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Unformatted text preview: l choose the paths C → B and B → A, since we know the potential
differences for each of these segments. Thus, the potential difference VA − VC can be written
as VA − VC = (VA − VB) + (VB − VC ). Since VB − VC = −(VC − VB) = −290 V, and
VA − VB = − (VB − VA) = 0 V, we have that VA − VC = − (VB − VA) − (VC − VB) = 0 V − 290 V = −290 V 42. REASONING AND SOLUTION The capacitance, voltage, and charge are related by
V= q
7.2 × 10−5 C
=
= 12 V
C
6.0 × 10−6 F (19.8) 43. SSM REASONING According to Equation 19.11b, the energy stored in a capacitor with
capacitance C and potential V across its plates is Energy = 1 CV 2 .
2
SOLUTION Therefore, solving Equation 19.11b for V, we have V= 2(Energy)
2(73 J)
=
= 1.1×103 V
–6
C
120 ×10 F 44. REASONING AND SOLUTION The capacitance is given by
C= kε 0 A
d = ( )( 5 8.85 × 10−12 F/m 5 × 10−6 m 2
−8 1 × 10 m )= 2 × 10−8F 45. REASONING
a. The energy used to produce the flash is stored in the capacitor as electrical energy. The
energy stored depends on the capacitance C of the capacitor and the potential difference V
between its plates; Energy = 1 CV 2 (Equation 19.11b).
2
b. The power of the flash is the energy consumed divided by the duration of the flash (see
Equation 6.10b). Chapter 19 Problems 1039 SOLUTION
a. The energy used to produce the flash is
Energy = 1 CV 2
2 = 1
2 (850 × 10−6 F ) ( 280 V )2 = 33 J b. The power developed by the flash is P= Energy
33 J
=
= 8500 W
Time
3.9 × 10−3 s 46. REASONING Equation 19.10 gives the capacitance as C = κε0A/d, where κ is the dielectric
constant, and A and d are, respectively, the plate area and separation. Other things being
equal, the capacitor with the larger plate area has the greater capacitance. The diameter of
the circle equals the length of a side of the square, so the circle fits within the square. The
square, therefore, has the larger area, and the capacitor with the square plates would have
the greater capacitance.
To make the capacitors have equal capacitances, the dielectric constant must compensate for
the larger area of the square plates. Therefore, since capacitance is proportional to the
dielectric constant, the capacitor with square plates must contain a dielectric material with a
smaller dielectric constant. Thus, the capacitor with circular plates contains the material
with the greater dielectric constant.
SOLUTION The area of the circular plates is Acircle = π ( 1 L)
2 2 , while the area of the 2 square plates is Asquare = L . Using these areas and applying Equation 19.10 to each
capacitor gives C= κ circleε 0π ( 1 L )
2 2 C= and d κ squareε 0 L2
d Since the values for C are the same, we have κ circleε 0π ( 1 L )
2 2 d κ circle = 4κ square π =
= κ squareε 0 L2
d
4 ( 3.00 ) π = 3.82 or κ circle = 4κ square π 1040 ELECTRIC POTENTIAL ENERGY AND THE ELECTRIC POTENTIAL 47. SSM REASONING AND SOLUTION Equation 19.10 gives the capacitance for a parallel plate capacitor filled with a dielectric of constant κ : C = κε 0 A / d . Solving for κ ,
we have
Cd (7.0 × 10−6 F)(1.0 ×10 –5 m)
κ=
=
= 5.3
ε 0 A (8.85 × 10 –12 F/m)(1.5 m 2 ) 48. REASONING
The energy stored in a capacitor is given by Energy = 1 CV 2
2
(Equation 19.11b), where C is the capacitance of the capacitor and V is the potential
difference across its plates. The only difference between the two capacitors is the dielectric
material (dielectric constant κ = 4.50) inside the filled capacitor. Therefore, the filled
capacitor’s capacitance C2 is greater than the capacitance C1 of the empty capacitor by a
factor of κ: C 2 = κ C1 (1) Because both capacitors store the same amount of energy, from Energy = 1 CV 2
2
(Equation 19.11b), we have that
1C V2
2 22 = 1 C1V12
2 (2) where V2 is the potential difference across the plates of the filled capacitor, and V1 = 12.0 V
is the potential difference across the plates of the empty capacitor.
SOLUTION Solving Equation (2) for V2, we obtain
V22 C1V12
=
C2 C1V12
V2 =
C2 or Substituting Equation (1) into Equation (3) yields
V2 = C1V12
=
C2 C1 V12 κ C1 = V12 κ = V1 κ = 12.0 V
= 5.66 V
4.50 (3) Chapter 19 Problems 1041 49. REASONING AND SOLUTION The total charge transferred is given by
q = CV = (2.5 × 10−8 F)(450 V)
The number of electrons transferred is, then,
Number of electrons = q
(2.5 × 10−8 F)(450 V)
=
= 7.0 × 1013
−19
e
1.60 × 10 C 50. REASONING The energy Q needed to melt ice is given by Equation 12.5 as Q = mLf,
where m is the mass and Lf is the latent heat of fusion for water. The energy needed to boil
away water is given by Q = mLv, where Lv is the latent heat of vaporization for water. The
latent heat of vaporization is greater than the latent heat of fusion (see Table 12.3).
Therefore, it requires more energy to boil away one kilogram of water than to melt one
kilogram of ice.
According to Equation 19.11b, the energy stored in a capacitor is Energy = 1 CV 2 ,
2 where C is the capacitance and V is the voltage across the p...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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