# hw2 - Physics 21 Fall, 2008 Solution to HW-2 21-1 Excess...

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Physics 21 Fall, 2008 Solution to HW-2 21-1 Excess electrons are placed on a small lead sphere with a mass of m =8 . 40 g so that its net charge is Q = 4 . 00 × 10 9 C. (a) Find the number of excess electrons on the sphere. (b)How many excess electrons are there per lead atom? The atomic number of lead is 82, and its molar mass is 207 g/mol. (a) The number of excess electrons n is the excess charge Q divided by the electronic charge e . n = 4 . 00 × 10 9 1 . 602 × 10 19 =2 . 5 × 10 10 (b) A mole is 6 . 02 × 10 23 atoms (Avogadro’s number). The mass of a mole of lead is 207 g, so the number of lead atoms in 8.40 g is 8 . 40 2 . 07 × 6 . 02 × 10 23 =2 . 44 × 10 22 . The number of excess electrons per atom is 2 . 5 × 10 10 2 . 44 × 10 22 10 12 The excess charge per lead atom is very small. 21-12 A charge of Q = 0 . 580 µ Cexertsanupwardforce of F =0 . 230 N on an unknown charge q a distance d = 0 . 350 m directly below it. (a) What is the unknown charge? (Give the magnitude and sign.) (b) What is the magnitude of the force that the unknown charge q exerts on the charge of 0 . 580 µ C? (c) What is the direction of the force that the unknown charge q exerts on the charge of 0 . 580 µ C? Q = -0.580 μ C q =? d = 0.350 m (a) The magnitude of the Coulomb force is F = 1 4 π± 0 | qQ | d 2 . Solving for | q | , we obtain | q | = 4 π± 0 Fd 2 | Q | = 4 π (8 . 854 × 10 12 )(0 . 230)(0 . 35) 2 0 . 580 × 10 6 =5 . 40 × 10 6 C=5 . 40 µ

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## This note was uploaded on 02/09/2009 for the course PHYS 021 taught by Professor Hickman during the Fall '08 term at Lehigh University .

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hw2 - Physics 21 Fall, 2008 Solution to HW-2 21-1 Excess...

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