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Unformatted text preview: lates. Since the voltage is the same for
both capacitors, the capacitor storing the greater energy has the greater capacitance.
Capacitor B contains more energy, since it can boil the water. Therefore, capacitor B must
have the greater capacitance. SOLUTION Using the relations Energy = 1 CV 2
2 (Equation 19.11b) and Q = mL (Equation 12.5), we find ( Energy )A = mLf 1
= 2 CAV 2 and 1
( Energy )B = mLv = 2 CBV 2 Dividing these two results gives
1
mLv 2 CBV 2
=
1C V2
mLf
2A Lv CB
=
Lf CA or or C B = CA Lv
Lf Taking the values for the latent heats from Table 12.3, we find CB = CA Lv
Lf ( = 9.3 ×10 −6 22.6 ×105 J/kg −5
F 33.5 × 104 J/kg = 6.3 × 10 F ) 1042 ELECTRIC POTENTIAL ENERGY AND THE ELECTRIC POTENTIAL 51. SSM REASONING According to Equation 19.11b, the energy stored in a capacitor with
a capacitance C and potential V across its plates is Energy = 1 CV 2 . Once we determine
2
how much energy is required to operate a 75W light bulb for one minute, we can then use
the expression for the energy to solve for V.
SOLUTION The energy stored in the capacitor, which is equal to the energy required to
operate a 75W bulb for one minute (= 60 s), is Energy = Pt = ( 75 W)(60 s) = 4500 J
Therefore, solving Equation 19.11b for V, we have V= 2(Energy)
=
C 2(4500 J)
= 52 V
3.3 F 52. REASONING
a. The conducting shells are equipotential surfaces, so the average magnitude E of the
∆V
electric field between them is given by E =
(Equation 19.7a, minus sign omitted),
∆s
where ∆V is the magnitude of the potential difference between the shells and ∆s is the
distance between them. This distance is equal to the radius router of the outer shell minus the
radius rinner of the inner shell.
κε A
b. Because this is not a parallelplate capacitor, we cannot use C = 0 (Equation 19.10)
d
to determine the capacitance C. Instead, we will make use of q = CV (Equation 19.8),
where q is the magnitude of the charge on one of the cylindrical shells and V is the
magnitude of the potential difference between them. The value of V is equal to the value ∆V
found in part (a).
SOLUTION a. Solving E = ∆V
(Equation 19.7a, minus sign omitted) for ∆V, we find
∆s ( )( ) ∆V = E∆s = E ( router − rinner ) = 4.2 ×104 V/m 2.50 ×10−3 m − 2.35 ×10−3 m = 6.3 V
b. Solving q = CV (Equation 19.8) for C, and noting that V, the potential difference
between the shells is identical to the potential difference ∆V = 6.3 V found in (a), we obtain C= q 1.7 × 10 −10 C
=
= 2.7 × 10 −11 F
V
6.3 V Chapter 19 Problems 1043 53. REASONING The charge q0 on the empty capacitor is related to its capacitance C0 and the
potential difference V across the plates by q0 = C0V (Equation 19.8). The charge q on the
capacitor filled with a dielectric is related to its capacitance C and the potential difference V
across the plates by q = C V. We know that the presence of the dielectric increases the
capacitance such that C = κ C0, where κ is the dielectric constant (see Equation 19.10 and
the discussion that follows). Since the magnitude of the surface charge on the dielectric is
equal to the difference in the charge on the plates with and without the dielectric, we have q − q0 = CV − C0V = (κ C0 ) V − C0V = C0V (κ − 1)
−6 SOLUTION Since C0 = 3.2 × 10
charge is ( F, V = 12 V, and κ = 4.5, the magnitude of the surface ) q − q0 = C0V (κ − 1) = 3.2 × 10−6 F (12 V )( 4.5 − 1) = 1.3 × 10−4 C 54. REASONING The charge q stored on the plates of a capacitor connected to a battery of
κε A
voltage V is q = CV (Equation 19.8). The capacitance C is C = 0 (Equation 19.10),
d
where κ is the dielectric constant of the material between the plates, ε0 is the permittivity of
free space, A is the area of each plate, and d is the distance between the plates. Once the
capacitor is charged and disconnected from the battery, there is no way for the charge on the
plates to change. Therefore, as the distance between the plates is doubled, the charge q must
remain constant. However, Equation 19.10 indicates that the capacitance is inversely
proportional to the distance d, so the capacitance decreases as the distance increases. In
Equation 19.8, as C decreases, the voltage V must increase in order that q remains constant.
The voltage increases as a result of the work done in moving the plates farther apart. In
solving this problem, we will apply Equations 19.8 and 19.10 to the capacitor twice, once
with the smaller and once with the larger value of the distance between the plates.
SOLUTION Using q = CV (Equation 19.8) and C = κε 0 A
d (Equation 19.10), we can express the charge on the capacitor as follows: ε AV κε A q = CV = 0 V = 0
d
d
where we have made use of the fact that κ = 1 , since the capacitor is empty. Applying this
result to the capacitor with smaller and larger values of the distance d, we have 1044 ELECTRIC POTENTIAL ENERGY AND THE ELECTRIC POTENTIAL q= ε 0 AVsmaller q= and dsmaller ε 0 AVlarger
d larger Since q is the same in each of these expressions, it follows that ε 0 AVsmaller
dsmaller = ε 0 AVlarger
dlarger Vsmaller or dsm...
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