Two identical conducting spheres, A and B, carry equal charge. They are initially separated by a distance much larger than their diameters, and the force between them is F. A third identical conducting sphere, C, is uncharged. Sphere C is first touched to A, then to B, and then removed. As a result, the force between A and B is equal to.
A. 0
B. F/16
C. F/4
D. 3F/8
E. F/2
(GR9677 #24)
Coulomb's Law, Fi = kqAqB/r2
qA = qB = Q
Fi = F = kQ2 /r2
When uncharged sphere C touches A, charge A distributes itself evenly:
qC = qA' = 1/2 qA = 1/2Q
When sphere C touches B:
qC' = qB' = 1/2(qC + qB) = 1/2(1/2Q + Q) = 3/4Q
Therefore,
Ff = kqA' qB'/r2
= k (1/2Q) (3/4Q)/r2
= 3/8 kQ2/r2
= 3/8 F
Answer: D
No comments :
Post a Comment