Question
In the figure below, particles 1 and 2 of charge q1 = q2 = +4.8O times 10-19 C are on a y axis at distance d = 21.0 cm from the origin. Particle 3 of charge q3 = +6.40 times 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 minimum and maximum? What are the (c) minimum and (d) maximum magnitudes? position of minimum force position of maximum force magnitude of minimum force magnitude of maximum force
Explanation / Answer
Obviously, at x=0, the force on the charge cancels to zero. So you have instantly an answer for (a) and (c). For the maximum, note that the y-components of the force exerted by 1 and 2 cancels out, so what's left is the x-component only. F = k q1 q3/r^2 force due to q1 For the x-component, F = k q1 q3/r^2 (cos(theta)) theta is the angle as viewed from q3 since cosine = adjacent over hypo cos(theta) = x/r = x/(sqrt(x^2+d^2) -hope you get this The x-component of force due to q1 is F = k q1 q3 x/r^3 now, Adding the contribution from the force due to q2, note q2=q1 F(net) = 2k q1 q3 x/r^3 This is the net force on q3. How will we know the maximum? Remember calculus? Relative extremum? Recall: To find the maximum of a function f(x). Set f'(x0)=0 and solve for x0. Then take f''(x0). If f''(x0)>0, then the at that point it, it curves up, thus we have a relative minimum. If ''(x0)