3 1 Find to thre signfeant digits the charge and the mass of

#3

1. Find to thre signfeant digits the charge and the mass of the Sgreestion Begin by looking up the mass of a neutral atom on the porodic tabte of te elements in Appendis C (a) an ionized bydrogen stom. represented as H th) a singly ionized sodium atom, Na (e) a chloride ion CI (di a doubly ionized calcium atom, Ca-Ca (c) the center of an ammonia molecule, modeled as an N ion (n quadruple iotuzed nitrogen atoma, N\"\", found in plasm\" inhot star () the nucleus of a nitrogen atom th) the molecular ion HO 2 Two identical conducting small spheres are placed with their centers 0 350 m apart One is given a charge of 12.0 nC and the other a charge of 15.0 nC (a) Find the electric force exerted by one sphere on the other (b) The spheres are connected by a conducting wire. Find the electric force each exerts on the other after they have come to equilibrium 3. Two small beads having positive charges g-13g and q: g are fixed at the opposite ends of a horizontal insulating rod of length d-1.50 m. The bead with charge gi is at the origin. As shown in the figure below, a third small, charged bead is free to slide on the rod (a) At what position x is the third bead in equilibrium (b) Can the equilibrium be stable? 4 Two charged particles are located on the x axis. The first is a charge at--The second is an unknown charge located at x +3a The net electric field these charges produce at the origin has a magnitude of 2k/a What are the two possible values of the unknown charge? 5. A uniformly charged ring of radius 10.0 cm has a total charge of 95 0 HC. Find the electric field on the axis of the ring at the following distances from the center of the ring (a) 1.00 cm (b) 5 00 cm (c) 30.0 cm (d) 100 cm 6. The figure below shows the electric field lines foe swn charged particles separated by a small distance

Solution

Part b: Stable equilibrium if charge is positive

Part a:

Force on Q due to q1 = k(13q)Q/X² acting to the right
Force on Q due to q2 = k(q)Q/(1.5-X)² acting to the left

When Q is in equilibrium these 2 opposite direction forces must have equal magnitudes:

k(13q)Q/X² = k(q)Q/(1.5-X)²
13/X² = 1/(1.5-X)²
3.60555128/X = 1/(1.5-X)
3.60555128 (1.5-X) = X
5.40832692 – 3.60555128X = X
X = 1.174m