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A thin wire of cross-sectional area A and Young's modulus Y is bent into a ring of radius R. A total charge Q is uniformly distributed on the ring and a point charge q0 is placed at its centre. Find the increase in the radius of the ring, delta_R.
- delta_R = q0*Q / (4*pi² * epsilon0 * R * A * Y)
- delta_R = q0*Q / (4*pi² * epsilon0 * R * A * Y)
- delta_R = q0*Q / (8*pi² * epsilon0 * R * A * Y)
- delta_R = q0*Q / (8*pi² * epsilon0 * R * A * Y)
Correct answer: delta_R = q0*Q / (8*pi² * epsilon0 * R * A * Y)
Solution
The central charge q0 exerts a radial outward force on each element of the ring, creating a tension T = q0*Q/(8*pi²*epsilon0*R²). Treating the ring as a stressed wire: Y = (T/A)/(delta_R/R), giving delta_R = T*R/(A*Y) = q0*Q/(8*pi²*epsilon0*R*A*Y).
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