chapt21 - Chapter 21: Electric Charge Problem 3 A particle...

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Chapter 21: Electric Charge Problem 3 A particle of charge +3.00 × 10 –6 C is 12.0 cm distant from a second particle of charge -1.50 × 10 C. Calculate the magnitude of the electrostatic force on each charge. Answer 2.81 N Solution The magnitude of the force on each charge is given by Coulomb’s law F = 1 4 πε o Q Q r 2 = 9 × 10 9 N-m 2 C 2 3 × 10 6 C ( ) 1.5 × 10 6 C ( ) 12 cm ( ) 2 = 2.81 N Problem 4 Identical isolated conducting spheres, 1 and 2, have equal charges and are separated by a distance large compared with their diameters (Fig. 21-22a). The electrostatic force acting on sphere 2 due to sphere 1 is F . Suppose now that a third identical sphere 3, having an insulating handle and initially neutral, is touched first to sphere 1 (Fig. 21-22b), then to sphere 2 (Fig. 21-22c), and finally removed (Fig. 21-22d). The force that now acts on sphere 2 has magnitude F’. What is the ratio F’/F? Figure 21-22 Answer 3/8
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21-2 Chapter 21: Electric Charge Solution Let q 1 and q 2 be the charges on spheres 1 and 2 respectively. Initially (Fig. 21-21a) the two charges are equal, say Q. The force between the two spheres will change because q 1 and q 2 change and the separation of the two remains constant. Therefore we need to determine what the new charges are in terms of the initial charge so that we can compare the magnitude of the new force to that of the initial force F = 1 4 πε o Q Q r 2 = 1 4 πε o QQ r 2 After step (b) of the procedure spheres 1 and 3 will split the charge that was on 1 initially and therefore each will have Q/2. That is q 1 = Q/2 and q 3 = Q/2. After step (c) of the procedure spheres 2 and 3 will share Q + Q/2. Note that sphere 3 is not discharged between steps (b) and (c) so it has Q/2 before touching sphere 2. Since the spheres are identical in size they will end up with identical amounts of charge, that is, half of Q + Q/2 = 3Q/4. Thus q 2 = 3Q/4 and q 3 = 3Q/4. This means that the magnitude of the force between spheres 1 and 2 after step (c) will be F' = 1 4 πε o Q 2 3Q 4 r 2 = 3 8 1 4 πε o QQ r 2 = 3 8 F The ratio F’/F is. F ' F = 1 4 πε o Q 2 3 Q 4 r 2 1 4 o QQ r 2 = 3 8 Problem 5 Of the charge Q initially on a tiny sphere, a portion q is to be transferred to a second, nearby sphere. Both spheres can be treated as particles. For what value of q/Q will the electrostatic force between the two spheres be maximized? Answer q/Q = 1/2 Let’s assume that Q is positive then the charge on the two spheres will be q and (Q – q). The magnitude of the force between the two spheres is F = 1 4 o q Q q ( ) r 2 The force is a function of q and we want to determine the value of q that will maximize the force. This is a Calc I max-min problem. Start by differentiating F with respect to q.
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Chapter 21: Electric Charge 21-3 dF dq = 1 4 πε o r 2 Q q ( ) q ( ) = 1 4 o r 2 Q 2 q ( ) The critical point and maximum occurs when the derivative is zero. The derivative is zero when Q – 2q = 0 or q/Q = 1/2.
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This note was uploaded on 08/03/2009 for the course PHYSICS 2049 taught by Professor C during the Spring '09 term at Vanderbilt.

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chapt21 - Chapter 21: Electric Charge Problem 3 A particle...

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