This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Chapter 21 1 The magnitude of the force that either charge exerts on the other is given by F = 1 4 j q 1 jj q 2 j r 2 ; where r is the distance between them. Thus r = s j q 1 jj q 2 j 4 F = s (8 : 99 10 9 N m 2 = C 2 )(26 : 10 6 C)(47 : 10 6 C) 5 : 70 N = 1 : 38 m : 5 The magnitude of the force of either of the charges on the other is given by F = 1 4 q ( Q q ) r 2 ; where r is the distance between the charges. You want the value of q that maximizes the function f ( q ) = q ( Q q ). Set the derivative df=dq equal to zero. This yields Q 2 q = 0, or q = Q= 2. 7 Assume the spheres are far apart. Then the charge distribution on each of them is spherically symmetric and Coulombs law can be used. Let q 1 and q 2 be the original charges and choose the coordinate system so the force on q 2 is positive if it is repelled by q 1 . Take the distance between the charges to be r . Then the force on q 2 is F a = 1 4 q 1 q 2 r 2 : The negative sign indicates that the spheres attract each other. After the wire is connected, the spheres, being identical, have the same charge. Since charge is conserved, the total charge is the same as it was originally. This means the charge on each sphere is ( q 1 + q 2 ) = 2. The force is now one of repulsion and is given by F b = 1 4 ( q 1 + q 2 ) 2 4 r 2 : Chapter 21 131 Solve the two force equations simultaneously for q 1 and q 2 . The first gives q 1 q 2 = 4 r 2 F a = (0 : 500 m) 2 (0 : 108 N) 8 : 99 10 9 N m 2 = C 2 = 3 : 00 10 12 C 2 and the second gives q 1 + q 2 = 2 r p 4 F b = 2(0 : 500 m) s : 0360 N 8 : 99 10 9 N m 2 = C 2 = 2 : 00 10 6 C : Thus q 2 = (3 : 00 10 12 C 2 ) q 1 and substitution into the second equation gives q 1 + 3 : 00 10 12 C 2 q 1 = 2 : 00 10 6 C : Multiply by q 1 to obtain the quadratic equation q 2 1 (2 : 00 10 6 C) q 1 3 : 00 10 12 C 2 = 0 : The solutions are q 1 = 2 : 00 10 6 C p ( 2 : 00 10 6 C) 2 + 4(3 : 00 10 12 C 2 ) 2 : If the positive sign is used, q 1 = 3 : 00 10 6 C and if the negative sign is used, q 1 = 1 : 00 10 6 C. Use q 2 = ( 3 : 00 10 12 ) =q 1 to calculate q 2 . If q 1 = 3 : 00 10 6 C, then q 2 = 1 : 00 10 6 C and if q 1 = 1 : 00 10 6 C, then q 2 = 3 : 00 10 6 C. Since the spheres are identical, the solutions are essentially the same: one sphere originally had charge 1 : 00 10 6 C and the other had charge +3...
View Full
Document
 Fall '98
 Heckman
 Physics, Charge, Force

Click to edit the document details