Consider a thin, circular disc of radius a, which rotates at a constant angular velocity o...

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Consider a thin, circular disc of radius a, which rotates at a constant angular velocity o about the z axis, which is perpendicular to the plane of the disc and passes through its centre. The centre of the disc is the origin of a Cartesian coordinate system. One side of the disc carries charge - the surface density of this charge is constant, and equal to Qo/(ra²), where the total charge Q, is positive. The medium outside the disc is vacuum. A point of observation, P, is situated at a perpendicular (cylindrical radial) distance R from the z axis, and at a distance z (z > 0) above the xy plane in which the disc lies. P is situated very far from the disc itself - that is, a << 1. R² + 2² (a) At time t = to, the rotation of the disc starts to slow down, and its angular velocity becomes a function of time, according to: Q(t): { CT approximately: Qoe-(1-b)/ where T is the time constant associated with the decrease of the disc's angular velocity with time. Assume now that the slowing of the rotation is very gradual, obeying √(R+ a)² + z² << CT (where c is vacuum light speed). Derive an expression for (t) (where t' represents retarded time) which is accurate to first order in R² + z² Use your result to prove that the vector potential at P will now be t< to tz to, A A(R, z, t) = Ho 167T Qoe-(-6)/ Ra² (R² +22)3/2 Î₁ for values of time which satisfy t> (to + √(R+ a)² + 2²/c). (b) Using the relevant information and assumptions from previous parts of the question, derive the scalar potential, , produced by the disc at point P. As part of your answer, explain why the rotation of the disc has no effect on this potential. Consider a thin, circular disc of radius a, which rotates at a constant angular velocity o about the z axis, which is perpendicular to the plane of the disc and passes through its centre. The centre of the disc is the origin of a Cartesian coordinate system. One side of the disc carries charge - the surface density of this charge is constant, and equal to Qo/(ra²), where the total charge Q, is positive. The medium outside the disc is vacuum. A point of observation, P, is situated at a perpendicular (cylindrical radial) distance R from the z axis, and at a distance z (z > 0) above the xy plane in which the disc lies. P is situated very far from the disc itself - that is, a << 1. R² + 2² (a) At time t = to, the rotation of the disc starts to slow down, and its angular velocity becomes a function of time, according to: Q(t): { CT approximately: Qoe-(1-b)/ where T is the time constant associated with the decrease of the disc's angular velocity with time. Assume now that the slowing of the rotation is very gradual, obeying √(R+ a)² + z² << CT (where c is vacuum light speed). Derive an expression for (t) (where t' represents retarded time) which is accurate to first order in R² + z² Use your result to prove that the vector potential at P will now be t< to tz to, A A(R, z, t) = Ho 167T Qoe-(-6)/ Ra² (R² +22)3/2 Î₁ for values of time which satisfy t> (to + √(R+ a)² + 2²/c). (b) Using the relevant information and assumptions from previous parts of the question, derive the scalar potential, , produced by the disc at point P. As part of your answer, explain why the rotation of the disc has no effect on this potential.

Answer rating: 100% (QA)

Lets start by addressing part a of your question You are given the expression for the angular velocity of the disc which changes over time t 0 for t t0 and 0 ett0 for t t0 Since the disc is slowing do... View the full answer