ok so now I am giving you a physics problem set, discuss about their solutions in a very well manner, like an instructor instructing his students, make sure to use very simple concepts to understand the problem and solution and take to advanced level for solving the actual problem provided, make it very detailed and well mannered

\subsection*{Problem}

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. The rod is initially held horizontally. 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 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 horizontal.
\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 the effect on the rod's rotational motion.
\end{enumerate}

\subsection*{Solution}

\textbf{(a) Electric Force Acting on the Sphere}

When the rod makes a small angle ( \theta ) with the horizontal, the position of the sphere can be described as:

[
x = L \cos\theta \approx L (1 - \tfrac{1}{2}\theta^2), \quad y = -L \sin\theta \approx -L \theta
]

The distance between the sphere and the fixed charge ( +Q ) is:

[
r = \sqrt{ (x - 0)^2 + (y + h)^2 } \approx \sqrt{ [L (1 - \tfrac{1}{2}\theta^2)]^2 + (-L \theta + h)^2 }
]

Since ( h \gg L ), we can approximate ( r \approx h - L \theta ).

The electric force acting on the sphere due to the fixed charge is:

[
F_e = \frac{1}{4\pi\varepsilon_0} \frac{q Q}{r^2}
]

Since ( q ) is negative and ( Q ) is positive, the force is attractive and acts along the line connecting the charges.

For small ( \theta ), the force in the vertical direction (which affects the motion) is:

[
F_y = F_e \cdot \frac{(y + h)}{r} \approx F_e \cdot \frac{h - L \theta + h}{h - L \theta} \approx F_e
]

Thus, the vertical component of the electric force is approximately equal to ( F_e ).

\textbf{(b) Deriving the Equation of Motion}

Applying Newton's second law in the vertical direction for the sphere:

[
m \ddot{y} = -m g + F_y
]

Since there is no gravitational field (as per the problem statement), ( g = 0 ), so:

[
m \ddot{y} = F_y = \frac{1}{4\pi\varepsilon_0} \frac{q Q}{r^2}
]

Substituting ( r \approx h ) and ( y \approx -L \theta ):

[
m (-L \ddot{\theta}) = \frac{1}{4\pi\varepsilon_0} \frac{q Q}{h^2}
]

Simplifying:

[
m L \ddot{\theta} + \frac{q Q}{4\pi\varepsilon_0 h^2} = 0
]

This is a linear second-order differential equation.

\textbf{(c) Calculating the Angular Frequency ( \omega )}

Rewriting the equation:

[
\ddot{\theta} + \left( \frac{q Q}{4\pi\varepsilon_0 m L h^2} \right) \theta = 0
]

This is the standard form of SHM, ( \ddot{\theta} + \omega^2 \theta = 0 ), where:

[
\omega = \sqrt{ \frac{q Q}{4\pi\varepsilon_0 m L h^2} }
]

Thus, the angular frequency of the SHM is:

[
\omega = \sqrt{ \frac{q Q}{4\pi\varepsilon_0 m L h^2} }
]

also make sure the answer you give me should be errorless in terms of math and latex both, for example math functions inside $ or , make sure to make it 0% error



\documentclass{article}
\usepackage{amsmath}
\usepackage{physics}
\usepackage{graphicx}
\usepackage{siunitx}
\usepackage{hyperref}
\begin{document}

\title{Advanced Electrostatics Problem Integrating SHM, Rigid Body Dynamics, and Electric Fields}
\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. The rod is initially held horizontally. 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 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 horizontal.
    \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 the 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 horizontal (measured from the horizontal), the position of the sphere at the end of the rod can be represented in Cartesian coordinates as:

\[
\begin{cases}
x = L \cos\theta \approx L \left( 1 - \dfrac{\theta^2}{2} \right) \\
y = -L \sin\theta \approx -L \theta
\end{cases}
\]

