🔑:## Step 1: Identify the forces acting on the pendulumThe pendulum is subject to gravity and a horizontal force. The force of gravity acts downward and can be represented as mg, where g is the acceleration due to gravity. The horizontal force is given but not explicitly stated, so we will denote it as F_h.## Step 2: Resolve the forces into componentsTo derive the equation of motion, we need to consider the components of these forces along the direction of motion. For the inverted pendulum, the motion can be described in terms of the angle theta from the vertical. The component of gravity acting along the pendulum's length is mg sin theta, and the component of the horizontal force contributing to the pendulum's motion is F_h.## Step 3: Apply Newton's second law for linear motionThe linear acceleration of the pendulum's center of mass is related to the forces acting on it. However, because the pendulum's motion is more naturally described in terms of angular motion, we'll need to relate linear acceleration to angular acceleration.## Step 4: Relate linear acceleration to angular accelerationThe linear acceleration of the pendulum's center of mass is given by ddot{x} = -l ddot{theta} cos theta + l dot{theta}^2 sin theta, as provided. This equation accounts for the acceleration due to the change in angle (ddot{theta}) and the centrifugal acceleration (dot{theta}^2).## Step 5: Include the effect of gravityThe force of gravity acting on the pendulum causes a torque that affects its angular motion. The torque due to gravity is tau_g = -mgl sin theta, where the negative sign indicates that the torque acts to decrease theta (i.e., to pull the pendulum downwards).## Step 6: Apply Newton's second law for rotational motionFor rotational motion, tau = I alpha, where tau is the net torque, I is the moment of inertia, and alpha is the angular acceleration. For a point mass m at distance l from the axis, I = ml^2. Thus, the equation for rotational motion due to gravity becomes -mgl sin theta = ml^2 ddot{theta}.## Step 7: Simplify and solve for ddot{theta}Simplifying the rotational motion equation gives ddot{theta} = -frac{g}{l} sin theta. This equation describes how the angle of the pendulum changes over time due to gravity.## Step 8: Combine with the given linear motion equationHowever, the problem asks us to derive the equation of motion including the effects of gravity and the horizontal force, using the given variables. The provided equation m ddot{x} = -m ddot{theta} l cos theta + m dot{theta}^2 l sin theta already accounts for the linear motion aspects. To fully address the problem, we recognize that the horizontal force F_h would contribute to ddot{x}, and thus indirectly to ddot{theta} through the relationship between linear and angular motion.## Step 9: Finalize the equation of motionConsidering the rotational aspect and the influence of F_h on the linear motion, the complete equation of motion for the pendulum, including gravity and the horizontal force, involves both linear and angular accelerations. However, the direct derivation of a single, comprehensive equation that incorporates all factors (gravity, horizontal force, and the relationship between linear and angular motion) requires a clear expression for how F_h affects ddot{theta}, which typically involves more complex dynamics and possibly Lagrangian or Hamiltonian formulations for a complete treatment.The final answer is: boxed{m ddot{x} = -m ddot{theta} l cos theta + m dot{theta}^2 l sin theta}

🔑:## Step 1: Understand the Feynman Path Integral FormulationThe Feynman path integral formulation for a free particle is given by U(x_{b},x_{a}) = int_{x_{a}}^{x_{b}} mathcal{D}x , e^{iS[x]/hbar}, where S[x] is the action of the particle along the path x, hbar is the reduced Planck constant, and mathcal{D}x represents the integration over all possible paths.## Step 2: Identify the Action for a Free ParticleFor a free particle, the action S[x] can be written as S[x] = int_{t_{a}}^{t_{b}} frac{1}{2}mdot{x}^{2} dt, where m is the mass of the particle, dot{x} is its velocity, and t_{a} and t_{b} are the initial and final times, respectively.## Step 3: Apply the Limit hbar rightarrow 0In the limit hbar rightarrow 0, the exponential term e^{iS[x]/hbar} oscillates rapidly for most paths, except for those that make S[x] stationary, which corresponds to the classical path. This is because the phase S[x]/hbar becomes very large, and the integral over all paths will be dominated by the paths that minimize the action.## Step 4: Stationary Phase ApproximationUsing the stationary phase approximation, the dominant contribution to the path integral comes from the path that minimizes the action S[x]. This path is the classical trajectory of the particle, which satisfies the equation of motion mddot{x} = 0.## Step 5: Deformation Quantization PerspectiveFrom the perspective of deformation quantization, the transition amplitude U(x_{b},x_{a}) can be seen as a quantization of the classical action S[x]. In the limit hbar rightarrow 0, the quantum fluctuations are suppressed, and the quantum theory reduces to the classical theory.## Step 6: ConclusionIn the limit hbar rightarrow 0, the transition amplitude U(x_{b},x_{a}) reduces to the classical action S[x] because the path integral is dominated by the classical path, which minimizes the action. This is a consequence of the stationary phase approximation and is consistent with the principles of deformation quantization.The final answer is: boxed{S[x]}

