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Unformatted text preview: A (200 V). 11. (c) The magnitude of the electric field between the plates is E = − ∆V
(Equation 19.7).
∆s
Since ∆s is the same for all four segments, E is proportional to the potential difference ∆V
in each segment. ∆V is 2 units for segment D, 1 unit for segment B, and 0 units for segments
A and C. 12. (e) The magnitude of the electric field between the plates is E = − 13. E = −7.5 V/m
14. E = −2.0 × 103 V/m
15. (a) The capacitance of a parallel plate capacitor is C = κε 0 A / d (Equation 19.10). Since the
area A increases by a factor of 4 and the spacing d between the plates increases by a factor
of 2, the capacitance increases by a factor of 4/2 = 2.
16. (b) The amount of charge on each plate is q = CV (Equation 19.8), where C = κε 0 A / d
(Equation 19.10). If the plate separation d increases, C decreases. If C decreases while V is
held constant, the amount of charge decreases. 17. (c) The energy stored in a capacitor is directly proportional to its capacitance (see Equation
19.11). The capacitance, on the other hand, is directly proportional to the dielectric constant
and the area of each plate and is inversely proportional to the plate separation (Equation
19.10). Therefore, inserting a dielectric, increasing the area of each plate, and decreasing the
plate separation increases the energy stored.
18. Energy = 6.0 × 10−4 J Chapter 19 Problems 1007 CHAPTER 19 ELECTRIC POTENTIAL ENERGY AND THE ELECTRIC POTENTIAL
PROBLEMS
1. REASONING When the electron moves from the ground to the cloud, the change in its
electric potential energy is ∆(EPE) = EPEcloud − EPEground. (Remember that the change in
any quantity is its final value minus its initial value.) The change in the electric potential
energy is related to the change ∆V in the potential by ∆(EPE) = q0∆V (Equation 19.4),
where q0 is the charge on the electron. This relation will allow us to find the change in the
electron’s potential energy.
SOLUTION The difference in the electric potentials between the cloud and the ground is
∆V = Vcloud − Vground = 1.3 × 108 V, and the charge on an electron is q0 = −1.60 × 10−19 C.
Thus, the change in the electron’s electric potential energy when the electron moves from
the ground to the cloud is ( )( ) ∆ ( EPE ) = q0 ∆V= −1.60 ×10−19 C 1.3 ×108 V = −2.1×10−11 J 2. REASONING AND SOLUTION The only force that acts on the αparticle is the
conservative electric force. Therefore, the total energy of the αparticle is conserved as it
moves from point A to point B:
1 mv 2 + EPE = 1 mv 2 + EPE
A
B 244B
2
2
3
144
244A
3 144
Total energy at
point A Total energy at
point B Since the αparticle starts from rest, vA = 0 m/s. The electric potential V is related to the
electric potential energy EPE by V = EPE/q (see Equation 19.3). With these changes, the
equation above gives the kinetic energy of the αparticle at point B to be
1 mv 2
B
2 = EPE A − EPE B = q (VA − VB ) Since an αparticle contains two protons, its charge is q = 2e = 3.2 × 10–19 C. Thus, the
kinetic energy (in electronvolts) is 1008 ELECTRIC POTENTIAL ENERGY AND THE ELECTRIC POTENTIAL 1 mv 2
B
2 ( ) = q (VA − VB ) = 3.2 × 10−19 C +250 V − ( −150 V ) 1.0 eV
= 1.28 × 10−16 J −19 1.6 × 10 3. 2 = 8.0 × 10 eV
J SSM REASONING The work WAB done by an electric force in moving a charge q0 from point A to point B and the electric potential difference VB − VA are related according to
−WAB
VB − VA =
(Equation 19.4). This expression can be solved directly for q0.
q0
SOLUTION Solving Equation 19.4 for the charge q0 gives
q0 = 4. −WAB VB − VA = WAB V A − VB = 2.70 × 10−3 J
= 5.40 × 10−5 C
50.0 V REASONING The maximum operation time is the energy used by the shaver divided by
the rate of usage, which is given. The energy that the shaver uses is the energy carried by
the charge that passes between the terminals of the battery. This energy is the charge times
the battery voltage. The charge can be determined from the number of particles specified
and the charge on each particle.
SOLUTION The energy used by the shaver is that obtained by the electric charge as it
passes from the positive terminal of the battery (terminal A), where the electric potential
energy EPEA is higher, to the negative terminal (terminal B), where the electric potential
energy EPEB is lower. The energy acquired by the charge is EPEA − EPEB. The rate at
which the shaver uses energy is the power P, which is given as 4.0 W. According to
Equation 6.10b, the power is the energy EPEA − EPEB divided by the time t, so that P = ( EPE A − EPE B ) / t . Solving for the time gives
t= EPE A − EPE B
P (1) According to Equation 19.4, this total energy is EPE A − EPE B = q0 (VA − VB ) , where q0 is
the charge and VA − VB is the electric potential difference between the battery terminals.
Substituting this expression into Equation (1) gives Chapter 19 Problems t= EPE A − EPE B
P = q0 (VA − VB )
P 1009 (2) The charge q0 is the number n of charged particles times the charge ( e = 1.6 ×10−19 C ) on...
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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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