Find the speed of the electron

Question

A uniformly charged thin ring has radius 13.5cm and total charge 23.0 nC . An electron is placed on the ring's axis a distance  27.0cm from the center of the ring and is constrained to stay on the axis of the ring. The electron is then released from rest.

Find the speed of the electron when it reaches the center of the ring.

Express your answer in meters per second.

v= m/s

Answer 1

Potential V₀ = kQ/(r² + z²)^1/2and potential at center of ring V = kQ/r where r = 13.5 cm = 0.135 m, Q = 23.0 nC = 2.3·10^{- 8} C, z = 27.0 cm = 0.27 m.

Then by theorem of change of kinetic energy we have mv²/2 = e(V - V₀) where m is mass of electron, e is charge of electron. From here v = [2e(V - V₀)/m]^1/2

Let evaluate V = 9·10⁹·2.3·10^-8/0.135 V = 1533.3 V;

THen evaluate V₀ = 9·10⁹·2.3·10^-8/(0.135² + 0.27²)^1/2 = 685.7 V.

So, v = [2·1.6·10^-19(1533.3 - 685.7)/9.11·10^{- 31}]^1/2 = 1.73·10⁷ m/s