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Two point charges Q1 and Q2 are fixed at points A and B separated by distance a. A light inextensible string of length 2a is attached at A and B. A small bead of charge q (same sign as Q1, Q2) slides freely on the string. Gravity is absent. Given Q1/Q2 = 4, in the equilibrium position the string segment from Q1 to the bead makes angle alpha with line AB. Find cos(alpha).
- sqrt(3)/24
- 21/24
- 1/4
- 9/24
Correct answer: 21/24
Solution
Force balance gives r1/r2 = sqrt(Q1/Q2) = 2, so r1 = 4a/3, r2 = 2a/3. Law of cosines in triangle ABP: r2² = a² + r1² - 2*a*r1*cos(alpha) yields cos(alpha) = (a² + r1² - r2²)/(2*a*r1) = 21a²/9 * 3/(8a²) = 21/24.
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