22. Electric Potential

1. Electric Potential Difference

2. Calculating Potential Difference

3. Potential Difference & the Electric Field

4. Charged Conductors

This parasailer landed on a 138,000-volt power line.

Why didn’t he get electrocuted?

He touches only 1 line – there’s no potential differences & hence no energy transfer involved.

22.1. Electric Potential Difference

Conservative force:

AB B AU U U ABWB

Ad F r

Electric potential difference electric potential energy difference per unit charge

B

Ad E rAB

AB

UV

q

BV if reference potential VA = 0.

[ V ] = J/C = Volt = V

For a uniform field:

AB ABV E r B A E r r

( path independent )

rAB E

Table 22.1. Force & Field, Potential Energy & Electric Potential

Quantity Symbol / Equation Units

B

AU d F r

UV

q

B

AV d E r

Force

Electric field

Potential energy difference

Electric potential difference

F N

E = F / q N/C or V/m

J

J/C or V

U F

V E

Potential Difference is Path Independent

Potential difference VAB depends only on positions of A & B.

Calculating along any paths (1, 2, or 3) gives VAB = E r.

GOT IT? 22.1

What would happen to VAB in the figure if

(a)E were doubled;

(b) r were doubled;

(c) the points were moved so the path lay at right

angles to E;

(d) the positions of A & B are interchanged.

doubles

doubles

becomes 0

reverses sign

The Volt & the Electronvolt

[ V ] = J/C = Volt = V U q V

E.g., for a 12V battery, 12J of work is done on every 1C charge that moves from its negative to its positive terminals.

Voltage = potential difference when no B(t) is present.

Electronvolt (eV) = energy gained by a particle carrying 1 elementary charge when it moves through a potential difference of 1 volt.

1 elementary charge = 1.61019 C = e

1 eV = 1.61019 J

Table 22.2. Typical Potential Differences

Between human arm & leq due to 1 mV heart’s electrical activity

Across biological cell membrane 80 mV

Between terminals of flashlight battery 1.5 V

Car battery 12 V

Electric outlet (depends on country) 100-240 V

Between lon-distance electric 365 kVtransmission line & ground

Between base of thunderstorm cloud & ground 100 MV

INTERNATIONAL VOLTAGE

Afghanistan 220 VAustralia 240 VBahamas 120 VBrazil 120/220 VCanada 120 VChina 220 VFinland 230 VGuam 110 VHongKong 220 VJapan 100 VMexico 127 VSpain 230 VUnited Kingdom 230 VUnited States 120 VVietnam 127/220 V

GOT IT? 22.2

(a) A proton ( charge e ),

(b) an particle ( charge 2e ), and

(c) a singly ionized O atom

each moves through a 10-V potential difference.

What’s the work in eV done on each?

10 eV

20 eV

10 eV

Example 22.1. X Rays

In an X-ray tube, a uniform electric field of 300 kN/C extends over a distance

of 10 cm, from an electron source to a target; the field points from the target

towards the source.

Find the potential difference between source & target and the energy gained

by an electron as it accelerates from source to target ( where its abrupt

deceleration produces X-rays ).

Express the energy in both electronvolts & joules.

ABV E r 300 / 0.10kN C m 30 kV

AB ABU e V 30 keV

30AB ABK U keV 154.8 10 J 4.8 fJ

Example 22.2. Charged Sheet

An isolated, infinite charged sheet carries a uniform surface charge density .

Find an expression for the potential difference from the sheet to a point a

perpendicular distance x from the sheet.

0 0xV E x 02x

E

Curved Paths & Nonuniform Fields

Staight path, uniform field: AB ABV E r

Curved path, nonuniform field:

0limAB i i

i

V

r

E rB

Ad E r

The figure shows three straight paths AB of the same length, each in a different electric field.

The field at A is the same in each.

Rank the potential differences ΔVAB.

GOT IT? 22.3

Largest ΔVAB .Smallest ΔVAB .

22.2. Calculating Potential Difference

Potential of a Point Charge

AB B AV V V B

Ad E r 2

ˆB

A

k qd

r r r

2

B

A

r

AB r

k qV dr

r

ˆ ˆ ˆd dr r r r r dr

1 1

B A

k qr r

For A,B on the same radial

For A,B not on the same radial, break the path into 2

parts,

1st along the radial & then along the arc.

Since, V = 0 along the arc, the above equation holds.

The Zero of Potential

Only potential differences have physical significance.

Simplified notation: RA A R AV V V V R = point of zero potential

VA = potential at A.

Some choices of zero potential

Power systems / Circuits Earth ( Ground )

Automobile electric systems Car’s body

Isolated charges Infinity

GOT IT? 22.4

You measure a potential difference of 50 V between two points a distance

10 cm apart in the field of a point charge.

If you move closer to the charge and measure the potential difference over

another 10-cm interval, will it be

(a) greater,

(b) less, or

(c) the same?

Example 22.3. Science Museum

The Hall of Electricity at the Boston Museum of Science contains a large Van de Graaff generator, a device that builds up charge on a metal sphere.

The sphere has radius R = 2.30 m and develops a charge Q = 640 C.

Considering this to be a single isolate sphere, find

(a) the potential at its surface,

(b) the work needed to bring a proton from infinity to the sphere’s surface,

(c) the potential difference between the sphere’s surface & a point 2R from its center.

QV R k

R 6

9640 10

9.0 10 /2.30

CVm C

m

(a) 2.50MV

W eV R(b) 2.50MeV 191.6 10 2.50C MV 134.0 10 J

,2 2R RV V R V R (c)2

Q Qk kR R

2

QkR

1.25MV

Example 22.4. High Voltage Power Line

A long, straight power-line wire has radius 1.0 cm

& carries line charge density = 2.6 C/m.

