22. electric potential 1.electric potential difference 2.calculating potential difference...
TRANSCRIPT
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.