Question

3 points) An electron is constrained to move along the central axis of a ring of radius R which is uniformly charged with total positive charge Q (a) Show that if the electron is near the centre of the ring (a small distance compared to R), it will experience a force of the form Fz--kz (where z is the distance from the centre of ring) (b) Since this force will make the electron follow simple harmonic oscillation, derive a formula for the frequern ula for the frequency of the oscillations. Hint: Note that for z<<R 22R? (2 +1R2 since 2 << 1

Answer 1

Electric field due to charged ring along its axis is E^vector = kQ/(R^2 + z^2) cos theta = k theta/(R^2 + z^2) times z/squareroot R^2 + z^2 k^vector therefore E_z = 1/4 pi to Q z/R^3 [1 + (z/R)^2]^3/2 since k = 1/4 pi to. therefore since z^2/R^2 < < 1 = > E_z = Q z/4 pi to R^3 Force on electron is F_z = -e E_z therefore F_z = -e Q/4 pi to R^3 z = -k z therefore F_z = -k z where k = e Q/4 pi to R^3 Now, d^2 z/dt^2 = -k z where m_e = mass of electron therefore d^2 z/dt^2 = -k/m_e z = -w^2 z where w = squareroot k/m_e therefore frequency of oscillation, f = 1/T = w/2 pi therefore f = 1/2 pi squareroot k/m_e = 1/2 pi squareroot e Q/4 pi to R^3 m_e therefore f = 1/2 pi squareroot e Q/4 pi to R^3 m_e leftarrow frequency of oscillation