Solution
Let the relaxed length be $l_0 = 0.2$ m and the spring constant $k = 200$ N/m.
The equilibrium length is $2l_0 = 0.4$ m. The extension is $x = 0.4 – 0.2 = 0.2$ m.
1. Find the Electrostatic Constant ($C$):
At equilibrium, the spring force equals the electrostatic repulsion:
$$F_s = F_e$$
$$kx = \frac{C}{r_{eq}^2}$$
$$200(0.2) = \frac{C}{(0.4)^2}$$
$$40 = \frac{C}{0.16} \implies C = 6.4 \, \text{N}\cdot\text{m}^2$$
(Here $C = \frac{q_1 q_2}{4\pi\epsilon_0}$).
2. Calculate Energy in Both States:
- Initial State (Equilibrium): Separation $r = 0.4$ m. $$U_i = U_{spring} + U_{elec}$$ $$U_i = \frac{1}{2}kx^2 + \frac{C}{r} = \frac{1}{2}(200)(0.2)^2 + \frac{6.4}{0.4}$$ $$U_i = 100(0.04) + 16 = 4 + 16 = 20 \, \text{J}$$
- Final State (Compressed to relaxed length): Separation $r = 0.2$ m. $$U_f = \frac{1}{2}k(0)^2 + \frac{6.4}{0.2}$$ $$U_f = 0 + 32 = 32 \, \text{J}$$
3. Work Done:
$$W = \Delta U = U_f – U_i = 32 – 20 = 12 \, \text{J}$$
Answer: (b) 12 J