Chapter 24: Capacitance and Dielectrics Capacitor: two conductors (separated by an insulator) usually oppositely charged a
+Q
b
-Q
Vab proportional to charge Q C = Q/ Vab (defines capacitance) units: 1F = 1 C/V
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The parallel plate capacitor σ Q A E= = ε0 ε0
+Q d -Q
A
A
Q Vab = Ed = d Aε 0 Q A C≡ = ε0 Vab d
Capacitance does not depend upon Q, V! => C depends upon geometric factors only How big is 1 Farad? (parallel plate example) p212c24: 2
Typical Capacitances ~ µF, nF, pF Example: A parallel plate capacitor has plates 2.00 m2 in area, separated by a distance of 5.00 mm. A potential difference of 10,000 V is applied across the capacitor. Determine the capacitance the charge on each plate the magnitude of the electric field in the region between the plates.
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A long cylindrical capacitor rb λ Vab = ln 2π ε 0 ra Q= λL
L
Q 2π ε 0 L C= = V ln rb ra C 2π ε 0 = L ln rb ra
rb
ra
coax :C ≈ 70 pF m
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A long cylindrical capacitor, small distance between cylinder walls 2π ε 0 L C= ln rb ra rb = ra + d rb ≈ ra = R > > d C=
[
2π ε 0 L
ln ( ra + d) ra 2
]
2π ε 0 L = d ln 1+ R 3
x x ln(1+ x) = x − + 2 3 2π ε 0 L 2π RLε 0 A C≈ = = ε0 d R d d
L
rb
ra
Capacitor looks approximately like parallel plates, in appropriate limit. p212c24: 5
Capacitors in circuits symbols analysis follow from conservation of energy (in terms of electric potential) conservation of charge
A 3 µF capacitor and a 6 µF capacitor are connected in series across an 18 V battery. Determine the equivalent capacitance, the charge on each capacitor and the potential difference across each capacitor. p212c24: 7
Capacitors in parallel a C1 V=Vab
+Q1 C2
+Q2
−Q1
−Q2
Q1 + Q2 = Q V1 = V2 = V Q1 = C1V1 Q2 = C2V2 Ceq ≡
b
Q Q1 + Q2 Q1 Q2 = = + V V V V
Q1 Q2 = + V1 V2 Ceq = C1 + C2
Q1 + Q2 = Q V1 = V2 = V
A 3 µF capacitor and a 6 µF capacitor are connected in parallel across an 18 V battery. Determine the equivalent capacitance, the potential difference across each capacitor and the charge on each capacitor. p212c24: 8
Combinations of combinations can be analyzed piecewise C2 C1 C3
Some configurations are not combinations that can be treated as combinations that can be analyzed as serial/parallel C1 C2
C5
C3 C4
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Energy stored in a capacitor When charged: Q = CV Charging q = Cv q C q= Q q Q2 W = ∫ dq = = U C 2C q= 0 dW = dq v = dq
dq
dq q
-q -dq
v = q/C
q -q dq
Q2 1 1 2 U= = QV = CV 2C 2 2
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Electric Field Energy Uniform field: parallel plate capacitor +
+
−
−
−
−
+
−
+
+
−
−
+
−
+
+
−
−
+
−
+
+
1 2 Q = CV U = CV 2 ε 0A C= volume= Ad V = Ed d u ≡ U / volume= energy density 1ε 0A = (Ed)2 /(Ad) 2 d 1 u = ε 0 E2 2
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In the circuit shown V = 48V, C1 = 9µF, C2 = 4µF and C3 = 8µF. (a)Determine the equivalent capacitance of the circuit, (b) determine the energy stored in the combination by calculating the energy stored in the equivalent capacitance, (c) calculate the charge on and potential difference across each capacitor and (d) calculate the energy stored in each individual physical capacitor.
C1 V C2
C3
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Dielectrics: insulating materials with other interesting properties In parallel plate capacitors For a charged, isolated capacitor −Q
+Q
−Q
+Q
V0
V
potential difference decreases same charge => capacitance increases C = Q/V > C0 = Q/V0 Dielectric Constant: K = C/C0 material property p212c24: 13
Effect of dielectric on Electric field parallel plates, constant charge Q = CV = C0V0 => V = V0 /V (reduced) => E = E0/K Material is polarized Effective surface charge distribution σ σ net σ − σ i 1 E0 = E= = σ i = σ 1− ε0 ε0 ε0 K ε = Kε 0
permittivity of dielectric
+σ
−σi
−σ
+σi
σ E= ε A A C = Kε 0 = ε d d 1 1 2 2 u = Kε 0 E = ε E 2 2
+σnet =σ −σi
−σnet =−(σ −σi) p212c24: 14
Example 25-8: Take a parallel plate capacitor whose plates have an area of 2000 cm2 and are separated by a distance of 1cm. The capacitor is charged to an initial voltage of 3 kV and then disconnected from the charging source. An insulating material is placed between the plates, completely filling the space, resulting in a decrease in the capacitors voltage to 1kV. Determine the original and new capacitance, the charge on the capacitor, the dielectric constant of the material, the permittivity of the dielectric, the original and new electric fields, the energy stored in the capacitor with and without the dielectric.
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−+
−+
−+
−+ −+
−+
−+
−+
−+
−+
−+
−+
−+
−+
−+
−+
−+
−+
How does an insulating dielectric material reduce electric fields by producing effective surface charge densities? Reorientation of polar molecules −+
−+
−+
−+
−+
−+
−+
−+
−+
−+
−+
−+
−+
−+
Induced polarization of non-polar molecules − +
− +
− +
− +
− +
− +
− +
− +
− +
− +
− +
− +
− +
− +
− +
− +
Dielectric Breakdown: breaking of molecular bonds/ionization of molecules. p212c24: 16
Polarization (approximately) proportional to applied Electric Field beyond linear approximation: nonlinear optics... Dielectric materials and Gauss’s Law Qenclosed ∫ KE ⋅ dA = ε 0 ∫ ε E ⋅ dA = Qenclosed = ∫ D ⋅ dA
Qenclosed = enclosed free charge D = ε E = Electric Displacement (trivia)
