Physics 2 Chapter 21 Problem 35 Chapter 21, Problem 035 Your answer ia partially
ID: 1636497 • Letter: P
Question
Physics 2 Chapter 21 Problem 35 Chapter 21, Problem 035 Your answer ia partially correct. Try again. In crystals of the salt cesum dlonde, ces am ns Cs form the eight corners of a cube andchlorine ion cr ss et the cube's center (see the figure). The edge length of the cute is L-0.32 nm. The Cs" iors are each deficient by one electrom (and thus each has a charge of te), and the Cl ion has one excess electrom (and thus has a charge ef e). (a) What is the magnitude of the net electrostatic force exerted on the d on by the eight Cs. ions at the corners of the cube? (b)If one of the Cs, ions missn9, the stal s said to have a defect; what s the magnitude of the electrostatic force exerted on the d-ion by the seven remaining Cs·ions? (e) Numbe' o (b) N Click if you would like to Show Work for this question: Unts |[ This 3.166-9 Qpen Show WorkExplanation / Answer
(a) Every cesium ion at a corner of the cube exerts a force of the same magnitude on the chlorine ion at the cube center. Each force is a force of attraction and is directed toward the cesium ion that exerts it, along the body diagonal of the cube. We can pair every cesium ion with another, diametrically positioned at the opposite corner of the cube. Since the two ions in such a pair exert forces that have the same magnitude but are oppositely directed, the two forces sum to zero and, since every cesium ion can be paired in this way, the total force on the chlorine ion is zero.
(b) Rather than remove a cesium ion, we superpose charge –e at the position of one cesium ion.
This neutralizes the ion, and as far as the electrical force on the chlorine ion is concerned, it is equivalent to removing the ion. The forces of the eight cesium ions at the cube corners sum to zero, so the only force on the chlorine ion is the force of the added charge.
The length of a body diagonal of a cube is
Sqrt(3).a
, where a is the length of a cube edge. Thus, the
distance from the center of the cube to a corner is
d = Sqrt(3).a/2
The force has a magnitude:
F = ke^2/d^2 = (8.99 x 10^9) (1.6 x 10^-19)^2 / (3/4)(0.32 x 10^-9)^2 = 3 x 10^-9 N