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Unformatted text preview: e particle travels horizontally, so that
hB = hA , with the result that
1
2
mvB
2 1
2 2
+ EPE B = mv A + EPE A Solving for vA gives
2
v A = vB + 2 ( EPE B − EPE A )
m According to Equation 19.4, the difference in electric potential energies EPE B − EPE A is
related to the electric potential difference VB − VA: EPE B − EPE A = q0 (VB − VA )
Substituting this expression into the expression for vA gives 1012 ELECTRIC POTENTIAL ENERGY AND THE ELECTRIC POTENTIAL vA = 8. 2
vB + 2q0 (VB − VA )
m = ( 0 m/s ) 2 + ( ) 2 −2.0 ×10−5 C ( −36 V )
4.0 ×10 −6 kg = 19 m/s REASONING Let the electrons accelerate to point B from a condition of rest at point A.
2
The speed vB of an electron at B is found from its kinetic energy KE B = 1 mvB
2
(Equation 6.2) at B. Because only the conservative electric and gravitational forces act on
the electron, its total mechanical energy (the sum of its kinetic energy and potential
energies) is conserved. Furthermore, if we assume that the electron only moves horizontally,
then its gravitational potential energy mgh is constant during its acceleration, and we
conclude that the sum of its kinetic energy KEB and electric potential energy EPEB at B is
equal to the sum of its kinetic energy KEA and electric potential energy EPEA at A:
1
2 2
2
mvB + EPE B = 1 mvA + EPE A
2 (1) The difference between the electric potential energy of the electron at B and at A, in turn,
depends upon the charge q = −e of the electron and the potential difference VB − VA that it
accelerates through:
EPE B − EPE A = q (VB − VA ) = − e (VB − VA ) (19.4) We note that, as the electron has a negative charge, it freely accelerates from a lower
potential VA toward a higher potential VB. Therefore the potential difference between B and
A must be VB− VA = +25 000 V.
SOLUTION The electron is initially at rest, so we have that vA = 0 m/s. Substituting this
2
value into Equation (1) and solving for vB , we obtain 1 mv 2
B
2 = EPE A − EPE B 2
vB = or 2 ( EPE A − EPE B )
m (2) Taking the square root of both sides of Equation (2), we find that vB = 2 ( EPE A − EPE B )
m = −2 ( EPE B − EPE A )
m (3) Note that, in order to match the expression on the left side of Equation 19.4, we have
exchanged the electric potential energy terms in the last step of Equation (3):
− ( EPE B − EPE A ) = ( EPE A − EPE B ) . Substituting Equation 19.4 into Equation (3), yields Chapter 19 Problems vf = −2 ( EPE B − EPE A )
m −2 ( −e ) (VB − VA ) = m 1013 2e (VB − VA ) = m Therefore, the speed of an electron just before it strikes the television screen is vB = 9. 2e (VB − VA )
m = 2 (1.6 × 10 −19 C ) ( +25 000 V ) = 9.4 × 10 7 m/s 9.11 × 10 −31 kg REASONING
a. The work WAB done by the electric force in moving a charge q from A to B is related to the potential difference VA − VB between the two points by Equation 19.4 as WAB =
q(VA − VB). Letting A = ground and B = cloud, the work done can be written as
Wgroundcloud = q(Vground − Vcloud) = −q(Vcloud − Vground) (1) b. According to the workenergy theorem (Equation 6.3), the work W done on an object of
2
mass m is equal to its final kinetic energy 1 mvf2 minus its initial kinetic energy 1 mv0 .
2
2
Setting W = Wgroundcloud and noting that v0 = 0 m/s, since the automobile starts from rest,
the workenergy theorem takes the form Wgroundcloud = 1 mvf2 . Solving this equation for vf
2
and substituting Equation (1) for Wgroundcloud, the final speed of the car is
2Wgroundcloud vf = m = ( −2q Vcloud − Vground ) (2) m c. If the work Wgroundcloud were converted completely into heat Q, this heat could be used to
raise the temperature of water. The relation between Q and the change ∆T in the temperature
of the water is Q = cm∆T (Equation 12.4), where c is the specific heat capacity of water and
m is its mass. Solving this expression for m and substituting Equation (1) for Wgroundcloud,
we have m= ( Wgroundcloud −q Vcloud − Vground
Q
=
=
c∆T
c∆T
c∆T ) (3) SOLUTION
a. The work done on the charge as it moves from the ground to the cloud is
Wgroundcloud = −q(Vcloud − Vground). Setting q = −25 C (the charge is negative) and
9 Vcloud − Vground = 1.2 × 10 V, we find that the work is 1014 ELECTRIC POTENTIAL ENERGY AND THE ELECTRIC POTENTIAL ( ) ( ) Wgroundcloud = −q Vcloud − Vground = − ( −25 C ) 1.2 ×109 V = 3.0 ×1010 J
b. From Equation (2) we have for the speed of the automobile that vf = ( −2q Vcloud − Vground
m )= ( −2 ( −25 C ) 1.2 × 109 V
1100 kg c. The mass of water that can be heated is given by m = ) = 7.4 ×103 m/s ( q Vcloud − Vground ) [see Equation c∆T
(3)]. Setting ∆T = 100 C° and using c = 4186 J/(kg⋅C°) from Table 12.2, we find that the
mass of water is
m= ( − q Vcloud − Vground
c ∆T )= ( − ( −25 C ) 1.2 × 109 V ) 4186 J/ ( kg ⋅ C° ) (100 C° ) = 7.2 × 104 kg 10. REASONING The particle slows down as it moves from the lower potential at A to the
higher pot...
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 Spring '13
 CHASTAIN
 Physics, The Lottery

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