Capacitors

PHY122 - Physics for the Life Sciences II
Lecture 4
The Electric Potential
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Session ID: PHY122S15
02/05/2015
Lecture 4
1
Electric Potential Energy
Because the electric force on a charge q in the
neighborhood of other point-like charges depends on
the relative distances only …
– (just like for gravity !!)
… we can assign a Potential Energy U to the charge q,
that will only depend on its spatial position
As for any potential energy, we are only concerned with
CHANGES in potential energy: ΔUelec …

 U
 F  dx  F  x  W conservative
i
i
i
i
;
W
forces only!!
 WNC   U i  K  Emech
i
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Lecture 4
i
 K
i
From:
F=ma; see
PHY121
2
Electric Work, Energy, & Potential
A charge q=2.0 μC moves in a uniform electric field
E =3.0×103 N/C (e.g. between the plates of a capacitor)
– calculate the change in (electric) potential energy ΔU when the
charge q is moved from point A to point B:
B
Work: Welec   Felec  dx
A


 Felec  AB  qE  AB  qEd
F is
constant!!
3
Welec  2.0 106  3.0 10  5.0 102 Nm  3.0  104 J
– Note: as we anticipated for
a conservative force, ….
– … the Work only depends
on the end points A and B !!
+ +Q
+
A
+
+ q
+
+
+
+
E
–
–
–
d=5 cm
–
Potential Energy U:
–
WA B  U
–
U  qEd  3.0 104 J
B
x
–
Potential V:
V  U q   Ed  150 J/C  150 V(olt)  E  3000 V/m  –
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Lecture 4
–Q
3
Moving “upstream” by 4.0 cm in a 2000 V/cm
electric field, the voltage changes by … Volt?
Rank
1
2
3
4
5
6
Responses
8000
-8000
500
80
-500
Other
Values: 8000, {7900,...
Value Matches: 221
70%
61%
60%
50%
40%
30%
20%
20%
12%
10%
4%
2%
2%
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Lecture 4
Ot
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‐5
00
80
50
0
‐8
00
0
80
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0%
4
Allesandro Volta
Inventor of the electro-chemical battery
– a pile of cells made of a zinc electrode and a copper
electrode, separated by card board saturated with
sulfuric acid or a brine mixture of salt and water.
– The electrolyte exists in the form 2H+ and SO42–. The
zinc, which is higher than both copper and hydrogen in
the electrochemical series, reacts with the negatively
charged sulfate (SO42–).
– The positively charged hydrogen ions (protons) capture
electrons from the copper, forming bubbles of hydrogen
gas, H2. This makes the zinc rod the negative electrode
and the copper rod the positive electrode.
– Essentially, the difference in electro-chemical ionic
binding energies provides the charge-pumping action !
– We now have two terminals, and the current will flow if
we connect them. The reactions in this cell are as
follows:
zinc Zn → Zn2+ + 2e –
sulfuric acid 2H+ + 2e – → H2
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Lecture 4
Alessandro Volta
(1745-1827)
5
12 V Lead-Acid Battery
– In the charged state, each cell contains electrodes of
elemental lead (Pb) and lead(IV) oxide (PbO2) in an
electrolyte of approximately 33.5% sulfuric acid (H2SO4).
– In the discharged state both electrodes turn into
lead(II) sulfate (PbSO4) and the electrolyte loses its
dissolved sulfuric acid and becomes primarily water.
– Anode Reaction:
Pb(s) + HSO4−(aq) → PbSO4(s) + H+(aq) + 2e−
– Cathode Reaction:
PbO2(s) + HSO4−(aq) + 3H+(aq) + 2e− → PbSO4(s) + 2H2O(l)
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Lecture 4
6
Equi-Potential Plane
What if A’ and B are both in the “plane” perpendicular to
the field E (the dotted black line)?
• In that case A'B x is zero and:
U  qE A'B x  0 J ; V  U q  0 J/C  0 V(olt)
This is true for any pair of A’, B points on the same plane:
• the Potential (Energy U/q) is the same everywhere on
–Q
–
this plane …
A
 Equipotential Plane
E
A’
– Equipotential surface is  E everywhere
– Field E is pointing along the “downward”
gradient of the potential
All this works for
any field configuration …
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–
q
–
–
–
x
Lecture 4
0V –
2V
5V
B
–
–
7
Select the correct answer ...
15%
6%
02/05/2015
Lecture 4
...
ab
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No
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…
3%
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...
...
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, t
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Fo
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A. For a point charge, the
equipotential surfaces
are cylinders …
B. For a planar charge, the
equipotential surfaces
are cylinders …
C. For a line charge, the
equipotential surfaces
are cylinders …
D. None of the above …
75%
8
Recap:
1
qQ
Coulomb Force (point charges q, Q): F 
4 0 r 2
Field: E ≡ F/q
examples:
1 Q
1 Q L
1 Q
1 Q
Epoint or 
; Eline or 
; Esingle 
; Eparallel plate 
2
4 0 r
2 0 r
2 0 A
0 A
sphere
cylinder
plane
capacitor
Energy of a test charge q in a uniform Electric Field E:
(e.g. between the plates of a charged capacitor)
U electric  Welec  

