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Unformatted text preview: al to its initial charge q2i plus the charge
transferred to it, which is equal to the product of the number N of electrons transferred and
the charge −e of each electron. Therefore,
q2f = q2i + N (−e) = q2i − Ne (1) Since the final charge on the rod is equal to half the total initial charge of the system, we can
substitute q2f = 1 ( q1i + q2i ) into Equation (1) and solve for N:
2
1
2 ( q1i + q2i ) = q2i − Ne or Ne = q2i 1 ( q1i + q2i ) =
2 1
2 ( q2i − q1i ) or N= ( q2i − q1i )
2e Therefore, the number of electrons that must be transferred to the rod is
N= 5. ( q2i − q1i ) = +2.0 ×10−6 C − ( −3.0 ×10−6 C ) = 1.6 ×1013
2e 2 (1.6 × 10−19 C ) SSM REASONING Identical conducting spheres equalize their charge upon touching.
When spheres A and B touch, an amount of charge +q, flows from A and instantaneously
neutralizes the –q charge on B leaving B momentarily neutral. Then, the remaining amount
of charge, equal to +4q, is equally split between A and B, leaving A and B each with equal
amounts of charge +2q. Sphere C is initially neutral, so when A and C touch, the +2q on A
splits equally to give +q on A and +q on C. When B and C touch, the +2q on B and the +q
on C combine to give a total charge of +3q, which is then equally divided between the
spheres B and C; thus, B and C are each left with an amount of charge +1.5q. Chapter 18 Problems 947 SOLUTION Taking note of the initial values given in the problem statement, and
summarizing the final results determined in the REASONING above, we conclude the
following:
a. Sphere C ends up with an amount of charge equal to +1.5 q .
b. The charges on the three spheres before they were touched, are, according to the problem
statement, +5q on sphere A, –q on sphere B, and zero charge on sphere C. Thus, the total
charge on the spheres is + 5 q – q + 0 = + 4 q . c. The charges on the spheres after they are touched are +q on sphere A, +1.5q on sphere B,
and +1.5q on sphere C. Thus, the total charge on the spheres is + q + 1.5 q + 1.5 q = +4 q .
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6. REASONING The conservation of electric charge states that, during any process, the net
electric charge of an isolated system remains constant (is conserved). Therefore, the net
charge (q1 + q2) on the two spheres before they touch is the same as the net charge after they
touch. When the two identical metal spheres touch, the net charge will spread out equally
over both of them. When the spheres are separated, the charge on each is the same.
SOLUTION
a. Since the final charge on each sphere is +5.0 µC, the final net charge on both spheres is
2(+5.0 µC) = +10.0 µC. The initial net charge must also be +10.0 µC. The only spheres
whose net charge is +10.0 µC are B (qB = − 2.0 µ C) and D (qD = +12.0 µ C)
b. Since the final charge on each sphere is +3.0 µC, the final net charge on the three spheres
is 3(+3.0 µC) = +9.0 µC. The initial net charge must also be +9.0 µC. The only spheres
whose net charge is +9.0 µC are
A (qA = − 8.0 µ C), C (qC = +5.0 µ C) and D (qD = +12.0 µ C)
c. Since the final charge on a given sphere in part (b) is +3.0 µC, we would have to add
−3.0 µC to make it electrically neutral. Since the charge on an electron is −1.6 × 10−19 C,
the number of electrons that would have to be added is −3.0 × 10−6 C
= 1.9 × 1013
−19
−1.6 × 10 C
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Number of electrons = 948 7. ELECTRIC FORCES AND ELECTRIC FIELDS REASONING
a. The number N of electrons is 10 times the number of water molecules in 1 liter of water.
The number of water molecules is equal to the number n of moles of water molecules times
Avogadro’s number NA: N = 10 n NA .
b. The net charge of all the electrons is equal to the number of electrons times the change on
one electron.
SOLUTION
a. The number N of water molecules is equal to 10 n NA , where n is the number of moles of
water molecules and NA is Avogadro’s number. The number of moles is equal to the mass
m of 1 liter of water divided by the mass per mole of water. The mass of water is equal to
its density ρ times the volume, as expressed by Equation 11.1. Thus, the number of
electrons is ρV
m
N = 10 n NA = 10 NA = 10 NA 18.0 g/mol 18.0 g/mol 3
−3
3 1000 g 1000 kg/m 1.00 × 10 m −1
23 = 10 1 kg 6.022 × 10 mol 18.0 g/mol ( )( ) ( ) = 3.35 × 10 electrons
26 b. The net charge Q of all the electrons is equal to the number of electrons times the change ( )( ) on one electron: Q = 3.35 × 1026 −1.60 × 10−19 C = −5.36 × 107 C .
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8. REASONING The magnitude F of the forces that point charges q1 and q2 exert on each other varies with the distance r separating them according to F = k q1 q2 (Equation 18.1),
r2
where k = 8.99×109 N·m2/C2.
We note that both charges are given in units of
microcoulombs (µC), rather than the base SI units of coul...
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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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