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22. Electric Potential. 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?. - PowerPoint PPT Presentation
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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
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
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 21 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
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
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 )
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.