

V =kq

R=8.99 !10

9 Nm2

C2( ) !1.6 !10"19

C

.529 !10"10m

V = 27.2J

C= 27.2Volts

U = qV= (-1.6 x 10-19 C) (27.2 V) = - 43.52 x 10-19 Jor - 27.2 eV where eV stands for electron volt.Note that the total energy E of the electron in the ground state of hydrogen is - 13.6 eVAlso U= 2E = -27.2 eV. This agrees with above formula.

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What is the electric potential due toseveral point charges?

For many point charges, the potential at a point in space isthe simple algebraic sum (Not a vector sum)

!"#

$%&

++=3

3

2

2

1

1

r

q

r

q

r

qkV

V =kqi

rii

!

r1

xr3

y

r2

q1q2

q3

24

Four Point charges

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What is the potential due to a dipole?

Two point charges that are opposite and equal

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Potential for a Continuous Distribution ofCharge

Point charge

For an element of charge

Integrate

r

kqV =

r

kdqdVdq = ,

!=r

kdqV

27

Chaper 24 Problem 22. With V = 0 at infinity, what is theelectric potential at P, the center of curvature of the uniformlycharged nonconducting rod?

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Chapter 24 Problem 26. What is the magnitude of the net electricpotential at the center?1. A thin rod of charge -3.0 µC that forms a full circle of radius 6.0 cm2. A second thin rod of charge 2.0 µC that forms a circular arc of radius4.0 cm, subtending an angle of 90° about the center of the full circle3. An electric dipole with a dipole moment that is perpendicular to aradial line and that has magnitude 1.28 multiplied by 10-21 C·m

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Figure 24-44 shows a thin plastic rod of length L anduniform positive charge Q lying on an x axis. With V =0 at infinity, find the electric potential at point P1 on theaxis, at distance d from one end of the rod.

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Potential due to a ring of charge• Direct integration. Since V is a scalar, it is easier to evaluate

V than E.• Find V on the axis of a ring of total charge Q. Use the formula

for a point charge, but replace q with elemental charge dq andintegrate.

Point charge

For an element of charge

r is a constant as we integrate.

This is simpler than finding E because Vis not a vector.

r

kqV =

r

kdqdVdq = ,

!=r

kdqV

Q)Rz(

kV

22+

=!!+

= dqRz

k

)( 22

!+

=

)( 22 Rz

kdq

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Potential due to a line charge

r

dqkdV =

We know that for an element of charge dqthe potential is

r

dxkVd ,So!

=

For the line charge let the charge density be λ.Then dq=λdx

22dxr ,But +=

22dx

dxkVd ,Then

+

!=

Now, we can find the total potential V produced by the rod at point P byintegrating along the length of the rod from x=0 to x=L

!!!+

"=

+

"==

L

022

L

022

L

0 dx

dxk

dx

dxk VdV !V = k" ln(x + x

2+ d

2)0

L

)dln)dLL(ln(k Vo,S 22!++"= or, V = k! ln

L + L2+ d

2

d

"

#$

%

&'

32

A new method to find E if the potential is known. If we know V, how do we find E?

So the x component of E is the derivative of V with respect to x, etc.–If V = a constant, then Ex = 0. The lines or surfaces on which Vremains constant are called equipotential lines or surfaces.–See example on next slide

!V = "

!E #d!s$

dV = !!E "d!s

dz

dVE

dy

dVE

dx

dVE

z

y

x

!=

!=

!=

dzEdyEdxEdV zyx !!!=

d!s = idx + jdy + kdz

!E = Exi + Ey j + Ezk

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Now find the electric field at point P1 on the axis, atdistance d from one end of the rod. Find the x and ycomponents Ex and Ey.

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Equipotential Surfaces

• Three examples

• What is the equipotential surface and equipotentialvolume for an arbitrary shaped charged conductor?

• See physlet 9.3.2 Which equipotential surfaces fit thefield lines?

