Question

A uniformly charged, thin ring has radius 15.0 cm and total charge +24.0 nC. An electron is placed on the ring’s axis a distance 30.0 cm from the center of the ring and is constrained to stay on the axis of the ring. The electron is then released from rest. (a) Describe the subsequent motion of the electron. (b) Find the speed of the electron when it reaches the center of the ring.

Answer 1

Answer:

A. The electron will begin to move along the axis, towards the centre and the instantaneous velocity because the force acting on it depends largely on acceleration and x until it reaches maximum velocity at centre.

B. Veloctiy (Vb) = 1.66m/s

Explanation:

Given the following data

x(a) = 0.3m

x(b) = 0

q = 1.6×10^-19

Q = 24nc

r = 0.15m

Required: the motion of the electron and the velocity (Vb)

1. At point A the electron will begin to move along the axis from point A to point B, the magnitude of the electric field will change while moving which depends on that and this will produce instantaneous force which will later change and the acceleration will change too while moving, the velocity would reach maximum value at point B

2. Potential energy and kinetic energy are given by

U(a) + K(a) = U(b) + K(b). . .1

Initial P.E and K.E are given as

U(a) = kQ/√x²(a) + a2

By substitution, we have

U(a) = 9×10^9 × (-1.9×10^-19)×24×10^-9/√(0.15)²+(0.3²)

U(a) = -1.03×10^-16

Final P.E and K.E are given as

U(b) = KQ/√x²(b) + a2

By substitution, we have

U(b) = 9×10^9×(-1.9×10^-19)×24×10^-9/√(0.15)²+(0)²

U(b) = -2.3×10^16

3. By substitution into equation 1 becomes

-1.03×10^-6 - 2.3×10^-16 + MV²(b)/2

V(b) = √2×1.27×10^-16/9.1×10^31

V(b) = 1.66×10^7m/s