Transcript 20. Electric Charge, Force, & Field

22. Electric Potential
1.
2.
3.
4.
Electric Potential Difference
Calculating Potential Difference
Potential Difference & the Electric Field
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:
U AB  U B  U A  WAB
B
   F  dr
A
( path independent )
Electric potential difference  electric potential energy difference per unit charge
VAB 
B
U AB
   E  dr
A
q
 VB
[ V ] = J/C = Volt = V
if reference potential VA = 0.
For a uniform field:
VAB  E  rAB
 E  rB  rA 
rAB
E
Table 22.1. Force & Field, Potential Energy
& Electric Potential
Quantity
Force
Electric field
Potential energy
difference
Symbol / Equation
F
Units
N
E=F/q
N/C or V/m
B
U    F  dr
A
J
F  U
Electric potential
difference
V 
U
q
B
V    E  dr
A
E  V
J/C or V
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
doubles
doubles
becomes 0
(a)E were doubled;
(b) r were doubled;
(c) the points were moved so the path lay at right
angles to E;
reverses sign
(d) the positions of A & B are interchanged.
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 = 1.61019 C = e
1 eV = 1.61019 J
Table 22.2. Typical Potential Differences
Between human arm & leq due to
heart’s electrical activity
1 mV
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
transmission line & ground
365 kV
Between base of thunderstorm cloud & ground
100 MV
INTERNATIONAL VOLTAGE
Afghanistan
Australia
Bahamas
Brazil
Canada
China
Finland
Guam
HongKong
Japan
Mexico
Spain
United Kingdom
United States
Vietnam
220 V
240 V
120 V
120/220 V
120 V
220 V
230 V
110 V
220 V
100 V
127 V
230 V
230 V
120 V
127/220 V
GOT IT? 22.2
10 eV
(a) A proton ( charge e ),
20 eV
(b) an  particle ( charge 2e ), and
10 eV
(c) a singly ionized O atom
each moves through a 10-V potential difference.
What’s the work in eV done on each?
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.
VAB    E  r  300 kN / C   0.10 m  30 kV
U AB  e VAB  30 keV
K AB  U AB  30 keV
 4.8 1015 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.
V0 x  E  x  0  
E

x
2 0
Curved Paths & Nonuniform Fields
Staight path, uniform field:
VAB  E  rAB
Curved path, nonuniform field:
 Ei  ri   E  dr
r  0
B
VAB   lim
i
A
GOT IT? 22.3
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.
Smallest ΔVAB .
Largest ΔVAB .
22.2. Calculating Potential Difference
Potential of a Point Charge
B
VAB  VB  VA    E  dr   
A
B
A
kq
rˆ  d r
2
r
For A,B on the same radial rˆ  d r  rˆ  rˆ dr
VAB   
rB
rA
kq
dr
r2
 dr
 1 1
 k q    
 rB rA 
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:
VRA  VA  VR  VA
R = point of zero potential
VA = potential at A.
Some choices of zero potential
Power systems / Circuits
Automobile electric systems
Isolated charges
Earth ( Ground )
Car’s body
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.
(a)
6
640

10
C
Q

9
V  R  k
  9.0 10 Vm / C 
R
2.30 m

(b) W  eV  R   2.50 MeV  1.6 1019 C
(c)
VR,2 R  V  2R V  R  k
 k
Q
 1.25 MV
2R
Q
Q
k
2R
R
 2.50 MV
  2.50 MV 
 4.0 1013 J
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?
rB
r
A
rA
VAB   E  d r   B

r


2  0

rB
rA
1
dr
r

rˆ  rˆ d r
2  0 r

r

ln B
2  0 rA
 2   9.0 109 V m / C  2.6 106 C / m  ln
 360 kV
22 m
0.01 m
Finding Potential Differences Using Superposition
Potential of a set of point charges:
V  P   k
i
Potential of a set of charge sources:
qi
rP  ri
V  P    Vi  P 
i
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.
V  P  k
 q 
q
k
r1
r2
1 1
r r
 kq     kq 2 1
r2 r1
 r1 r2 
r12  r 2  a2  2r a cos
+q: hill
r22  r 2  a2  2r a cos
r22  r12  4 r a cos
  r2  r1  r1  r2 
r >> a 
V  P  k q
r2  r1  2 a cos
r2  r1
2 qa cos 
p cos 

k

k
r2
r2
r2
V=0
q: hole
p = 2qa
= dipole moment
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.
2
3
Hence, W = 0 for all 3 paths.
P

1
q
Work along path 2 is 0 since V = 0 on it.
q
Continuous Charge Distributions
Superposition:
V   dV
V r    k
 k
dq
r
 r d 3r
r  r
 k
 dV
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.
V  x   k

k
Same r for all dq
dq
r
k
x a
2
Q
x
k

2
x a
2
Q
x
a
2
 dq
Example 22.7.
Charged Disk
A 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

a
0
disk
sheet

2k Q
a2
k
x r
2
2
dq
 Q 
2 r dr

2 
2
2

a
x r 

2Q
k 2
a
point charge

k

a
0

r
x2  r 2
2Q
k 2
a
dr
x2  a2  x
x r
2
a
0

2kQ 
 2k Q
a

x

 a  2 x
 a2 

0

kQ


x
2
x
a
x
a
22.3. Potential Difference & the Electric Field
Equipotential = surface on which V = const.
W = 0 along a path  E
 V = 0 between any 2 points on a surface  E.
Equipotential  Field lines.
Steep hill
Close contour
Strong E
V<0
V=0
V>0
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
rB
VAB   E  d r
rA

dV  E  d r
  Ei dxi
i

Ei  
V
 xi

i
V
d xi
 xi
E   V
=  ( Gradient of V )
V
V
V 
E  
i
j
k

x

y

z


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:
V  x 
2k Q
a2
2k Q 
V
 2 
Ex  
a 
x

Ey  Ez  0

x2  a2  x
x
x2  a2


1


dangerous conclusion
x>0
x<0
Tip: Field & Potential
Values of E and V aren’t directly related.
V flat, Ex = 0
V falling, Ex > 0
V rising, Ex < 0
Ex  
V
x
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.
Their potentials are
V1  k
Same V
Q1
R1
V2  k
Q2
R2
If we connect them with a thin wire,
there’ll be charge transfer until V1 = V2 , i.e.,
In terms of the surface charge densities
we have
j 
1R1   2 R2
Qj
4 R2j

 Smaller sphere has higher field at surface.
E1 R1  E2 R2
Q1 Q2

R1 R 2
Conceptual Example 22.1. An Irregular Condutor
Sketch some equipotentials & electric field lines for an isolated egg-shaped conductor.
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.
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.
Strongest field :
E
70 V
7  10 3 m
 10 kV / m
Weakest field :
E
7 mm
12 mm
70 V
12  103 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 )
Electrostatic precipitators:
Removes pollutant particles (up to
99%) using gas ions produced by
Corona discharge across power-line insulator.
Corona discharge.
Laser printer / Xerox machines:
Ink consists of plastic toner particles that adhere to charged regions on lightsensitive drum, which is initially charged uniformy by corona discharge.