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A disk of radius R carrying uniform positive surface charge density sigma lies in the xy-plane centered at the origin. Its potential on the z-axis is V(z) = (sigma/(2*epsilon0))*(sqrt(R² + z²) - z). A positive charge q starts at rest at z = z0 > 0 on the axis. Besides the Coulomb force, it experiences a constant downward force F = -c*k_hat (c > 0). Define beta = 2*c*epsilon0/(q*sigma). Which statement(s) is/are correct?

1. For beta = 1/4 and z0 = (25/7)R, the particle reaches the origin.
2. For beta = 1/4 and z0 = (3/7)R, the particle reaches the origin.
3. For beta = 1/4 and z0 = R/sqrt(3), the particle returns back to z = z0.
4. For beta > 1 and any z0 > 0, the particle always reaches the origin.

Correct answer: For beta = 1/4 and z0 = (25/7)R, the particle reaches the origin.

Solution

Total potential energy: U(z) = q*V(z) + c*z = (q*sigma/(2*epsilon0))[sqrt(R²+z²) - z] + c*z = (q*sigma/(2*epsilon0))[sqrt(R²+z²) - z + beta*z], using beta = 2*c*epsilon0/(q*sigma). Starting from rest, the particle reaches the origin if U(0) <= U(z0) and no barrier higher than U(z0) lies in (0, z0). With beta = 1/4: U(z) proportional to f(z) = sqrt(R²+z²) - (3/4)z. Evaluating f at z0 = (25/7)R: sqrt(R² + (625/49)R²) = sqrt((49+625)/49)R = sqrt(674/49)R approx 3.71R; minus (3/4)(25/7)R = (75/28)R approx 2.68R, gives f(z0) approx 1.03R. f(0) = R. Since f(z0) > f(0) for this case the particle (released from rest, total energy E = U(z0)) can reach the origin where U is lower. The condition is satisfied for z0 = (25/7)R, so that option is correct. (This matches the known JEE Advanced answer A.)

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