Question

Two identical conducting balls each of radii $r$, and mass $m$, connected by a light conducting spring of stiffness $k$ and un-deformed length $l_0$ ($l_0 >> r$) are at rest in free space. If a uniform electric field of strength $E$ directed along the spring is switched on, the balls may oscillate.

(a) What should value of the stiffness $k$ of the spring be to make oscillations possible? (b) Determine the amplitude and period of oscillations.

Answer 1

Answer: (a) $k > 2\pi\epsilon_0 E^2 r$, (b) $A = \frac{2\pi\epsilon_0 E^2 r l_0}{k - 2\pi\epsilon_0 E^2 r}$, $T = 2\pi \sqrt{\frac{m}{k - 2\pi\epsilon_0 E^2 r}}$

Answer 2

The problem describes two identical conducting balls connected by a light conducting spring. This means the entire system (two balls + spring) forms a single conductor. When a uniform electric field E is switched on, directed along the spring, charges will be induced on the surfaces of the balls. The ball on the side from which the electric field lines enter will acquire a net negative charge, and the ball on the side where the lines exit will acquire a net positive charge. Let these induced charges be -Q and +Q respectively.

The system is initially at rest, meaning the spring is at its un-deformed length l_0. When the field E is switched on, the positive ball experiences an electric force F_E = QE in the direction of E, and the negative ball experiences an electric force F_E = QE opposite to E. These forces act to pull the balls apart, stretching the spring.

1. Determine the induced charge Q: Since the balls are conducting and connected by a conducting spring, they form an equipotential surface. Let the distance between the centers of the balls be l. The un-deformed length of the spring is l_0, so l = l_0 + x, where x is the extension of the spring. Let the origin be the midpoint between the centers of the balls. The left ball is at -l/2 and the right ball is at l/2. The potential at the center of the left ball V_L is due to its own charge -Q, the charge +Q on the right ball, and the external electric field E. Similarly for the right ball V_R. Given l_0 >> r, the balls are far apart. We can approximate the potential at the center of each ball due to the other ball as if they were point charges. The potential due to the external field E at a position y is -Ey (if E is along +y axis).

So, $V_L = \frac{1}{4\pi\epsilon_0} \frac{-Q}{r} + \frac{1}{4\pi\epsilon_0} \frac{Q}{l} - E(-\frac{l}{2})$ $V_R = \frac{1}{4\pi\epsilon_0} \frac{Q}{r} + \frac{1}{4\pi\epsilon_0} \frac{-Q}{l} - E(\frac{l}{2})$

Since $V_L = V_R$ (equipotential): $\frac{-Q}{r} + \frac{Q}{l} + \frac{El}{2} = \frac{Q}{r} - \frac{Q}{l} - \frac{El}{2}$ $\frac{El}{2} + \frac{El}{2} = \frac{Q}{r} + \frac{Q}{r} - \frac{Q}{l} - \frac{Q}{l}$ $El = \frac{2Q}{r} - \frac{2Q}{l}$ $El = 2Q \left( \frac{1}{r} - \frac{1}{l} \right)$ $El = 2Q \frac{l-r}{rl}$ $Q = \frac{E l^2 r}{2(l-r)}$

Using $K_e = \frac{1}{4\pi\epsilon_0}$: $Q = \frac{E l^2 r}{2(l-r)} \times 4\pi\epsilon_0 = \frac{2\pi\epsilon_0 E l^2 r}{l-r}$

Given $l_0 >> r$, it implies $l = l_0+x$ will also be much greater than $r$. So, we can approximate $l-r \approx l$. Therefore, the induced charge $Q$ is approximately: $Q \approx \frac{2\pi\epsilon_0 E l^2 r}{l} = 2\pi\epsilon_0 E l r$

2. Equilibrium Position: The electric force on each ball is $F_E = QE$. This force tends to stretch the spring. The restoring force from the spring is $F_s = kx$. At equilibrium, $F_E = F_s$. Let the equilibrium extension be $x_{eq}$. The length of the spring at equilibrium is $l_{eq} = l_0 + x_{eq}$. $Q E = k x_{eq}$ Substitute $Q = 2\pi\epsilon_0 E l_{eq} r$: $(2\pi\epsilon_0 E l_{eq} r) E = k x_{eq}$ $2\pi\epsilon_0 E^2 r l_{eq} = k x_{eq}$ Substitute $l_{eq} = l_0 + x_{eq}$: $2\pi\epsilon_0 E^2 r (l_0 + x_{eq}) = k x_{eq}$ $2\pi\epsilon_0 E^2 r l_0 + 2\pi\epsilon_0 E^2 r x_{eq} = k x_{eq}$ $2\pi\epsilon_0 E^2 r l_0 = (k - 2\pi\epsilon_0 E^2 r) x_{eq}$ $x_{eq} = \frac{2\pi\epsilon_0 E^2 r l_0}{k - 2\pi\epsilon_0 E^2 r}$

(a) What should value of the stiffness k of the spring be to make oscillations possible? For oscillations to be possible, there must be a stable equilibrium position. This means the equilibrium extension $x_{eq}$ must be real and positive (since the force is stretching). For $x_{eq}$ to be positive, the denominator must be positive (as the numerator is positive). $k - 2\pi\epsilon_0 E^2 r > 0$ $k > 2\pi\epsilon_0 E^2 r$ This is the condition for oscillations to be possible. If $k \le 2\pi\epsilon_0 E^2 r$, the electric force would be too strong, and the balls would keep moving apart without an equilibrium point, or the spring would be compressed.

(b) Determine the amplitude and period of oscillations.

