# CH23 - PART 4 ELECTROMAGNETISM CHAPTER 23 ELECTRIC CHARGE,...

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PART 4 ELECTROMAGNETISM CHAPTER 23 ELECTRIC CHARGE, FORCE, AND FIELD ActivPhysics can help with these problems: Activities 11.1–11.8 Section 23 - 2: Electric Charge Problem 1. Suppose the electron and proton charges differed by one part in one billion. Estimate the net charge you would carry. Solution Nearly all of the mass of an atom is in its nucleus, and about one half of the nuclear mass of the light elements in living matter (H, O, N, and C) is protons. Thus, the number of protons in a 65 kg average-sized person is approximately 1 2 27 28 65 167 10 2 10 () ( . ) kg kg = ×× ¼ , which is also the number of electrons, since an average person is electrically neutral. If there were a charge imbalance of qq e proton electron −= 10 9 , a person’s net charge would be about ±× × × 21 0 1 0 28 9 16 10 32 19 .. , ×= ± C C or several coulombs (huge by ordinary standards). Problem 2. A typical lightning flash delivers about 25 C of negative charge from cloud to ground. How many electrons are involved? Solution The number is Qe == =×= × 25 10 156 10 19 20 C . Problem 3. Protons and neutrons are made from combinations of the two most common quarks, the u quark and the d quark. The u quark’s charge is + 2 3 e while the d quark carries 1 3 e . How could three of these quarks combine to make (a) a proton and (b) a neutron? Solution (a) The proton’s charge is 1 2 3 2 3 1 3 eeee =+− , corresponding to a combination of uud quarks; (b) for neutrons, 0 = 2 3 1 3 1 3 −− corresponds to udd . (See Chapter 39, or Chapter 45 in the extended version of the text.) Problem 4. A 2-g ping-pong ball rubbed against a wool jacket acquires a net positive charge of 1C µ . Estimate the fraction of the ball’s electrons that have been removed. Solution If half the ball’s mass is protons, their number (equal to the original number of electrons) is 1 g = m p . The number of electrons removed is 1 C = e , so the fraction removed is . . . 1 1 10 10 1 104 62 4 19 11 C g g = = e gm p = (a hundred billionth).

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CHAPTER 23 541 Section 23-3: Coulomb’s Law Problem 5. If the charge imbalance of Problem 1 existed, what would be the approximate force between you and another person 10 m away? Treat the people as point charges, and compare the answer with your weight. Solution The magnitude of the Coulomb force between two point charges of 3.2 C (see solution to Problem 1), at a distance of 10 m, is kq r 22 9 2 2 2 8 91 0 3 2 1 0 9 2 21 0 == = × (/ ) ( . ) . NmC C m N . This is approximately 1.45 million times the weight of an average-sized 65 kg person. Problem 6. Find the ratio of the electrical force between a proton and an electron to the gravitational force between the two. Why doesn’t it matter that you aren’t told the distance between them? Solution At all distances (for which the particles can be regarded as classical point charges), the Coulomb force is stronger than the gravitational force by a factor of: F F ke r r Gm m pe elec grav = F H G I K J F H G I K J 2 2 2 = ×⋅ × × × × −− ) ( . ) 10 / )( . )( . ) ..
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## This note was uploaded on 09/28/2008 for the course PY 208N taught by Professor Dougherty during the Spring '08 term at N.C. State.

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CH23 - PART 4 ELECTROMAGNETISM CHAPTER 23 ELECTRIC CHARGE,...

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