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
Electric potential
difference
Symbol / Equation
F
Units
N
E=F/q
N/C or V/m
B
U    F  dr
A
F  U
U
V 
q
B
V    E  dr
A
E  V
J
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
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
 dr
 1 1
kq
d r  k q    
2
r
 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)
Q
V  R  k
R
  9.0 10 Vm / C 
9
 640 10
2.30 m

(b) W  e V  R   2.50 MeV  1.6 1019 C
(c)
VR,2 R  V  2R   V  R   k
 k
Q
 1.25 MV
2R
6
Q
Q
k
2R
R
C
 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
22 m
 2   9.0 10 V m / C  2.6 10 C / m  ln
0.01 m
9
 360 kV
6
Finding Potential Differences Using Superposition
Potential of a set of point charges:
qi
rP  ri
V  P   k
i
Potential of a set of charge sources:
V  P    Vi
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  a 2  2r a cos 
+q: hill
r22  r 2  a 2  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
 k
dq
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.
dq
V  x   k
r

k
Same r for all dq

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   a2 
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

E y  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 
 1 R1   2 R2
Qj
4 R 2j

 Smaller sphere has higher field at surface.
E1 R1  E2 R2
Q1 Q2

R1 R 2
Isolated conductor with irregular shape.
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
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.