Amplitude: The system starts from rest at the un-deformed length ($x=0$). When the field is switched on, the system is displaced from its new equilibrium position ($x_{eq}$). Since it starts from rest at an extremum of its motion (relative to the equilibrium point), the amplitude of oscillation A is equal to the magnitude of the displacement from the equilibrium position to the starting position. $A = |0 - x_{eq}| = x_{eq}$ $A = \frac{2\pi\epsilon_0 E^2 r l_0}{k - 2\pi\epsilon_0 E^2 r}$

Period of oscillations: Let $x$ be the extension of the spring from its un-deformed length $l_0$. The instantaneous length is $l = l_0 + x$. The net force on one ball (say, the right one) is $F_{net} = F_E - F_s$. $F_{net} = QE - kx$ Substitute $Q = 2\pi\epsilon_0 E r l$: $F_{net} = (2\pi\epsilon_0 E r l) E - kx$ $F_{net} = 2\pi\epsilon_0 E^2 r (l_0 + x) - kx$ $F_{net} = 2\pi\epsilon_0 E^2 r l_0 + 2\pi\epsilon_0 E^2 r x - kx$ $F_{net} = 2\pi\epsilon_0 E^2 r l_0 - (k - 2\pi\epsilon_0 E^2 r) x$

This net force acts on one ball of mass $m$. So, applying Newton's second law: $m \frac{d^2x}{dt^2} = 2\pi\epsilon_0 E^2 r l_0 - (k - 2\pi\epsilon_0 E^2 r) x$ We know that $2\pi\epsilon_0 E^2 r l_0 = x_{eq} (k - 2\pi\epsilon_0 E^2 r)$. $m \frac{d^2x}{dt^2} = x_{eq} (k - 2\pi\epsilon_0 E^2 r) - (k - 2\pi\epsilon_0 E^2 r) x$ $m \frac{d^2x}{dt^2} = - (k - 2\pi\epsilon_0 E^2 r) (x - x_{eq})$

Let $y = x - x_{eq}$. Then $\frac{d^2y}{dt^2} = \frac{d^2x}{dt^2}$. $m \frac{d^2y}{dt^2} = - (k - 2\pi\epsilon_0 E^2 r) y$ $\frac{d^2y}{dt^2} = - \frac{k - 2\pi\epsilon_0 E^2 r}{m} y$

This is the equation for Simple Harmonic Motion (SHM) of the form $\frac{d^2y}{dt^2} = -\omega^2 y$. The angular frequency $\omega$ is given by: $\omega^2 = \frac{k - 2\pi\epsilon_0 E^2 r}{m}$ $\omega = \sqrt{\frac{k - 2\pi\epsilon_0 E^2 r}{m}}$

The period of oscillation $T$ is $T = \frac{2\pi}{\omega}$: $T = 2\pi \sqrt{\frac{m}{k - 2\pi\epsilon_0 E^2 r}}$

The final answer is $\boxed{Answer: (a) k > 2\pi\epsilon_0 E^2 r, (b) A = \frac{2\pi\epsilon_0 E^2 r l_0}{k - 2\pi\epsilon_0 E^2 r}, T = 2\pi \sqrt{\frac{m}{k - 2\pi\epsilon_0 E^2 r}}}$.

Explanation of the solution:

1. Induced Charge: The two conducting balls connected by a conducting spring form a single conductor. When placed in a uniform electric field, charges are induced on the balls. Due to the large separation ($l_0 >> r$), the potential difference between the centers of the balls due to the external field is $El$, and the potential due to the induced charges is $2Q(\frac{1}{r} - \frac{1}{l}) / (4\pi\epsilon_0)$. Equating these potentials (since the system is equipotential) and approximating $l-r \approx l$ for $l>>r$, the induced charge $Q$ on each ball is found to be $2\pi\epsilon_0 E r l$.
2. Equilibrium Position: The electric force $F_E = QE$ tends to stretch the spring, while the spring's restoring force $F_s = kx$ opposes this. At equilibrium, $F_E = F_s$. Substituting the expression for $Q$ and solving for the equilibrium extension $x_{eq}$ gives $x_{eq} = \frac{2\pi\epsilon_0 E^2 r l_0}{k - 2\pi\epsilon_0 E^2 r}$.
3. Condition for Oscillations (Part a): For oscillations to be possible, a stable equilibrium must exist, meaning $x_{eq}$ must be a positive, finite value. This requires the denominator to be positive: $k - 2\pi\epsilon_0 E^2 r > 0$, or $k > 2\pi\epsilon_0 E^2 r$.
4. Amplitude (Part b): The balls start at rest with the spring at its un-deformed length ($x=0$). When the field is switched on, this becomes the initial position for the oscillation. Since the system starts from rest, this is an extreme point of the oscillation. The amplitude $A$ is therefore the magnitude of the displacement from this starting point to the equilibrium position: $A = |x_{eq}| = \frac{2\pi\epsilon_0 E^2 r l_0}{k - 2\pi\epsilon_0 E^2 r}$.
5. Period (Part b): To find the period, we write the net force on one ball as a function of its displacement $x$ from the un-deformed length. This force is $F_{net} = QE - kx$. Substituting $Q = 2\pi\epsilon_0 E r (l_0+x)$ and using Newton's second law ($F_{net} = m \frac{d^2x}{dt^2}$), we obtain the differential equation for motion. By rearranging it into the standard SHM form $\frac{d^2y}{dt^2} = -\omega^2 y$ (where $y = x - x_{eq}$), we identify the angular frequency $\omega = \sqrt{\frac{k - 2\pi\epsilon_0 E^2 r}{m}}$. The period is then $T = 2\pi/\omega = 2\pi \sqrt{\frac{m}{k - 2\pi\epsilon_0 E^2 r}}$.