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2
2
d = 2v0t + apt 2 (1) We can solve Equation (1) for the initial speed v0 of the proton, but, first, we must
determine the time t and the acceleration ap of the proton . Since the proton strikes the
negative plate at the same instant the electron strikes the positive plate, we can use the
motion of the electron to determine the time t. Chapter 19 Problems For the electron, 1
2 1051 d = 1 aet 2 , where we have taken into account the fact that the electron is
2 released from rest. Solving this expression for t we have t = d / ae . Substituting this expression into Equation (1), we have
d = 2v0 d ap + d ae ae (2) The accelerations can be found by noting that the magnitudes of the forces on the electron
and proton are equal, since these particles have the same magnitude of charge. The force on
the electron is F = eE = eV / d , and the acceleration of the electron is, therefore, ae = F
eV
=
me
me d (3) Newton's second law requires that me ae = mp ap , so that ap
ae = me
mp (4) Combining Equations (2), (3) and (4) leads to the following expression for v0, the initial
speed of the proton:
1 m eV
v0 = 1– e 2 mp me SOLUTION Substituting values into the expression above, we find
v0 = 1 9.11× 10 –31 kg (1.60 ×10 –19C)(175 V)
= 2.77 × 106 m/s
1– –31 2 1.67 ×10 –27 kg 9.11×10 kg 68. REASONING The only force acting on each particle is the conservative electric force.
Therefore, the total energy (kinetic energy plus electric potential energy) is conserved as the
particles move apart. In addition, the net external force acting on the system of two particles
is zero (the electric force that each particle exerts on the other is an internal force). Thus, the
total linear momentum of the system is also conserved. We will use the conservation of
energy and the conservation of linear momentum to find the initial separation of the
particles. 1052 ELECTRIC POTENTIAL ENERGY AND THE ELECTRIC POTENTIAL SOLUTION For two points, A and B, along the motion, the conservation of energy is
1 m v2 + 1 m v2
A
2 1 1,4 2 2 2,
144 2444A
3
Initial kinetic energy
of the two particles kq1q2 + rA
{ = 1 m v2 + 1 m v2
B
2 1 1,4 2 2 2, B
144 2444
3
Final kinetic energy
of the two particles Initial electric
potential energy + kq1q2
rB
{
Final electric
potential energy Solving this equation for 1/rA and setting v1,A = v2,A = 0 m/s, since the particles are initially
at rest, we obtain 1
1
1
=
+
rA
rB
kq1q2 2
2
( 12 m1v1,B + 1 m2v2,B )
2 (1) This equation cannot be solved for the initial separation rA, because the final speed v2,B of
the second particle is not known. To find this speed, we will use the conservation of linear
momentum:
m1v1, A + m2v2, A = m1v1,B + m2v2, B
144
244
3
144
244
3
Initial linear momentum Final linear momentum Setting v1,A = v2,A = 0 and solving for v2,B gives
v2,B = − m1
m2 v1, B = − 3.00 × 10−3 kg
6.00 × 10−3 kg (125 m/s ) = − 62.5 m/s Substituting this value for v2,B into Equation (1) yields
1
1
1
=
+
rA
0.100 m
8.99 × 109 N ⋅ m 2 / C2 8.00 × 10−6 C ( )( ( ) ) 2 × 1 3.00 × 10−3 kg (125 m/s ) + 2
rA = 1.41 × 10 −2 m 2 1
2 ( 6.00 × 10−3 kg ) ( −62.5 m/s )2 CHAPTER 20 ELECTRIC CIRCUITS
ANSWERS TO FOCUS ON CONCEPTS QUESTIONS
1. 1.5 A
2. (c) Ohm’s law states that the voltage V is directly proportional to the current I, according to
V = IR, where R is the resistance. Thus, a plot of voltage versus current is a straight line that
passes through the origin.
3. (e) Since both wires are made from the same material, the resistivity ρ is the same for each.
L
The resistance R is given by R = ρ
(Equation 20.3), where L is the length and A is the
A
crosssectional area of the wire. With twice the length and onehalf the radius (onefourth
L
2
the crosssectional area), the second wire has
=
= 8 times the resistance as the first
A 1/ 4
wire. 4. 250 C°
5. (a) Power P is the current I times the voltage V or P = IV (Equation 20.6a). However, since
Ohm’s law applies to a resistance R, the power is also P = I 2 R (Equation 20.6b) and
V2
(Equation 20.6c). Therefore, all three of the changes specified leave the power
P=
R
unchanged.
6. 27 W
7. 0.29 A
8. (d) According to Ohm’s law, the voltage across the resistance R1 is V1 = IR1 . The two
resistors are connected in series, and their equivalent resistance is, therefore, R1 + R2.
V
According to Ohm’s law, the current in the circuit is I =
. Substituting this
R1 + R2 V
expression into the expression for V1 gives V1 = R + R R1 . 1
2 1054 ELECTRIC CIRCUITS 9. (b) The series connection has an equivalent resistance of RS = R + R = 2 R . The parallel
connection has an equivalent resistance that can be determined from Therefore, it follows that RP = 1 R . The ratio of these values is
2 RS
RP = 1
112
=+=.
RP R R R 2R
= 4.
1R
2 10. (e) Since the two resistors are connected in parallel across the battery terminals, the same
voltage is applied to each. Thus, according to Ohm’s law, the current in each resistor is
I
V / R1 R2
=
inversely proportional to the resistance, so that 1 =
.
I 2 V / R2 R1 11. 0.019 A
12. (c) I...
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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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