Here, we have used the small-angle approximations:

\[
\sin\theta \approx \theta, \quad \cos\theta \approx 1 - \dfrac{\theta^2}{2}
\]

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

The displacement vector from the sphere to the point charge is:

\[
\begin{cases}
\Delta x = x - 0 = L \cos\theta \\
\Delta y = y - (-h) = -L \theta + h = h - L \theta
\end{cases}
\]

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

\begin{align*}
r &= \sqrt{ (\Delta x)^2 + (\Delta y)^2 } \\
&= \sqrt{ \left[ L \cos\theta \right]^2 + \left( h - L \theta \right)^2 } \\
&\approx \sqrt{ \left[ L \left( 1 - \dfrac{\theta^2}{2} \right) \right]^2 + \left( h - L \theta \right)^2 }
\end{align*}

Expanding and simplifying the expression, and considering \( h \gg L \), we can approximate \( r \) as:

\begin{align*}
r &\approx \sqrt{ L^2 \left( 1 - \theta^2 + \dfrac{\theta^4}{4} \right) + h^2 - 2 h L \theta + L^2 \theta^2 } \\
&\approx \sqrt{ h^2 - 2 h L \theta + L^2 } \\
&\approx h \sqrt{ 1 - \dfrac{2 L \theta}{h} + \dfrac{L^2}{h^2} }
\end{align*}

Since \( \dfrac{L}{h} \ll 1 \), we can expand the square root using the binomial approximation:

\[
\sqrt{1 + \varepsilon} \approx 1 + \dfrac{\varepsilon}{2} - \dfrac{\varepsilon^2}{8}, \quad \text{for } |\varepsilon| \ll 1
\]

Let \( \varepsilon = -\dfrac{2 L \theta}{h} + \dfrac{L^2}{h^2} \), then:

\begin{align*}
r &\approx h \left[ 1 + \dfrac{1}{2} \left( -\dfrac{2 L \theta}{h} + \dfrac{L^2}{h^2} \right) \right] \\
&\approx h - L \theta + \dfrac{L^2}{2 h}
\end{align*}

Since \( \dfrac{L^2}{2 h} \) is negligible compared to \( h \) and \( L \theta \), we can further simplify:

\[
r \approx h - L \theta
\]

The electric force \( \vec{F} \) acting on the sphere due to the fixed charge \( +Q \) is given by Coulomb's law:

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

Here, \( q \) is negative and \( Q \) is positive, so \( q Q < 0 \), and the force is attractive towards the charge \( +Q \).

The unit vector \( \hat{r} \) points from the sphere to the charge \( +Q \). The components of \( \vec{F} \) are:

\[
\begin{cases}
F_x = F \cdot \dfrac{\Delta x}{r} \\
F_y = F \cdot \dfrac{\Delta y}{r}
\end{cases}
\]

But since \( \Delta x = L \cos\theta \approx L \), and \( \Delta y = h - L \theta \approx h \), and \( h \gg L \), we have \( \dfrac{\Delta x}{r} \ll 1 \). Therefore, the horizontal component \( F_x \) is negligible compared to the vertical component \( F_y \).

Thus, the vertical component of the electric force acting on the sphere is approximately:

\[
F_y \approx \dfrac{1}{4\pi\varepsilon_0} \dfrac{q Q}{r^2}
\]

Substituting \( r \approx h - L \theta \), we expand \( r^{-2} \) using Taylor series:

\begin{align*}
\dfrac{1}{r^2} &= \dfrac{1}{\left( h - L \theta \right)^2} \\
&= \dfrac{1}{h^2 \left( 1 - \dfrac{L \theta}{h} \right)^2 } \\
&\approx \dfrac{1}{h^2} \left( 1 + 2 \dfrac{L \theta}{h} \right)
\end{align*}

Therefore, the vertical component of the electric force is:

\begin{align*}
F_y &\approx \dfrac{1}{4\pi\varepsilon_0} \dfrac{q Q}{h^2} \left( 1 + 2 \dfrac{L \theta}{h} \right) \\
&= \dfrac{1}{4\pi\varepsilon_0} \dfrac{q Q}{h^2} + \dfrac{1}{4\pi\varepsilon_0} \dfrac{2 q Q L \theta}{h^3}
\end{align*}

Since the first term is constant, it does not contribute to the restoring force. The second term depends on \( \theta \) and provides the restoring force necessary for SHM.