Felec ds   Felec D/ /  qED/ /
path D
uniform
field
Potential difference ΔV :
(“voltage difference” between terminals)
Vcapacitor
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U electric

q
Lecture 4
9
Equipotential Surfaces
(Equi)potential surfaces denote surfaces in space of equal
potential.
– the surfaces are not necessarily planes! e.g. spheres for point charge
Equipotential lines indicate lines of constant potential that
intersect a given surface
– e.g. lines of equal height on a topo-map
– Note: VG=UG/m=gh, with g=constant!
Where the equipotential lines are
close together, the field is large!
– Electric Field: E=V/d (magnitude),
with d=separation between lines
(along the slope) …
– Cfr: Gravitational Field: EG=gh /d=gsinθ, and FG=mEG=mgsinθ !
02/05/2015
Lecture 4
slope angle
10
Capacitance of a Capacitor
The Capacitance of a capacitor is a measure
of the amount of charge Q that a capacitor
will hold for a given potential difference ΔV
between the plates
Q
C
– Capacitance C:
V
– Unit: C/V=F(arad)
– For a Parallel-Plate capacitor:
Q
Q
d  CPP 
V  Ed 
0 A
V
0 A
Michael Faraday
(1791 – 1867)
A
 0
Q
d
Qd
Capacitors
(pF – mF) and
Capacitor
Symbols
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Lecture 4
11
A 1.0 mF capacitor has a 10.0 V potential
difference across its terminals;
Q
C
its charge is … mC
V
Rank
1
2
3
4
5
6
78%
Responses
10
0.1
10000
1E-05
0.01
Other
16%
Values: 10, {9.9, 10...
Value Matches: 278
02/05/2015
2%
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Lecture 4
2
3
1%
4
1%
5
3%
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12
Energy, Potential, Capacitance of a Point Charge
A charge q moves in the field of a point charge Q
– calculate the change in (electric) potential going from A to B:
B
r
rB
B
qQ
1 
qQ
1
K
Work: Welec   Felec ds  KqQ  2 r  ds  KqQ  2 dr  K
rB
r
r
rA
rA
A
rA
– Note: as we anticipated for
a conservative force, ….
– … the Work only depends
on the end points A and B !!
Equipotential
surfaces
Potential Energy U:
WA B  U  U B  U A
qQ
qQ
thus: U A  K
 U (r)  K
rA
r
Potential V:
V (r)  U (r) q  K Q r
1 1
Q
 V  KQ     C 

V
 rB rA 
Lecture 4
02/05/2015
A
q
rA
rB
B
Q
13
Potential & Capacitance of a Conducting Sphere …
Because of the symmetry, the electric field of a conducting
sphere with charge Q and that of a point charge Q are
identical OUTSIDE of the sphere
dQ’
– the sphere itself is, of course, an equipotential surface.
To charge a sphere, we must bring charge dQ’ onto the
sphere
which already has some charge Q’:
Q
2
2
Q'
Q
Q
1
1
1
1
 CV 2
U  K
dQ '  K
 QV 
2
2 C
2
R
R 2
0
E=0
– self-energy!
E=KQ’/r2
V=constant
R
V=KQ’/r
Similarly for a charged capacitor:
Q’
Q
2
1 Q
Q'
1
1
U 
dQ ' 
 Q V 2 C V 2
2 0 A R 2
 A
0 0
A
2
1
1
2
– the stored energy: U   0  Ed    0 E  
Ad
2
2
d
Volume !
U
1
  0 E 2 generally true!
– the energy density: u 
Lecture 4
Volume 2
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14
Example Problem
A system of two uncharged metal spheres, spaced 20.0 cm
(center to center) apart, has a capacitance of C=26.0 pF.
– How much work would it take to move an amount of charge
q=16.0 nC from one sphere to the other?
d
Solution:
W=?
q
– Clearly, a case of Work – Energy conversion …
– the final energy Uf after the transfer of charge is:
U  U f  U i 
1
CV 2
2
16 10 C 

