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HW5Solutions

# HW5Solutions - HW5 Solutions Notice numbers may change...

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HW5 Solutions Notice numbers may change randomly in your assignments and you may have to recalculate solutions for your specific case. Tipler 21.P.018 Estimate the force required to bind the two protons in the He nucleus together. (Hint: Model the protons as point charges. Assume the diameter of the He nucleus to be approximately 10 -15 m.) Solution: Since the nucleus is in equilibrium: F binding = F electrostatics " F binding = kq 2 r 2 = 9 # 10 9 # 1.6 # 10 \$ 19 ( ) 2 10 \$ 15 ( ) 2 = 0.23 kN Tipler 21.P.023 What is the total charge of all the protons in 1.50 kg of carbon? Solution: Carbon has Z = 6 protons hence Q = 6n C e, where nc = the number of atoms in m = 0.20 kg of carbon. Since N A = Avogadro number = number of Carbon atoms in a mole = number of atoms in M = 12 g of Carbon-12 = 6.022 x 10 23 atoms/mol: N A : M = n C : m " n C = mN A M " Q = 6 e mN A M = 7.23 # 10 7 C Tipler 21.P.028 A 2.0 µ C point charge and a 4.0 µ C point charge are a distance L apart. Where should a third point charge be placed so that the electric force on that third charge is zero? Solution:

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q 3 q 2 q 4 x L-x
Tipler 21.P.035 Five identical point charges, each having charge Q , are equally spaced on a semicircle of radius R as shown in the figure below. Find the force (in terms of k , Q , and R ) on a charge q located equidistant from the five other charges. (Use the following variables as necessary: Q , R , q , k , i for , and j for .) Solution:

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Tipler 21.P.069 A rigid 1.00 m long rod is pivoted about its center. A charge q 1 = 5.00 10 -7 C is placed on one end of the rod, and a charge q 2 = - q 1 is placed a distance d = 10.0 cm directly below it.

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HW5Solutions - HW5 Solutions Notice numbers may change...

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