Assuming no other charges are present,

what’s the potential difference between the wire & the ground, 22 m below?

B

A

r

AB rV d E r

0

ˆ ˆ2

B

A

r

rdr

r

r r

0

1

2

B

A

r

rdrr

0

ln2

B

A

r

r

9 6 222 9.0 10 / 2.6 10 / ln

0.01

mV m C C m

m

360 kV

Finding Potential Differences Using Superposition

Potential of a set of point charges: i

i P i

qV P k

r r

Potential of a set of charge sources: ii

V P V P

Example 22.5. Dipole Potential

An electric dipole consists of point charges q a distance 2a apart.

Find the potential at an arbitrary point P, and approximate for the casewhere

the distance to P is large compared with the charge separation.

1 2

qqV P k k

r r

1 2

1 1kq

r r

2 2 2

1 2 cosr r a r a

2 2 22 2 cosr r a r a

2 22 1 4 cosr r r a

2 1 1 2r r r r

r >> a 2 1 2 cosr r a

2 12

r rV P k q

r

2

2 cosqak

r

2

cospk

r

p = 2qa

= dipole moment

2 1

2 1

r rkq

r r

+q: hill

q: hole

V = 0

GOT IT? 22.5

The figure show 3 paths from infinity to a point P on a dipole’s

perpendicular bisector.

Compare the work done in moving a charge to P on each of the paths.

V is path independent

work on all 3 paths are the same.

Work along path 2 is 0 since V = 0 on it.

Hence, W = 0 for all 3 paths.P

1

23

q q

Continuous Charge Distributions

Superposition:

V dV dqkr

dV

kr

3d rV k

r

rr r

Example 22.6. Charged Ring

A total charge Q is distributed uniformly around a thin ring of radius a.

Find the potential on the ring’s axis.

dqV x k

r 2 2

kdq

x a

2 2

kQ

x a

Same r for all dq

Qk x ax

Example 22.7. Charged DiskA charged disk of radius a carries a charge Q distributed uniformly over its surface.

Find the potential at a point P on the disk axis, a distance x from the disk.

V x dV 2 2

kdq

x r

2 22

2k Qx a x

a

22 202

a k Qr dr

ax r

2 2 20

2 aQ rk dra x r

2 22

0

2 aQk x ra

sheet

point charge

disk

20

2 2

2

k Q kQa x x x a

a a

k Qx a

x

22.3. Potential Difference & the Electric Field

W = 0 along a path E

V = 0 between any 2 points on a surface E.Equipotential Field lines.

Equipotential = surface on which V = const.

V > 0V < 0 V = 0

Steep hillClose contourStrong E

GOT IT? 22.6

The figure show cross sections through 2 equipotential surfaces.

In both diagrams, the potential difference between adjacent

equipotentials is the same.

Which could represent the field of a point charge? Explain.

(a). Potential decreases as r 1 , so the spacings between

equipotentials should increase with r .

Calculating Field from Potential

B

A

r

AB rV d E r

dV d E r i ii

E dx ii i

Vd x

x

ii

VE

x

V E = ( Gradient of V )

V V V

x y z

E i j k

E is strong where V changes rapidly ( equipotentials dense ).

Example 22.8. Charged Disk

Use the result of Example 22.7 to find E on the axis of a charged disk.

Example 22.7: 2 22

2k QV x x a x

a

2 2 2

21

k Q x

a x a

x

VE

x

x > 0x < 0

0y zE E dangerous conclusion

Tip: Field & Potential

Values of E and V aren’t directly related.

x

VE

x

V falling, Ex > 0

V flat, Ex = 0

V rising, Ex < 0

22.4. Charged Conductors

In electrostatic equilibrium,

E = 0 inside a conductor.

E// = 0 on surface of conductor.

W = 0 for moving charges on / inside conductor. The entire conductor is an equipotential.

Consider an isolated, spherical conductor of radius R and charge

Q.

Q is uniformly distributed on the surface

E outside is that of a point charge Q.

V(r) = k Q / R. for r R.

Consider 2 widely separated, charged conducting spheres.

11

1

QV k

R 2

22

QV k

R

Their potentials are

If we connect them with a thin wire,

there’ll be charge transfer until V1 = V2 , i.e.,1 2

1 2

Q Q

R R

24j

jj

Q

R

In terms of the surface charge densities

1 1 2 2R R we have

Smaller sphere has higher field at surface.

1 1 2 2E R E R

Same V

Ans.

Surface is equipotential | E | is larger where curvature of surface is large.

More field lines emerging from sharply curved regions.

From afar, conductor is like a point charge.

Conceptual Example 22.1. An Irregular Condutor

Sketch some equipotentials & electric field lines for an isolated egg-shaped conductor.

Conductor in the Presence of Another Charge

Making the Connection

The potential difference between the conductor and the outermost equipotential shown in figure is 70 V.

Determine approximate values for the strongest & weakest electric fields in the region, assuming it is drawn at the sizes shown.

7 mm12 mm

Strongest field :

3

70

7 10

VE

m

10 /kV m

Weakest field :

3

70

12 10

VE

m

5.8 /kV m

Application: Corona Discharge, Pollution Control, and Xerography

Air ionizes for E > MN/C.

Recombination of e with ion Corona discharge ( blue glow )

Corona discharge across power-line insulator.

Electrostatic precipitators:

Removes pollutant particles (up to

99%) using gas ions produced by

Corona discharge.

Laser printer / Xerox machines: Ink consists of plastic toner particles that adhere to charged regions on light-sensitive drum, which is initially charged uniformy by corona discharge.