I will give you a physics problem and its solution in latex, what you have to do is, you gotta explain me the entire solution like a teacher explaining his students, use visualisation tactics, explaining each and every step in the solution to its utmost detail, elaborating calculation process and everything, make it seem like the problem solution discussion took 2 hours
Ok so here is the problem and its solution: \documentclass{article}
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\begin{document}

\title{Analysis of a Charged Pendulum Exhibiting Simple Harmonic Motion}
\author{}
\date{}
\maketitle

\section*{Problem Statement}

A uniform rigid rod of length \( L \) and mass \( M \) is hinged at one end and can rotate freely in a vertical plane about the hinge without friction. A small sphere of mass \( m \) and negative charge \( -q \) is attached at the free end of the rod. At a point directly below the hinge at a distance \( h \) (with \( h \gg L \)), there is a fixed positive point charge \( +Q \). The rod is displaced by a small angle \( \theta \) from the vertical and released from rest.

\begin{enumerate}
    \item[(a)] Determine the expression for the electric force acting on the sphere when the rod makes a small angle \( \theta \) with the vertical.
    \item[(b)] Using small-angle approximations, derive the equation of motion for the sphere and show that it undergoes simple harmonic motion (SHM).
    \item[(c)] Calculate the angular frequency \( \omega \) of the SHM.
    \item[(d)] Analyze the torque acting on the rod due to the electric force and determine its effect on the rod's rotational motion.
\end{enumerate}

\section*{Solution}

\subsection*{(a) Expression for the Electric Force Acting on the Sphere}

When the rod makes a small angle \( \theta \) with the vertical, the position of the sphere at the end of the rod can be approximated using small-angle approximations.

Using the hinge as the origin:

\begin{align*}
x &= L \sin \theta \approx L \theta \quad (\text{since } \sin \theta \approx \theta \text{ for small } \theta) \\
y &= -L \cos \theta \approx -L (1 - \tfrac{\theta^2}{2}) \approx -L \quad (\text{since } \theta^2 \text{ is negligible})
\end{align*}

The fixed positive point charge \( +Q \) is located at \( (0, -h) \).

The displacement components from the sphere to the charge \( +Q \) are:

\begin{align*}
\Delta x &= x - 0 = L \theta \\
\Delta y &= y - (-h) = -L + h = h - L
\end{align*}

Since \( h \gg L \), we can approximate:

\[
\Delta y \approx h
\]

The distance \( r \) between the sphere and the fixed charge is:

\[
r = \sqrt{ (\Delta x)^2 + (\Delta y)^2 } \approx \sqrt{ (L \theta)^2 + h^2 }
\]

Using the binomial approximation for small \( \varepsilon \):

\[
\sqrt{ h^2 + (L \theta)^2 } = h \sqrt{ 1 + \left( \dfrac{ L \theta }{ h } \right)^2 } \approx h \left( 1 + \dfrac{1}{2} \left( \dfrac{ L \theta }{ h } \right)^2 \right)
\]

Therefore:

\[
r \approx h \left( 1 + \dfrac{ L^2 \theta^2 }{ 2 h^2 } \right)
\]

The electric force magnitude acting on the sphere due to \( +Q \) is given by Coulomb's law:

\[
F = \dfrac{1}{4\pi \varepsilon_0} \dfrac{ |q| Q }{ r^2 }
\]

Substituting \( r \):

\begin{align*}
F &= \dfrac{1}{4\pi \varepsilon_0} \dfrac{ |q| Q }{ \left( h \left( 1 + \dfrac{ L^2 \theta^2 }{ 2 h^2 } \right) \right)^2 } \\
&= \dfrac{1}{4\pi \varepsilon_0} \dfrac{ |q| Q }{ h^2 \left( 1 + \dfrac{ L^2 \theta^2 }{ 2 h^2 } \right)^2 }
\end{align*}

Using the binomial approximation \( (1 + x)^{-2} \approx 1 - 2x \) for small \( x \):

\[
\left( 1 + \dfrac{ L^2 \theta^2 }{ 2 h^2 } \right)^{-2} \approx 1 - 2 \left( \dfrac{ L^2 \theta^2 }{ 2 h^2 } \right) = 1 - \dfrac{ L^2 \theta^2 }{ h^2 }
\]

Therefore, the force simplifies to:

\[
F \approx \dfrac{1}{4\pi \varepsilon_0} \dfrac{ |q| Q }{ h^2 } \left( 1 - \dfrac{ L^2 \theta^2 }{ h^2 } \right)
\]

The direction of the force is along the line connecting the sphere and the charge \( +Q \). The horizontal component of the force (\( F_x \)) is responsible for the torque about the hinge.

