Electric Potential

qA

C

B

rA

Br

path independence

a

a

R r

V Q

4 rQ

4 R

Today…• Conservative Forces and Energy Conservation

– Total energy is constant and is sum of kinetic and potential

• Introduce Concept of Electric Potential– A property of the space and sources as is the Electric Field– Potential differences drive all biological & chemical reactions,

as well as all electric circuits

• Calculating Electric Potentials put V(infinity)=0– Charged Spherical Shell– N point charges– Example: electric potential of a charged sphere

• Electrical Breakdown– Sparks– Lightning!!

Text Reference: Chapter 24.1- 4Examples: 24.1,2 and 4-6

Conservation of Energy of a particle from phys 111

• Kinetic Energy (K)– non-relativistic

• Potential Energy (U)– determined by force law

• for Conservative Forces: K+U is constant– total energy is always constant

• examples of conservative forces– gravity; gravitational potential energy – springs; coiled spring energy (Hooke’s Law):

U(x)=kx2

– electric; electric potential energy (today!)

• examples of non-conservative forces (heat)– friction– viscous damping (terminal velocity)

2

2

1mvK

),,( zyxU

Example: Gravitational Force is conservative(and attractive)

• Consider a comet in a highly elliptical orbit

• At point 1, particle has a lot of potential energy, but little kinetic energy

Total energy = K + U

is constant!• At point 2, particle has little potential

energy, but a lot of kinetic energy

Morepotentialenergy

Lesspotentialenergy

pt 1 pt 2

0

U(r)

U(r1)

( )GMm

U rr

U(r2)

Electric forces are conservative, too• Consider a charged particle traveling

through a region of static electric field:

• A negative charge is attracted to the fixed positive charge

• negative charge has more potential energy and less kinetic energy far from the fixed positive charge, and…

• more kinetic energy and less potential energy near the fixed positive charge.

• But, the total energy is conserved

+

-

• We will now discuss electric potential energy and the electrostatic potential….

Electric potential and potential energy• Imagine a positive test charge, Qo, in an external

electric field, E(x,y,z) <- it’s a vector field

• What is the potential energy, U(x,y,z) of the charge in this field?– Must define where in space U(x,y,z) is zero, perhaps at

infinity (for charge distributions that are finite)

– U(x,y,z) is equal to the work you have to do to take Qo from where U is zero to point (x,y,z)

•Define V(x,y,z) = U(x,y,z) / Qo (U = QV)

•U depends on what Qo is, but V is independent of Qo (which can be + or -)

•V(x,y,z) is the electric potential in volts associated with E(x,y,z). (1V = 1 J/c)

–V(x,y,z) is a scalar field

Electric potential difference

• Suppose charge q0 is moved from pt A to pt B through a region of space described by electric field E.

A B

q0 E

Felec

Fwe supply = -Felec

• To move a charge in an E-field, we must supply a force just equal and opposite to that experienced by the charge due to the E-field.

• Since there will be a force on the charge due to E, a certain amount of work WAB≡WAB will have to be done to accomplish this task.

1

• Remember: work is force times distance

• To get a positive test charge from the lower potential to the higher potential you need to invest energy - you need to do work

• The overall sign of this: A positive charge would “fall” from a higher potential to a lower one

• If a positive charge moves from high to low potential, it can do work on you; you do “negative work” on the charge

Electric potential difference, cont.

Lecture 5, ACT 1

• A single charge ( Q = -1C) is fixed at the origin. Define point A at x = + 5m and point B at x = +2m.– What is the sign of the potential difference

between A and B? (Δ VAB VB - VA )

x-1C AB

(a) ΔVAB < (b) Δ VAB = (c) Δ VAB >

Since , ΔVAB <0 !!

•The simplest way to get the sign of the potential difference is to imagine placing a positive charge at point A and determining which way it would move. Remember that it will always “fall” to lower potential.•A positive charge at A would be attracted to the -1C charge; therefore NEGATIVE work would be done to move the charge from

A to B.•You can also determine the sign directly from the definition:

Δ VAB is Independent of Path

• The integral is the sum of the tangential (to the path) component of the electric field along a path from A to B.

• This integral does not depend upon the exact path chosen to move from A to B.

• Δ VAB is the same for any path chosen to move from A to B (because electric forces are conservative).

A B

q0E

Does it really work?• Consider case of constant field:

– Direct: A - B

• Long way round: A - C – B

• So here we have at least one example of a case in which the integral is the same for BOTH paths.

• In fact, it works for all paths see Appendix.

A

CB

Eh r

dl

2

Lecture 5, ACT 2

• A direct calculation of the work done to move a positive charge from point A to point B is not easy.

