Question

An ideal spring, with spring constant k, is attached to a particle of charge q and mass m in such a way that it can only move along the c-axis. The spring's equilibrium is used to set r = 0. The system is then put in an electric field E = EX. 1. Where would the new equilibrium be? 2. If the system was at rest and equilibrium initially, what is the maximum speed of the system after the electric field is turned on? 3. Under the same conditions what is the maximum distance from the origin the mass can reach?

Answer 1

1) The net force on the mass after the electric field is turned on is

. ...(1)

F =ma-ka qE -kx + qE

At equilibrium F = 0

Therefore the point of equilibrium is given by

=0 -k Teq +q E

Hence

eqE/k

2) Solving equation (1) gives

¢1 cos(wt) + c2 sin(wt) (t)qE/k

where c1 and c2 are constants and

is the angular frequency of oscillation

Vk/m

Given x = 0 at t = 0. Hence

qEk C1

and

$\dot x = 0$ at t = 0

Hence

= 0

Therefore

qE cos(wt) (t)E

The speed is

qEu sin(wt (t)

Max speed is when sin(wt) = 1

Therefore

$\dot x(t)|_{max}=\frac{q Ew}{k}=\frac{qE\sqrt{k/m}}{\,k}=\frac{qE}{\sqrt{m\,k}}$

3) Max displacement is when cos(wt) = -1

qE x(t)

Е 2qЕ (t) — Де