Step 1: Determine Induced Charge and Force

Let the separation between the centers of the spheres be $L$. Since the spheres are connected by a conductor, they must be at the same potential. The external field $E$ creates a potential difference $EL$ between the positions of the spheres.

Charges $+q$ and $-q$ are induced on the spheres to counteract this. The potential difference due to these charges is approximately (for $L \gg r$):

$$ \Delta V_{ind} = \frac{q}{4\pi\epsilon_0 r} – \frac{-q}{4\pi\epsilon_0 r} = \frac{q}{2\pi\epsilon_0 r} $$

For equilibrium of potential (net $\Delta V = 0$):

$$ \frac{q}{2\pi\epsilon_0 r} = EL \implies q = 2\pi\epsilon_0 r EL $$

The electrostatic force pulling the spheres apart is $F_{elec} = qE$. Substituting $q$:

$$ F_{elec} = (2\pi\epsilon_0 r EL)E = (2\pi\epsilon_0 r E^2) L $$

Let $\alpha = 2\pi\epsilon_0 r E^2$. Then $F_{elec} = \alpha L$. This force acts like a “negative spring” (force proportional to displacement, but pushing away).

Step 2: Condition for Oscillation (Part a)

The equation of motion for the relative separation $L$ (with reduced mass $\mu = m/2$) involves the spring force $F_{spring} = -k(L-l_0)$ and the electric force $F_{elec} = \alpha L$.

$$ \frac{m}{2} \ddot{L} = F_{elec} + F_{spring} = \alpha L – k(L – l_0) $$ $$ \frac{m}{2} \ddot{L} = -(k – \alpha)L + k l_0 $$

For oscillations to occur, the net coefficient of $L$ must be negative (restoring force). Thus:

$$ k – \alpha > 0 \implies k > 2\pi\epsilon_0 r E^2 $$

Step 3: Amplitude and Period (Part b)

Period: The effective spring constant is $K_{eff} = k – \alpha$. The angular frequency is:

$$ \omega = \sqrt{\frac{K_{eff}}{\mu}} = \sqrt{\frac{k – \alpha}{m/2}} = \sqrt{\frac{2(k – 2\pi\epsilon_0 r E^2)}{m}} $$ $$ T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{m}{2(k – 2\pi\epsilon_0 r E^2)}} $$

Amplitude: The system is released from the undeformed length $l_0$. The new equilibrium position $L_{eq}$ is where acceleration is zero:

$$ (k-\alpha)L_{eq} = k l_0 \implies L_{eq} = l_0 \frac{k}{k-\alpha} $$

The amplitude $A_{rel}$ of the relative coordinate oscillation is the distance from the release point ($l_0$) to the equilibrium point ($L_{eq}$):

$$ A_{rel} = |L_{eq} – l_0| = l_0 \left( \frac{k}{k-\alpha} – 1 \right) = l_0 \frac{\alpha}{k-\alpha} $$

The question asks for the amplitude of oscillation of the balls. In the center of mass frame, each ball moves by half the relative displacement. So the amplitude of each ball is $A = A_{rel}/2$.

$$ A = \frac{1}{2} \frac{\alpha l_0}{k-\alpha} = \frac{1}{2} \frac{(2\pi\epsilon_0 r E^2) l_0}{k – 2\pi\epsilon_0 r E^2} = \frac{\pi\epsilon_0 r l_0 E^2}{k – 2\pi\epsilon_0 r E^2} $$