Solution to Question 45
Physics Principle: Electrostatic Potential Energy and Unstable Equilibrium.
Setup:
Fixed charge \(Q_1 = -4.0 \mu\text{C}\) at height \(y=3.0 \text{ m}\).
Moving particle \(Q_2 = +1.0 \mu\text{C}\), mass \(m=0.9 \text{ g}\), at height \(y=2.2 \text{ m}\).
Goal: Reach the ground (\(y=0\)).
Step 1: Analyze forces and equilibrium
Distance \(r\) is measured from \(Q_1\). Start \(r = 3.0 – 2.2 = 0.8 \text{ m}\).
Forces on \(Q_2\): Upward attraction \(F_e\) and Downward gravity \(mg\).
Equilibrium occurs where \(F_e = mg\):
$$\frac{k |Q_1 Q_2|}{r_{eq}^2} = mg$$
$$r_{eq}^2 = \frac{(9 \times 10^9)(4 \times 10^{-12})}{(9 \times 10^{-4})(10)} = \frac{36 \times 10^{-3}}{9 \times 10^{-3}} = 4$$
$$r_{eq} = 2.0 \text{ m}$$
At the start (\(r=0.8\text{ m}\)), \(r < r_{eq}\), so the upward electric force is stronger than gravity. The particle is in a potential well. To reach the ground (\(r=3.0\text{ m}\)), it must "cross the hill" at \(r_{eq} = 2.0\text{ m}\). This is a point of unstable equilibrium (maximum potential energy). Once it passes \(r=2.0\), gravity dominates and pulls it down.
Step 2: Conservation of Energy
We need just enough initial kinetic energy to reach \(r = 2.0 \text{ m}\) with zero velocity. $$KE_i + U(r_i) = KE_{top} + U(r_{eq})$$ $$\frac{1}{2}mv^2 + \left( -\frac{k|Q_1 Q_2|}{r_i} + mgh_i \right) = 0 + \left( -\frac{k|Q_1 Q_2|}{r_{eq}} + mgh_{eq} \right)$$
Let’s define Gravitational PE relative to the fixed charge level for simplicity (so \(h = -r\)). Or standard ground ref \(h = 3-r\).
At \(r_i = 0.8\): \(U_e = -\frac{0.036}{0.8} = -0.045\), \(U_g = mg(2.2) \approx 0.0198\). Total \(U_i = -0.0252 \text{ J}\).
At \(r_{eq} = 2.0\): \(U_e = -\frac{0.036}{2.0} = -0.018\), \(U_g = mg(1.0) = 0.009\). Total \(U_f = -0.009 \text{ J}\).
Change in Energy needed: $$\frac{1}{2}mv^2 = U_f – U_i = -0.009 – (-0.0252) = 0.0162 \text{ J}$$ $$v^2 = \frac{2 \times 0.0162}{9 \times 10^{-4}} = \frac{0.0324}{0.0009} = 36$$ $$v = 6 \text{ m/s}$$