35

a) Uniform E field

V = constant in y and zdirections

b) Point charge

(concentric shells)

c) Electric Dipole

(ellipsoidal concentric shells)

E = Ex ,Ey = 0,Ez = 0

Ex = !dV

dx

V = !Exd

Blue lines are the electric field lines

Orange dotted lines represent the equipotential surfaces

x

36

Electric Potential Energy U of a system ofcharges

How much work is required to set up the arrangement of Figure 24-46 if q = 3.20 pC, a = 54.0 cm, and the particles are initially infinitely far apart and at rest?

q1q2

q3q4

W = U

U = k(q1q2

r12

+q1q3

r13

+q1q4

r14

+q2q3

r23

+q2q4

r24

+q3q4

r34

)

U = k(!qq

a+qq

2a!qq

a!qq

a+qq

2a!qq

a)

U = k(!4q

2

a+2q

2

2a)

37

38

Dielectric Breakdown: Application of Gauss’s Law

If the electric field in a gas exceeds a certain value, the gasbreaks down and you get a spark or lightning bolt if the gasis air. In dry air at STP, you get a spark when

E ! 3"104V

cm

12

V = constant on surface of conductor Radius r2

r1

39

This explains why:• Sharp points on conductors have the highest electric fields and cause

corona discharge or sparks.

• Pick up the most charge with charge tester from the pointy regions of thenon-spherical conductor.

• Use non-spherical metal conductor charged with teflon rod. Showvariation of charge across surface with charge tester.

Van de Graaff+ + + +

- - - -

Radius R

1

2

V = constant on surface of conductor

Cloud

40

How does a conductor shield the interior from anexterior electric field?

• Start out with a uniform electric fieldwith no excess charge on conductor.Electrons on surface of conductor adjustso that:

1. E=0 inside conductor2. Electric field lines are perpendicularto the surface. Suppose they weren’t?3. Does E = just outside the conductor4. Is σ uniform over the surface?5. Is the surface an equipotential?

6. If the surface had an excess charge, how would your answers change?

0!

"

41

What is the electric potential of a uniformly charged circular disk?

We can treat the disk as a set of ring charges.The ring of radius R’ and thickness dR’ has anarea of 2πR’dR’ and it’s charge is dq = σdA = σ(2πR’)dR’ where σ=Q/(πR2), the surface chargedensity. The potential dV at point P due to thecharge dq on this ring given by

Integrating R’ from R’=0 to R’=R

q

dV =kdq

(z2+ (R ')

2)

dV =k!2"R 'dR '

(z2_(R ')

2)

!

V = k!2"R 'dR '

(z2+ (R ')

2)0

R

#

!V = 2k"# ( z2+ R

2$ z)

42

Chapter 24 Problem 19

The ammonia molecule NH3 has a permanent electric dipolemoment equal to 1.31 D, where 1 D = 1 debye unit = 3.34multiplied by 10-30 C·m. Calculate the electric potential due to anammonia molecule at a point 44.0 nm away along the axis of thedipole. (Set V = 0 at infinity.)

43

Chapter 24 Problem 55

Two metal spheres, each of radius 1.0 cm, have a center-to-centerseparation of 2.2 m. Sphere 1 has charge +2.0 multiplied by 10-8 C.Sphere 2 has charge of -3.8 multiplied by 10-8 C. Assume that theseparation is large enough for us to assume that the charge on eachsphere is uniformly distributed (the spheres do not affect each other).Take V = 0 at infinity.

(a) Calculate the potential at the point halfway between the centers.(b) Calculate the potential on the surface of sphere 1.(c) Calculate the potential on the surface of sphere 2.

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Chapter 24 Problem 57

A metal sphere of radius 11 cm has a net charge of 2.0 multipliedby 10-8 C.

(a) What is the electric field at the sphere's surface?(b) If V = 0 at infinity, what is the electric potential at the sphere'ssurface?(c) At what distance from the sphere's surface has the electricpotential decreased by 500 V?