The horizontal component is:

\[
F_x = F \cdot \sin \phi
\]

Where \( \phi \) is the angle between the line connecting the charges and the vertical axis. For small \( \theta \):

\[
\sin \phi \approx \dfrac{ \Delta x }{ r } \approx \dfrac{ L \theta }{ h }
\]

Substituting back:

\[
F_x \approx F \cdot \dfrac{ L \theta }{ h } = \left( \dfrac{1}{4\pi \varepsilon_0} \dfrac{ |q| Q }{ h^2 } \left( 1 - \dfrac{ L^2 \theta^2 }{ h^2 } \right) \right) \dfrac{ L \theta }{ h }
\]

Neglecting \( \theta^2 \) terms (since \( \theta \) is small):

\[
F_x \approx \dfrac{1}{4\pi \varepsilon_0} \dfrac{ |q| Q L \theta }{ h^3 }
\]

\subsection*{(b) Deriving the Equation of Motion and Showing SHM}

The torque \( \tau \) about the hinge due to the horizontal component \( F_x \) is:

\[
\tau = - L \cdot F_x = - L \left( \dfrac{1}{4\pi \varepsilon_0} \dfrac{ |q| Q L \theta }{ h^3 } \right) = - \dfrac{ |q| Q L^2 \theta }{ 4\pi \varepsilon_0 h^3 }
\]

The negative sign indicates that the torque acts in the direction opposite to the angular displacement \( \theta \), providing a restoring torque.

Using Newton's second law for rotational motion:

\[
I \ddot{ \theta } = \tau
\]

Where \( I \) is the moment of inertia of the rod-sphere system about the hinge:

\[
I = I_{\text{rod}} + I_{\text{sphere}} = \dfrac{1}{3} ML^2 + mL^2 = \left( \dfrac{1}{3} M + m \right) L^2
\]

Substituting \( \tau \) and \( I \) into the equation of motion:

\[
\left( \dfrac{1}{3} M + m \right) L^2 \ddot{ \theta } = - \dfrac{ |q| Q L^2 \theta }{ 4\pi \varepsilon_0 h^3 }
\]

Simplifying:

\[
\ddot{ \theta } + \omega^2 \theta = 0
\]

Where the angular frequency \( \omega \) is:

\[
\omega = \sqrt{ \dfrac{ |q| Q }{ 4\pi \varepsilon_0 \left( \dfrac{1}{3} M + m \right) h^3 } }
\]

Thus, the motion of the sphere is simple harmonic with angular frequency \( \omega \).

\subsection*{(c) Calculating the Angular Frequency}

Given the expression for \( \omega \):

\[
\omega = \sqrt{ \dfrac{ |q| Q }{ 4\pi \varepsilon_0 \left( \dfrac{1}{3} M + m \right) h^3 } }
\]

This represents the angular frequency of the SHM for the charged pendulum system.


\subsection*{(d) Torque }

\subsection*{1. Calculate the Horizontal Component of the Electric Force \( F_x \)}

The electric force \( \vec{F} \) acts along the line connecting the sphere and \( +Q \). The horizontal component is:

\[
F_x = F \cdot \dfrac{ \Delta x }{ r }.
\]

Using \( r \approx h \) and \( \dfrac{ \Delta x }{ r } \approx \dfrac{ L \theta }{ h } \):

\[
F_x \approx F \cdot \dfrac{ L \theta }{ h } = \left( \dfrac{1}{4\pi \varepsilon_0} \dfrac{ |q Q| }{ h^2 } \left( 1 - \dfrac{ L^2 \theta^2 }{ h^2 } \right ) \right ) \cdot \dfrac{ L \theta }{ h } = \dfrac{1}{4\pi \varepsilon_0} \dfrac{ |q Q| L \theta }{ h^3 } \left( 1 - \dfrac{ L^2 \theta^2 }{ h^2 } \right ).
\]

Neglecting higher-order terms involving \( \theta^3 \):

\[
F_x \approx \dfrac{1}{4\pi \varepsilon_0} \dfrac{ |q Q| L \theta }{ h^3 }.
\]

\subsection*{2. Calculate the Torque \( \tau \)}

The torque about the hinge due to the horizontal component of the electric force is:

\[
\tau = - L F_x.
\]

The negative sign indicates that the torque acts to reduce \( \theta \), providing a restoring torque.

Substituting \( F_x \):

\[
\tau = - L \left( \dfrac{1}{4\pi \varepsilon_0} \dfrac{ |q Q| L \theta }{ h^3 } \right ) = - \dfrac{1}{4\pi \varepsilon_0} \dfrac{ |q Q| L^2 \theta }{ h^3 }.
\]

\section*{Final Expression for the Torque}

The torque acting on the rod due to the electric force is:

\[
\tau = - \dfrac{1}{4\pi \varepsilon_0} \dfrac{ |q Q| L^2 }{ h^3 } \theta.
\]

This torque provides a restoring effect, proportional to the angular displacement \( \theta \), and acts to bring the rod back to its equilibrium position.


\end{document}

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