Question

You are working for a manufacturing company. Your supervisor has an idea for controlling the position of a small bead by using electric fields. The physical setup is shown in the figure below 1 Fixed beads ng Movable bead + nq -pa mg Three fixed beads are secured along the y-axis, with a bead of charge +ng 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 insulating wire laid along the x-axis, a movable bead with charge -pa is placed. By adjusting m and n, the movable bead can be moved to different equilibrium positions x along the wire. Your supervisor asks you to set up the system and test it. In particular, he asks for answers othe following questions. (a) If n 3 in a particular test, what is the value of m that will place the movable bead in equilibrium at x-a? m= (b) In a second test with n 3, what is the value of m that will place the movable bead in equilibrium at x- 2a? (c) If n-3 and the largest you can make m is m 7, what is the closest equilibrium position of the bead to the origin? (Give your answer in terms of a.) (d) You decide to impress your supervisor by finding the range of values of m/n that will place the bead in equilibrium somewhere along the x-axis. Note that m, n, and p are not necessarily integers. (Give your answer as an inequality. Use the following as necessary: m, n, p.)

Answer 1

a) For n = 3 & x = a By force equilibrium F^1 = 2F cos theta k npq^2/a^2 = 2, k m p q^2/(a squareroot 2)^2 times 1/ squareroot 2 np = mp/ squareroot 2 doublerightarrow m = squareroot 2 n so n = 3 m = 3 squareroot 2 n = 3 & n = 2a By force equilibrium F^1 = 2 F cos theta [tan theta = a/2a] k n p q^2/(2 a)^2 = 2 times k m p q^2/(squareroot 5 a)^2 - 2a/ squareroot 5 a n/4 = 2/5 m times 2/ squareroot 5 doublerightarrow m = 5 squareroot 5 n/16 m = 15 squareroot 5/16 if n = 3 n = 3, m = 7 then by force equilibrium k n p q^2/x^2 = 2 times k m p q^2/(squareroot x^2 + a^2)^2 times x/ squareroot x^2 + a^2 3/x^2 = 2.7 x/(x^2 + a^2)^3/2 doublerightarrow 3/14(x^2 + a^2)^3/2 = x^3