• Neither the magnitude nor the direction of the field is constant along the straight line from A to B.

• But, you DO NOT have to do the direct calculation.

• Remember: potential difference is INDEPENDENT OF THE PATH!!

• Therefore we can take any path we wish.

2AA positive charge Q is moved from A to B along the path shown. What is the sign of the work done to move the charge from A to B?

A B

(a) WAB < 0 (b) WAB = 0 (c) WAB > 0

Choose a path along the arc of a circle centered at the charge. Along this path at every point!!

Electric Potential: where is it zero?• So far we have only considered potential differences.

• Define the electric potential of a point in space as the potential difference between that point and a reference point.

• a good reference point is infinity ... we often set V = 0

• the electric potential is then defined as:

• for a point charge at origin, integrate in from infinity along some axis, e.g., the x-axis

• here “r” is distance to origin

lineintegral

( ) ( )r

V r V dl E

• THE POTENTIAL IS A FUNCTION OF THE SPACE !!!!!• The Potential does not depend on the “test” charge at all.• Like ACT 1, except here a positive source charge creates the field.

Lecture 5, ACT 2Two test charges are brought separately to the vicinity of positive charge Q.

– charge +q is brought to pt A, a distance r from Q.

– charge +2q is brought to pt B, a distance 2r from Q.

– Compare the potential at point A (VA) to that at B (VB):

Q

QA

qr

B

2q2r

+

+

2B

(a) VA < VB (b) VA = VB (c) VA > VB

A positive “test charge” would move (“fall”) from point A towards point B.Therefore, VA > VB

In fact, since point B is twice as far from the charge as point A, we calculate that VA = 2VB!!

Potential from charged spherical shell

a

a

Q4 a

a r

V

Q4 r

• Potential

• r > a:

• r < a:

• E-field (from Gauss' Law)

Er = 0• r < a:

Er = 1

4 0 Q

r2 • r >a:

What does the result mean?

• This is the plot of the radial component of the electric field of a charged spherical shell:

Notice that inside the shell, the electric field is zero. Outside the shell, the electric field falls off as 1/r2.

The potential for r>a is given by the

integral of Er. This integral is simply the

area underneath the Er curve.

a

a

a r

Er

R

Q4 a

a r

V

Q4 r

R

Potential from N charges

The potential from a collection of N charges is just the algebraic sum of the potential due to each charge separately.

xr1

r2 r3

q1

q3

q2

3

Lecture 5, ACT 3• Which of the following charge distributions produces V(x) = 0

for all points on the x-axis? (we are defining V(x) 0 at x = )

(a)

x

+2C

-2C

+1C

-1C

(b)

x

+2C

-1C

+1C

-2C(c)

x

+2C

-1C

-2C

+1C

The key here is to realize that to calculate the total potential at a point, we must only make an ALGEBRAIC sum of the individual contributions. Therefore, to make V(x)=0 for all x, we must have the +Q and -Q contributions cancel, which means that any point on the x-axis must be equidistant from +2C and -2C and also from +1C and -1C. This condition is met only in case (a)!

Sparks• High electric fields can ionize nonconducting

materials (“dielectrics”)

• Breakdown can occur when the field is greater than the “dielectric strength” of the material.– E.g., in air,

6 6max 3 10 N/C 3 10 V/m 30 kV/cmE

Insulator ConductorDielectric

Breakdown

Ex.

doorknobV

fingerV

2mmd

Arc discharge equalizes the potential

What is ΔV?

maxV E d 30 kV/cm • 0.2 cm

6 kV

Note: High humidity can also bleed the charge off reduce ΔV.

Lecture 5, ACT 4 Two charged balls are each at the same potential V. Ball 2 is

twice as large as ball 1.

As V is increased, which ball will induce breakdown first?

(a) Ball 1 (b) Ball 2 (c) Same Time

r1

Ball 1

r2

Ball 2

surface 2

QE k

r

QV k

r

Smaller r higher E closer to breakdownsurface

VE

r

Ex. 100 kVV 3

6

100 10 V0.03m 3cm

3 10 V/mr

High Voltage Terminals must be big!

Lightning!

Factoids: ~ 200 M voltsV

~ 40,000 ampI

~ 30mst12~ 10 WP

_

+

__

+ +

Collisions produce charged particles.

The heavier particles (-) sit near the bottom of the cloud; the lighter particles (+) near

the top.

Stepped Leader

Negatively charged electrons

begin zigzagging downward.

AttractionAs the stepped

leader nears the ground, it

draws a streamer of

positive charge upward.

