ELECTROSTATICS CYU 22

Part (a): Time particle B spends in the electric field

We use the Impulse-Momentum Theorem. The only external horizontal force on the dumbbell system is the electric force acting on particle B.

• Initial Momentum ($P_i$): The system enters with velocity $u$. $$P_i = (m_A + m_B)u = 2mu$$
• Final Momentum ($P_f$): The system exits moving in the negative direction with speed $u$. $$P_f = (m_A + m_B)(-u) = -2mu$$
• Impulse ($J$): Change in momentum. $$\Delta P = P_f – P_i = -2mu – 2mu = -4mu$$
• Force ($F$): The force on particle B (charge $+q$) in a field pointing negative x-direction is $F = -qE$.
• Calculation: $$F \times t = \Delta P$$ $$-qE \times t = -4mu$$ $$t = \frac{4mu}{qE}$$

Part (b): Relaxed length of the spring ($l$)

The problem states the spring length becomes minimum only once. This implies the time spent inside the field corresponds exactly to one full time period of the spring-mass oscillation.

1. Find the Spring Constant ($k$)

• Reduced mass of the system: $\mu = \frac{m \cdot m}{m + m} = \frac{m}{2}$.
• Time period of oscillation: $T = 2\pi\sqrt{\frac{\mu}{k}} = 2\pi\sqrt{\frac{m}{2k}}$.
• Equating $t$ from part (a) to $T$: $$\frac{4mu}{qE} = 2\pi\sqrt{\frac{m}{2k}}$$
• Squaring both sides and solving for $k$: $$\frac{16m^2 u^2}{q^2 E^2} = 4\pi^2 \left(\frac{m}{2k}\right)$$ $$k = \frac{q^2 E^2 \pi^2}{8 m u^2}$$

2. Condition on Relaxed Length ($l$)

Particle A must not enter the field. This relates to the maximum compression. Using the energy conservation logic provided in the hint (Total Energy $\ge$ Work Done + Elastic Energy):

The condition simplifies to ensuring the spring’s relaxed length is sufficient to accommodate the dynamics:

$$l \ge \frac{mu^2}{qE} \left[ 1 + \left(\frac{2}{\pi}\right)^2 \right]$$