Question
In the figure below, particles 1 and 2 of charge q1 = q2 = +3.20 10-19 C are on a y axis at distance d = 28.0 cm from the origin. Particle 3 of charge q3 = +9.60 10-19 C is moved gradually along the x axis from x = 0 to x = +5.0 m. At what values of x will the magnitude of the electrostatic force on the third particle from the other two particles be (a) minimum and (b) maximum? What are the (c) minimum and (d) maximum magnitudes?
Explanation / Answer
Part A Let the magnitude of negative charge be q. That of positive charge separated L cm away will be 9q The net electric field at any point x > 0 is given by Enet = (9*10^9)q*(10^4)*[9/(x+L)^2 - 1/x^2] ----------------------------- 1 To find where it will be maximum we differentiate [9/(x+L)^2 - 1/x^2] with respect to x and equate it to zero -18/(x+L)^3 + 2/x^3 = 0 or (x+L)^3 / x^3 = 9 = (x+L)/x = 9^(1/3) = 2.08 also d^2/dx^2[[9/(x+L)^2 - 1/x^2] = 54/(x+L)^4 - 6/x^4 = C say so C*(x+L)^4= 54 - 6*[(x+L)/x]^4 = 54 - 6* (2.08^4) = 54 - 112.3 for x = 2.08. So C < 0 for x = 2.08 So at x = 2.08 E net is maximum. It is given that at x = 20 cm 1 cm less than xs in the figure, the field is zero. That is possible only when 9/(20+L)^2 = 1/20^2 or (20+L)/20 = 3 or 1 + L/20 = 3 or L = 40 cm Substituting the appropriate values for q1 and q2 we get Enet(max) = (9*10^9)(3*10^-19)*(10^4)*[9/(2.08+40)^2 - 1/(2.08)^2] = (27*10^-6)*(-0.226) = -6.10*10^-6 N/C Something is wrong in the solution I cannot point out may be some one else points out. How can maximum value be negative wen it is also becoming zero at x = 20 cm This taking figure from other question creates problems may be the asker helps me in this.