Chapter 21: Electric Charge

21-9 THINK Since opposite charges attract, the initial charge configurations must be of opposite signs. Similarly, since like charges repel, the final charge configurations must carry the same sign.

EXPRESS We assume that the spheres are far apart. Then the charge distribution on each of them is spherically symmetric and Coulomb’s law can be used. Let q1 and q2 be the original charges. We choose the coordinate system so the force on q2 is positive if it is repelled by q1. Then the force on q2 is

1 q1q2 q1q2 Fa   2  k 2 4 0 r r

9 2 2 where k =1/ 4pe 0 = 8.99醋 10 N m C and r = 0.500 m. The negative sign indicates that the spheres attract each other. After the wire is connected, the spheres, being identical, acquire 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 (q1 + q2)/2. The force is now repulsive and is given by

q1 q2 q1 q2 2 1 d2 id2 i bq1  q2 g Fb  2  k 2 . 4 0 r 4r

We solve the two force equations simultaneously for q1 and q2.

ANALYZE The first equation gives the product

2 r2 F (0.500m) ( 0.108 N) q q = -a = - = -3.00 10-12 C 2 , 1 2 k 8.99醋 109 N m 2 C 2 and the second gives the sum

F 0.0360N q  q  2r b  2 0.500m  2.00 106 C 1 2 k b g8.99 109 N  m2 C2

where we have taken the positive root (which amounts to assuming q1 + q2  0). Thus, the product result provides the relation -(3.00 10-12 C 2 ) q2 = q1 which we substitute into the sum result, producing 12 2 3.00 10 C 6 q1   2.00 10 C. q1

Multiplying by q1 and rearranging, we obtain a quadratic equation

2 6 12 2 q1  c2.00 10 Chq1  3.00 10 C  0. The solutions are 2 2.00 106 C  c2.00 106 Ch 4c3.00 1012 C2 h q  . 1 2

–6 If the positive sign is used, q1 = 3.00  10 C, and if the negative sign is used, -6 q1 = -1.00 10 C .

–12 –6 -6 (a) Using q2 = (–3.00  10 )/q1 with q1 = 3.00  10 C, we get q2 = -1.00 10 C .

–6 -6 (b) If we instead work with the q1 = –1.00  10 C root, then we find q2 = 3.00 10 C .

LEARN Note that 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 +3.00  10–6 C.

What happens if we had not made the assumption, above, that q1 + q2  0? If the signs of the charges were reversed (so q1 + q2 < 0), then the forces remain the same, so a charge of +1.00  10–6 C on one sphere and a charge of –3.00  10–6 C on the other also satisfies the conditions of the problem.

21-19 THINK Our system consists of two charges in a straight line. We’d like to place a third charge so that all three charges are in equilibrium.

EXPRESS If the system of three charges is to be in equilibrium, the force on each charge must be zero. The third charge q3 must lie between the other two or else the forces acting on it due to the other charges would be in the same direction and q3 could not be in equilibrium. Suppose q3 is at a distance x from q, and L – x from 4.00q. The force acting on it is then given by 1 骣qq4 qq F =琪 3 - 3 3 琪 2 2 4pe0 桫x (L- x)

where the positive direction is rightward. We require F3 = 0 and solve for x.

ANALYZE (a) Canceling common factors yields 1/x2 = 4/(L – x)2 and taking the square root yields 1/x = 2/(L – x). The solution is x = L/3. With L = 9.00 cm, we have x = 3.00 cm. (b) Similarly, the y coordinate of q3 is y = 0.

(c) The force on q is 2 -1骣qq3 4.00q Fq =琪 2 + 2 . 4pe0 桫x L

The signs are chosen so that a negative force value would cause q to move leftward. We require Fq = 0 and solve for q3:

4qx2 4q 4 q= - = - q � 3 - = - 0.444 3 L2 9 q 9 where x = L/3 is used.

LEARN We may also verify that the force on 4.00q also vanishes:

1骣 4q24qq 1骣 4 q 24(- 4 9)q2 1骣 4 q 2 4 q 2 F =琪 +0 =琪 + = - = 0 . 4q 琪 22 琪 2 2琪 2 2 4pe0桫L(L- x) 4 p e 0桫 L( 4 9) L 4 p e 0 桫 L L