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

To derive the equation of motion, we consider the dynamics of the rod and the sphere. Since the rod is rigid and rotates about the hinge without friction, we can apply Newton's second law for rotational motion.

The torque \( \tau \) acting on the rod due to the electric force is:

\[
\tau = L \cdot F_y
\]

Using the approximation for \( F_y \) from part (a):

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

Again, the first term is constant torque, which would tend to provide a constant angular acceleration, but since the rod is released from rest, only the \( \theta \)-dependent term contributes to SHM.

The net torque causing rotational motion is:

\[
\tau_{\text{net}} = \dfrac{1}{4\pi\varepsilon_0} \dfrac{2 q Q L^2 \theta}{h^3}
\]

The moment of inertia \( I \) of the system (rod plus sphere) about the hinge is:

\[
I = I_{\text{rod}} + I_{\text{sphere}} = \dfrac{1}{3} M L^2 + m L^2
\]

If \( m \) is small compared to \( M \), we may approximate \( I \approx \dfrac{1}{3} M L^2 \). However, for generality, we'll keep both terms.

Newton's second law for rotation:

\[
I \ddot{\theta} = -\tau_{\text{net}}
\]

Substitute the expression for \( \tau_{\text{net}} \):

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

Rewriting:

\[
\ddot{\theta} + \left( \dfrac{2 q Q L^2 }{4\pi\varepsilon_0 I h^3} \right) \theta = 0
\]

This is a linear second-order differential equation of the form:

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

Which describes simple harmonic motion (SHM) with angular frequency:

\[
\omega = \sqrt{ \dfrac{2 q Q L^2 }{4\pi\varepsilon_0 I h^3} }
\]

Since \( q Q < 0 \) (attractive force), and \( I > 0 \), \( \omega^2 \) is positive, confirming that the motion is indeed SHM.

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

Using the expression derived above for \( \omega \):

\[
\omega = \sqrt{ \dfrac{2 | q | Q L^2 }{4\pi\varepsilon_0 I h^3} }
\]

Simplify the expression:

\[
\omega = \sqrt{ \dfrac{ | q | Q L^2 }{2\pi\varepsilon_0 I h^3} }
\]

Substituting \( I = \dfrac{1}{3} M L^2 + m L^2 = \left( \dfrac{1}{3} M + m \right) L^2 \):

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

Thus, the angular frequency of the SHM is:

\[
\omega = \sqrt{ \dfrac{ | q | Q }{2\pi\varepsilon_0 \left( \dfrac{1}{3} M + m \right) h^3} }
\]

\subsection*{(d) Analyzing the Torque and Its Effect on Rotational Motion}

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

\[
\tau = L \cdot F_y
\]

As previously derived, the torque is:

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

Since the torque is proportional to \( \theta \), it provides a restoring torque that tends to bring the rod back towards the horizontal position. This torque results in angular acceleration \( \ddot{\theta} \) as given by:

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

This negative sign indicates that the torque opposes the angle \( \theta \), consistent with SHM.

The effect of this torque on the rod's rotational motion is to cause the rod to oscillate about the horizontal position with angular frequency \( \omega \).

