Question

My Notes Ask Your You are working for a manufacturing coriparny. Your supervisor has an idea fur controlling the position ofa stll bead by using alectric flelds. The physical setup is shown In tha figure helaw. Fixed beads Movable bead Three fixed beads are secured along the y axis, with a bead of charge .nq at the origin and charges of mq a distance a above and below the origin. The values of m and n can be set by an operator On an inaulating wire laid along the x-axis, a movable bead with charge -po is placed, By acjusting m and n, the movable bead can be moved to ditterent equilibrium positions x along the wire. Your supervisor azks you to set up the system and test it. In particular, he aks for answers to the following questions (a) f n-9 In a particular test, what is the value of m that will place the movable bead In equl iorium at x-a? (b) In a second test with n-9, what is the value orm that will place the movable bead in equilibrium at x-2a? (G) f n-9 and the largast you can make ms m -9, what is the dseat equbrium position of the bead to the orgin? (Give your swern eris of d.] (d) You decide to impress your suerr by finding the range of values of m'n that will place the bead in equilibriunm somewhere along the x ads. Note that m, n, and p are not necessarly intagers. [Glve your answer as an inequality Use the following as necessary:m, P.)

Answer 1

The expression of coulomb's force is F = Here, q and q, are the charges, r is the distance between these charges, k is coulomb's constant Like charges repel each other and unlike charges attract each other (a) For the charge -pg to be at equilibrium, the force acting on this charge due to other three charges will be zero. ma 3 +ng 45° Ea mg 2 Figure: System of charges The force on .charge - рg due to +ng is, #(ра) (пg) (-i) The force on charge -pq due to -mq (upper charge) is, k(pg)(mg) cos 45°f- sin 45°j) charge p due to -mq (lower charge) is k(pg)(mg)cos45°f +sin45°j) (a) The force on C

The net force should be zero on charge -pq R+F+0 k(pg)(ma) (Za (cos 450f +sin 45°7)+(pq)(ng)(-i) = 0 (a) cos 45°-sin 45°)+(pg)(ma) knpa (cos 457-sin 450) bnpq cos 45°i+sin 45° p 0 kpa m 71 (cos 45°f+sin45°j cos 45-sin 45°j)+. 0 _ l(E) a 2m 45)-ni =0 COs (m(c0 45)-n)=0 -11 m cos 45° -n= 0 -n 0 m m=ny Since, n 9 n=92 (b) For the charge -pg to be at equilibrium, the force acting on this charge due to other three charges will be zero. The force on charge -p due to +nq is k(pa) (ng) (-) = (2a) The angle between charge -pg and -mą (upper charge) is = fan 2a 26.56°

The force charge p due to -mq (upper charge) is, on k(pg)(mg) (50) (cos 26.56-sin 26.56°j Here,5a is the distance between charges -pq due to -mq . charge -p due to -mq (lower charge) is k(pq)(mg)cos26.56°f +sin 26.56°j) (5a) The force on Here, 5a is the distance between charges pq due to -mq The net force should be zero on charge -pq (cos 26.56+sin 26.56°) *(pq)(ng) (-i) 0 (2a) k(pa) (mg) k(pg)(ma) cos 26.56-sin 26.56°j (Sa) onpq cos 26.56°-sin 26.56°)+ npq (cos 26.56sin 26.56°7)+pq(4 =0 (a) a) (2a)2 kpq V5 (cos 26.56i +sin 26.56°j)+^(-i)|=0 26.56-sin 26.56°j). (5) m 0 cos 2m 26.56 0 cos --1 4 45 2m cos 26.56 5 0 4 2m cos 26.56°) 5 4 5 m= 77 8cos 26.56° m 0.698n Since, n9 0.698 (9) = 6.288 m= m 6.3

(c) Consider that the distance of charge -pq and +nq is x. For the charge -pg to be at equilibrium, the force acting on this charge due to other three charges will be zero. The force on charge -p due to +nq is #(ра) (пg) (-i) The force on charge p due to -mq (upper charge) is, k(pa)(mg)cosei- sin ej) Here, Va xis the distance between charges -pq due to -ma . charge pdue to -mq (lower charge) is *(ра) (mа) (сos ef +sin @) The force on +x Here, vax distance between charges -p due to -mą. 1S The net force should be zero on charge pa kpmgcos0i -sinei pq ) ( mg)cosei- (va k(pg (ng) (-)-0 + sin 6 (x) +r kmpq kmpa (cosei-sinej)+ (-)-0 cos ei+sin koipq a x 2 kompg(Cosei-p (i)=0 x Refer the above figure and use trigonometry identities cose Vxa 2kmpa Solve cos (x) +x

2kmp (i)= 0 (x)* vx+a 2m n (a +x) (a+x) 3/2 2m n 3/2 2m 2/3 2m n 2/3 2m -1 1/2 a2 2/3 2m -1 77 2/3 2m -1 Since, 9 and m= 19 2/3 2(19) x = 0.788a

(d) The distance on the x-axis for the equilibrium is: a 12 2/3 2(19) is very-very large as compared to value of n such that: m >>n. This If the value of m implies that: x->0 And as m - - then, x 0o 1 m <ーく+0 2 Thus, the range of m/n is: