Question

Two identical spheres each having charge Q =-11 nC and mass mOF 10 milligrams are held fixed at their positions 5.0 cm apart. A particle with charge q-+1.0 nC and mass mo-0.40 milligrams is released from rest at a position 5.0 cm from each sphere (so initially the particle and the spheres form an equilateral triangle). Find the speed of the positively charged particle when it passes a point exactly in between the two spheres. (k - 9.0 x 109 N-m2/c2)

Answer 1

Suppose at any time 't' charge + q is reaches at point A such that DA = y Force between + q at charge Q at point B - F_1 = k qQ/AB^2 equation (1) Similarly force of attraction between charge + q and charge Q situated at C - F_2 = k q.Q/AC^2 equation (2) Since AB = AC so from (1) and (2) - F_1 = F_2 = k middot qQ/AB^2 equation (3) \r\n If we divide F_1^vector and F_2^vector into two components that we see that F_1 Cos theta and F_2 Cos theta are equal and opposite so cancel each other. While vertical components are in the same direction. So net force on + q - F = F_1 Sin theta + F_2 Sin theta F = 2 F_1 Sin theta (Because F_1 = F_2) F = 2 times k qQ/AB^2 times AD/AB But AD = y and AB = squareroot y^2 + (d/2)^2 Therefore F = 2 k qQ