\subsection*{Conclusion}

Through the application of electrostatic force analysis, small-angle approximations, and rotational dynamics, we have shown that:

\begin{enumerate}
    \item The electric force acting on the sphere when the rod makes a small angle \( \theta \) with the horizontal is approximately:

    \[
    F_y \approx \dfrac{1}{4\pi\varepsilon_0} \dfrac{q Q}{h^2} + \dfrac{1}{4\pi\varepsilon_0} \dfrac{2 q Q L \theta}{h^3}
    \]

    \item The sphere (and rod) undergoes simple harmonic motion described by the equation:

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

    \item The angular frequency of the SHM is:

    \[
    \omega = \sqrt{ \dfrac{ | q | Q }{2\pi\varepsilon_0 \left( \dfrac{1}{3} M + m \right) h^3} }
    \]

    \item The torque due to the electric force provides a restoring torque proportional to \( \theta \), leading to SHM in the rotational motion of the rod.
\end{enumerate}

\end{document}

now in the same way, discuss about the solutions as if it is a conceptual discussion after finding out the solution, like some post solution discussion, you understand? make it very easy to understand ok? and the latex should have 0% error



\documentclass{article}
\usepackage{amsmath}
\usepackage{physics}
\usepackage{graphicx}
\usepackage{siunitx}
\usepackage{hyperref}
\begin{document}

\title{Advanced Electrostatics Problem Integrating SHM, Rigid Body Dynamics, and Electric Fields}
\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. The rod is initially held horizontally. 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 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 horizontal.
    \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 the 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 horizontal, the position of the sphere at the end of the rod can be expressed in Cartesian coordinates:

\[
\begin{cases}
x = L \cos\theta \approx L \left( 1 - \dfrac{\theta^2}{2} \right) \\
y = -L \sin\theta \approx -L \theta
\end{cases}
\]

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

The displacement vector from the sphere to the charge \( +Q \) is:

\[
\begin{cases}
\Delta x = x - 0 = L \cos\theta \\
\Delta y = y - (-h) = -L \theta + h
\end{cases}
\]

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

\begin{align*}
r &= \sqrt{ (\Delta x)^2 + (\Delta y)^2 } \\
&\approx \sqrt{ \left[ L \left( 1 - \dfrac{\theta^2}{2} \right) \right]^2 + \left( h - L \theta \right)^2 } \\
&\approx \sqrt{ L^2 \left( 1 - \theta^2 + \dfrac{\theta^4}{4} \right) + h^2 - 2 h L \theta + L^2 \theta^2 }
\end{align*}

Since \( h \gg L \), terms containing \( L^2 \) are much smaller compared to \( h^2 \) and can be neglected:

\[
r \approx \sqrt{ h^2 - 2 h L \theta } = h \sqrt{ 1 - \dfrac{2 L \theta}{h} }
\]

Using the binomial expansion for small \( \dfrac{L \theta}{h} \):

\[
\sqrt{ 1 - \dfrac{2 L \theta}{h} } \approx 1 - \dfrac{L \theta}{h}
\]

Thus, the distance simplifies to:

\[
r \approx h - L \theta
\]

The electric force \( \vec{F} \) acting on the sphere is given by Coulomb's law:

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

Since \( q \) is negative and \( Q \) is positive, the force is attractive and directed along \( \hat{r} \), the unit vector pointing from the sphere to the charge.

For small \( \theta \), the direction of \( \vec{F} \) is nearly vertical, so we can approximate the vertical component of the force as:

\[
F_y \approx \dfrac{1}{4\pi\varepsilon_0} \dfrac{q Q}{r^2}
\]

Substituting \( r \approx h - L \theta \):

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

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

Using the approximation:

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

The vertical component of the electric force becomes:

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

The torque \( \tau \) about the hinge due to \( F_y \) is:

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

The constant term \( \dfrac{1}{4\pi\varepsilon_0} \dfrac{q Q L}{h^2} \) does not affect the oscillatory motion; the term proportional to \( \theta \) provides the restoring torque:

\[
\tau_{\text{restoring}} = \dfrac{1}{4\pi\varepsilon_0} \dfrac{2 q Q L^2 \theta}{h^3}
\]

