In the figure below, particles 1 and 2 of charge q1 = q2 = +4.80 10-19 C are on
ID: 2162158 • Letter: I
Question
In the figure below, particles 1 and 2 of charge q1 = q2 = +4.80 10-19 C are on a y axis at distance d = 27.0 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
In this setup the equal/opposite y forces on q3 cancel and the equal/same direction x forces add. So to have the minimum force we simply place q3 directly between q1 and q2 (x = 0), and F = 0, answers A and C. As q3 moves along x, the total force magnitude F, which is the vector sum of the x forces Fx from q1 and q2, first increases from 0 as the force angle moves from the vertical, but then passes through a maximum and decreases as the effect of distance becomes dominant. The x value for maximum force must be found by setting the derivative of Fx = 0 and solving for x. F due to either q1 or q2 = k(q1+q2)q3/r^2 = k(q1+q2)q3/(x^2+y^2). Fx due to either q1 or q2 = F*x/r = F*x/sqrt(x^2+y^2) = k(q1+q2)q3*x/(x^2+y^2)^(3/2). To obtain the derivative of Fx we first replace the invariant term k(q1+q2)q3 by 1, yielding d(Fx)/dx = d(x/(x^2+y^2)^(3/2))/dx. As solved in the ref. below, the derivative of x/(x^2+y^2)^(3/2) is (y^2-2x^2)/(y^2+x^2)^(5/2). Setting the numerator to zero we have y^2-2x^2 = 0 ==> x = y/sqrt(2) = 0.240416305603426 m, answer B. Finally we solve for total force F = 2Fx, using the Fx equation given above. The result is 2.29822690936678E-26 N, answer D.