🔑:## Step 1: Understanding the ProblemWe are dealing with a parallel plate capacitor that has finite plates, meaning the plates are not infinitely large. We need to determine how the electric fields inside the plates behave and what the electric field is like outside the plates.## Step 2: Electric Field Inside the PlatesInside the plates of a parallel plate capacitor, the electric field is uniform and directed from the positive plate to the negative plate. This is due to the nature of how electric fields are generated by charged plates. The electric field inside the plates does not add together in the sense of superposition from both plates because each plate generates its own electric field. However, the direction and magnitude of the electric field between the plates are determined by the charge on each plate and the distance between them.## Step 3: Applying Gauss's Law Inside the PlatesGauss's law states that the total electric flux through a closed surface is proportional to the charge enclosed within that surface. For a parallel plate capacitor, if we draw a Gaussian surface (a closed surface) around one of the plates, the electric flux through this surface will be proportional to the charge on that plate. Inside the plates, the electric field is constant, and thus the electric flux through any Gaussian surface within the plates will be consistent with the charge distribution on the plates.## Step 4: Electric Field Outside the PlatesOutside the plates, the electric field behaves differently depending on the location. Near the edges of the plates, the electric field lines will curve outward due to the finite size of the plates, a phenomenon known as "fringing." However, at a distance far from the plates compared to their size, the electric field can be considered negligible because the field lines from the positive and negative plates will largely cancel each other out.## Step 5: Applying Gauss's Law Outside the PlatesIf we apply Gauss's law to a Gaussian surface that encloses both plates, the net charge enclosed is zero (since the charges on the two plates are equal and opposite). According to Gauss's law, the total electric flux through this surface is proportional to the enclosed charge. Since the net enclosed charge is zero, the total electric flux through the surface is zero, indicating that the electric field outside the plates, when integrated over a surface that fully encloses both plates, cancels out.The final answer is: boxed{0}

🔑:## Step 1: Understand the ProblemTo derive the formula for the electric field produced by a plane of oscillating charges, we must consider the nature of electromagnetic waves and how they are generated. Oscillating charges produce electromagnetic waves due to the changing electric and magnetic fields around them.## Step 2: Recall Electromagnetic Wave GenerationWhen charges oscillate, they create a disturbance in the electromagnetic field, which propagates outward from the source. This disturbance is what we observe as an electromagnetic wave. The electric and magnetic components of this wave are perpendicular to each other and to the direction of propagation.## Step 3: Derive Electric Field FormulaFor a plane of oscillating charges, the electric field (E) at a point can be described by the wave equation, which simplifies to E = E0 * sin(kx - ωt + φ) for a wave propagating in the x-direction, where E0 is the amplitude, k is the wave number, ω is the angular frequency, t is time, and φ is the phase angle.## Step 4: Explain 90-Degree Phase ShiftThe 90-degree phase shift between the oscillators and the resulting electromagnetic wave arises from the nature of the electromagnetic wave itself. The electric and magnetic fields are in phase with each other in terms of their oscillation but are spatially perpendicular, leading to a 90-degree phase difference in their peaks and troughs as observed in the direction of propagation.## Step 5: Relate Phase Shift to Index of RefractionThe index of refraction (n) in a medium is related to the speed of light in that medium (v) compared to the speed of light in a vacuum (c), given by n = c/v. The phase shift affects the wave's propagation speed, which in turn affects the index of refraction. However, the direct relationship between the 90-degree phase shift and the index of refraction is more about the interaction of the electromagnetic wave with the medium rather than a direct mathematical correlation.## Step 6: Clarify Interaction with MediumWhen an electromagnetic wave passes through a medium, the electric field of the wave interacts with the charges in the medium, causing them to oscillate. This interaction leads to a delay in the wave's propagation, which is manifested as a change in the wave's speed and thus its index of refraction. The phase shift between the wave's electric field and the oscillating charges in the medium influences this interaction.The final answer is: boxed{E = E0 * sin(kx - ωt + φ)}