Applying Newton's second law for rotation:

\[
I \ddot{\theta} = -\tau_{\text{restoring}}
\]

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

\[
I = I_{\text{rod}} + I_{\text{sphere}} = \dfrac{1}{3} M L^2 + m L^2
\]

Substituting the expression for \( \tau_{\text{restoring}} \):

\[
I \ddot{\theta} + \dfrac{2 q Q L^2 }{4\pi\varepsilon_0 h^3} \theta = 0
\]

Simplifying:

\[
\ddot{\theta} + \left( \dfrac{2 q Q L^2 }{4\pi\varepsilon_0 I h^3} \right) \theta = 0
\]

This is the standard form of SHM, indicating that the system undergoes simple harmonic motion.

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

By comparing to the SHM equation \( \ddot{\theta} + \omega^2 \theta = 0 \), we identify:

\[
\omega^2 = \dfrac{2 q Q L^2 }{4\pi\varepsilon_0 I h^3}
\]

Therefore, the angular frequency is:

\[
\omega = \sqrt{ \dfrac{ | q | Q L^2 }{2\pi\varepsilon_0 I h^3} }
\]

Substituting \( I \):

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

\subsection*{(d) Analyzing the Torque and Its Effect on Rotational Motion}

The restoring torque is proportional to \( \theta \):

\[
\tau_{\text{restoring}} = -\kappa \theta, \quad \text{where } \kappa = \dfrac{2 q Q L^2 }{4\pi\varepsilon_0 h^3}
\]

This torque tends to bring the rod back to its equilibrium position. The negative sign indicates that it opposes the displacement.

\section*{Discussion}

\subsection*{Conceptual Understanding}

This problem beautifully intertwines concepts from electrostatics and rotational dynamics to explore simple harmonic motion.

\paragraph{1. Electric Force Variation}

As the rod rotates slightly, the charged sphere moves closer to or farther from the fixed charge. The distance \( r \) changes linearly with \( \theta \) for small angles. This causes the electric force to vary, introducing a restoring effect essential for SHM.

\paragraph{2. Small-Angle Approximations}

By assuming \( \theta \) is small, we simplify trigonometric functions:

\[
\sin\theta \approx \theta, \quad \cos\theta \approx 1 - \dfrac{\theta^2}{2}
\]

These approximations linearize the equations, making the complex problem more tractable.

\paragraph{3. Restoring Torque and SHM}

A key characteristic of SHM is a restoring force (or torque) proportional to displacement:

\[
F_{\text{restoring}} \propto -\theta \quad \text{or} \quad \tau_{\text{restoring}} \propto -\theta
\]

In this case, the torque due to the electric force fulfills this condition, leading to SHM.

\paragraph{4. Moment of Inertia's Role}

The moment of inertia \( I \) determines how the system's mass distribution affects its rotational motion. A larger \( I \) (more mass or mass farther from the axis) results in a lower angular frequency \( \omega \).

\subsection*{Practical Implications}

Understanding such systems has broader applications:

- **Oscillations in Electric Fields**: Charged particles in electric fields can exhibit oscillatory behavior, important in particle traps and accelerators.
- **Design of Sensors and Devices**: Electrostatic forces are used in micromechanical systems (MEMS) for precision devices.

\subsection*{Key Takeaways}

- The interplay between electric forces and mechanical motion can lead to interesting dynamics like SHM.
- Small-angle approximations are powerful tools for simplifying and solving complex physics problems.
- The concepts of torque and moment of inertia are crucial in rotational dynamics and influence the motion's characteristics.

\subsection*{Further Exploration}

To deepen your understanding, consider:

- **Energy Analysis**: Explore the potential energy in the system due to the electric force and how it relates to the SHM.
- **Nonlinear Effects**: Investigate what happens when the small-angle approximation is not valid, leading to nonlinear differential equations.
- **Damping Effects**: Introduce resistance (like air friction) and study damped oscillations.

\end{document}