Flowing Charge

As the leader and the

streamer come together, powerful

electric current begins flowing

Contact!Intense wave of positive charge, a “return stroke,” travels upward

at 108 m/s

Summary

Reading Assignment: Appendix B;Chapter 24.3,5 and 25.1

examples: 24.7,11,13,15 and 25.1

• If we know the electric field E,

allows us to calculate the potential function V everywhere

(define VA = 0 above)

• Potential due to n charges:

• Equipotential surfaces are surfaces where the potential is constant.

Appendix A: Independent of Path?

• Consider any path from A to B as being made up of a succession of arc plus radial parts as above. The work along the arcs will always be 0, leaving just the sum of the radial parts. All inner sums will cancel, leaving just the initial and final radii as above... Therefore it's general!

q

A

B

r1

r2

Appendix A: Independent of Path?• We want to evaluate potential difference from A to B

q

A

B

rA

Br

E

• What path should we choose to evaluate the integral?

•If we choose straight line, the integral is difficult to evaluate.

q

A

C

B

rA

Br

E

• Magnitude different at each pt along line.• Angle between E and path is different at each

pt along line.

• If we choose path ACB as shown, our calculation is much easier!

From A to C, E is perpendicular to the path. i.e

From C to B, E is parallel to the path. i.e

Appendix A: Independent of Path?• Evaluate potential difference from

A to B along path ACB.

q

A

C

B

rA

Br

E

Evaluate the integral:

by definition:

Appendix A: Independent of Path?

• How general is this result?

• Consider the approximation to the straight path from A->B (white arrow) = 2 arcs (radii = r1 and r2) plus the 3 connecting radial pieces.

qA

B

rA

Br

C

q

A

B

r1

r2• For the 2 arcs + 3 radials path:

This is the same result as above!!The straight line path is better

approximated by Increasing the number of arcs and radial pieces.

Appendix B

a b

cunchargedconductor

solid spherewith totalcharge Q

r

III

IIIIV

Calculate the potential V(r) at the point shown (r<a)

Calculating Electric PotentialsCalculate the potential V(r)

at the point shown (r < a)• Where do we know the potential,

and where do we need to know it?

• Determine E(r) for all regions in between these two points

• Determine V for each region by integration

• Check the sign of each potential difference V... and so on ...

(from the point of viewof a positive charge)

V=0 at r = ... we need r < a ... a b

cunchargedconductor

solid spherewith totalcharge Q

r

III

IIIIV

V > 0 means we went “uphill”V < 0 means we went “downhill”

Calculating Electric Potentials

a b

c

r

III

IIIIV

ld

E

Calculate the potential V(r) at the point shown (r < a)

• Look at first term:

• Line integral from infinity to c has to be positive, pushing against a force:

• What’s left?

Line integral is going “in” which is just the opposite of what usually is done

- controlled by limitsc

Q

r

Q

o

c

o 4

1

4

1

Calculating Electric Potentials

a b

c

r

III

IIIIV

ld

E

Calculate the potential V(r) at the point shown (r < a)

• Look at third term:

• Line integral from b to a, again has to be positive, pushing against a force:

Line integral is going “in” which is just the opposite of what usually is done

- controlled by limits

• What’s left?

Previous slidewe have calculated this already

ab

abQ

r

Q

o

a

bo

)(

4

1

4

1

Calculating Electric PotentialsCalculate the potential V(r)

at the point shown (r < a)

a b

c

r

III

IIIIV

ld

E

• Look at last term:

• Line integral from a to r, again has to be positive, pushing against a force.

• But this time the force doesn’t vary the same way, since “r ’ ” determines the amount of source charge

Qa

rQ

3

3

• What’s left to do?

• ADD THEM ALL UP!

• Sum the potentials

2 2

3 2

1 11

4 2 4 2

r

o oa

Q r Q r

a a a

This is the charge that is inside “r” and sources field

Calculating Electric PotentialsCalculate the potential V(r) at the point shown (r < a)

ab

c

r

III

IIIIV

ld

E

ld

• Add up the terms from I, III and IV:

I III IV

The potential difference from infinity to a if the conducting shell weren’t there

An adjustment to account for the fact that the conductor is an equipotential, V= 0 from c → b

Potential increase from moving into the sphere

Calculating Electric Potentials Summary

ab

c

r

III

IIIIVThe potential as a function of r for all

4 regions is:

I r > c:

II b < r < c:

III a < r < b:

IV r < a:

Let’s try some numbers

ab

c

r

III

IIIIVQ = 6C a = 5cm b = 8cm c = 10cm

I r > c: V(r = 12cm) = 449.5 kV

II b < r < c: V(r = 9cm) = 539.4 kV

III a < r < b: V(r = 7cm) = 635.7 kV

IV r < a: V(r = 3cm) = 961.2 kV