Physics Solution – Question 2

Solution to Question 2

1. System Analysis and Diagram

We consider a rigid insulating rod of length $l$ pivoted at point O. A mass $m$ with charge $-q$ is at the bottom end, and a fixed charge $+Q$ is located at a height $h$ above O.

To determine the condition for stable equilibrium, we displace the rod by a small angle $\theta$ from the vertical and analyze the restoring torques. For stability, the net torque must tend to return the rod to the vertical position ($\theta = 0$).

Figure 1: Forces acting on the charge $-q$ when displaced by a small angle $\theta$.

2. Torque Calculation

There are two torques acting about the pivot O:

Torque due to Gravity ($\tau_g$): Tends to bring the rod back to the vertical (Restoring Torque).
Torque due to Electric Force ($\tau_e$): Tends to pull the rod further away from the vertical towards $Q$ (Destabilizing Torque).

Restoring Torque ($\tau_g$):

The perpendicular distance from the pivot to the line of action of gravity is $l \sin\theta$.

$$ \tau_g = mg(l \sin\theta) $$

For small angles ($\theta \to 0$), $\sin\theta \approx \theta$:

$$ \tau_g \approx mgl\theta $$

Destabilizing Torque ($\tau_e$):

The electrostatic attractive force $F_e$ is given by Coulomb’s law. The distance between $+Q$ and $-q$ is approximately $(h+l)$ for small $\theta$.

$$ F_e = \frac{1}{4\pi\epsilon_0} \frac{Qq}{(h+l)^2} $$

To find the torque, we need the perpendicular distance ($d_{\perp}$) from the pivot O to the line joining $-q$ and $+Q$. Consider the triangle formed by points $Q$, $O$, and the displaced position of $-q$. The area of this triangle can be written in two ways: $$ \text{Area} = \frac{1}{2} \cdot h \cdot (l \sin\theta) $$ $$ \text{Area} = \frac{1}{2} \cdot \text{distance}(Q, -q) \cdot d_{\perp} $$

Equating these areas and using distance $(Q, -q) \approx (h+l)$:

$$ h l \sin\theta \approx (h+l) d_{\perp} $$ $$ d_{\perp} \approx \frac{hl\theta}{h+l} $$

Therefore, the electrostatic torque is:

$$ \tau_e = F_e \cdot d_{\perp} = \left( \frac{Qq}{4\pi\epsilon_0 (h+l)^2} \right) \left( \frac{hl\theta}{h+l} \right) $$ $$ \tau_e = \frac{Qqhl\theta}{4\pi\epsilon_0 (h+l)^3} $$

3. Stability Condition

For the equilibrium to be stable, the restoring torque must be greater than the destabilizing torque for a small displacement.

$$ \tau_g > \tau_e $$

Substituting the expressions derived:

$$ mgl\theta > \frac{Qqhl\theta}{4\pi\epsilon_0 (h+l)^3} $$

Canceling common terms $l$ and $\theta$ (since $\theta > 0$):

$$ mg > \frac{Qqh}{4\pi\epsilon_0 (h+l)^3} $$

Solving for the mass $m$:

$$ m > \frac{Qqh}{4\pi\epsilon_0 g (h+l)^3} $$

The range of values of mass $m$ for stable equilibrium is: $$ m > \frac{Qqh}{4\pi\epsilon_0 g (h+l)^3} $$