q

 4.92 μJ
0 
12
2C 2  26 10 F
2
9
2
– The net work expended equals the energy now stored in this
capacitor: U  1 CV 2  4.92 μJ
2
– Note, the UNITS check out fine:
02/05/2015
Lecture 4
C2
 VC   J/C  C  J
CV
15
Example: Energy of a System of Charges
• Consider two charges Q1= +5μC, Q2= –8 μC, 5 cm apart;
Their energy is:
Q
U  U12  Q1V2  Q1 K 2  5.0 106  9.0 109
r12
8.0  10
5.0  10
6
2
 7.2 J
• Now, calculate the potential energy of a third charge
Q3=+6 μC, 3 cm from Q1, 4 cm from Q2:
Q3 =+6 μC
 Q1 Q2 
U  U 31  U 32  Q3V1  Q3V2  Q3 K   
 r31 r32 
6
8.0106 
6
9  5.010
 6.0 10  9.0 10 
2 
2   1.80 J
3.0

10
4.0

10


Note: the potential energy of Q3
infinitely far away is ZERO
because V(r)  1/r !
5 cm
Q1=+5μC
Q2 = –8 μC
• The TOTAL energy of the SYSTEM of ALL 3 charges is:
U  U12  U 31  U 32  Q1V2  Q3V1  Q3V2  7.2  1.8  9.0 J
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Lecture 4
17
Example: Fusion in the Sun
A proton has a diameter of approximately d=1.6×10–15 m .
When protons in the Sun collide to this distance, fusion
may happen: p + p → D+ + e+ + v (D + = pn)
What is temperature T of the Sun’s thermonuclear core?
– Energy of the two protons required to approach each other to
distance d: U  eV  e Ke d  e  9.0  105 J/C
– Thus, assuming the protons collided head-on with equal initial
speeds:
eV
1
2
 9.2 106 m/s
K i  2  m p v  U f  eV  v 
2
m




p
– The protons form a plasma (ionized gas), and its temperature T is a
measure of the average kinetic energy K of the protons:
eV  K 
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3
1
m p v 2  2 k BT
2
 T
Lecture 4
mp v2
3k B
2eV

 3.4 109 K
3k B
19
Dielectrics
+Q
E
+
–
A Dielectric material is an insulator
which contains polarizable molecules …
+
–
+
–
– Typically, the polarization field Epol
partially opposes the external field E
– the magnitude of the polarization
field is proportional to the external field …
+
–
+
–
+
–
–Q
+
–
Hence, inserting a dielectric material
Epol
+
–
between the plates of a capacitor
DECREASES the NET electric field between the plates:
E’≡E/κ with κ the dielectric constant for the material
– DEcreases the potential ΔV=Ed by the same factor κ: ΔV’= ΔV/κ
Q
A
– and INcreases the capacitance by factor κ :
C'
Dielectrics are taken into account
by the substitution:  0   0
02/05/2015
Lecture 4
V '
  0
d
20
Summary Electro-Statics:
qQ
Coulomb Force (for point charges): F 
4 0 r 2
Field:
E ≡ F/q
1 Q
1 Q
1 Q

E
;
E

;
examples: Epoint or 
single
capacitor
2 0 A
 0 A
4 0 r 2
plane
sphere
Potential: Vpoint or 
sphere
Q
1 Qd
; Vcapacitor 
4 0 r
 0 A
Q
;
Capacitance: C 
V
Energy:
1
1
Dielectrics: ε0 → κε0
1
1 Q2 1
2
 K
 CV ; U charged  CV 2 ;
2
2 R 2
capacitor
U  qV ; U charged
sphere
1
Energy density: u   0 E 2 ;
2
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Lecture 4
22