Figure \(4-11\) shows a parallel-plate capacitor whose upper plate of mass \(m\) is supported by a spring insulated from ground. A voltage \(V\) is applied to the upper plate. There is zero tension in the spring when the length of the air gap is \(x_{0}\). Set \(m=0.1\) kilogram, \(x_{0}=0.01\) meter, \(k=150\) newtons per meter, \(V=\) 100 volts, and \(A=0.01\) square meter. Assume that the mass of the spring is negligible, and neglect edge effects. a) Set the various potential energies of the complete system equal to zero when the tension in the spring is zero, at \(x=x_{0}\). Show that the potential energy is then $$ W=m g\left(x-x_{0}\right)+\frac{1}{2} k\left(x-x_{0}\right)^{2}-\frac{1}{2} \epsilon_{0} A V^{2}\left(\frac{1}{x}-\frac{1}{x_{0}}\right) $$ Draw a curve of \(W\) as a function of \(x\) between \(x=10^{-6}\) and \(x=10^{-2}\) meter. Use a logarithmic scale for the \(x\)-axis. b) Show that, at equilibrium, $$ m g+k\left(x-x_{0}\right)+\frac{1}{2} \epsilon_{0} A V^{2} / x^{2}=0 $$ You can arrive at this result in two different ways. There is stable equilibrium at \(x_{e q}=3.46 \times 10^{-3}\) meter and unstable equilibrium at \(x=2.93 \times 10^{-5}\) meter. c) If the system is brought to the point of stable equilibrium, and then disturbed slightly, it oscillates. For small oscillations, the restoring force \(F\) is proportional to the displacement and we have a simple harmonic motion: $$ F=m\left(\frac{d^{2} x}{d t^{2}}\right)_{e q}=-\left(\frac{d W}{d x}\right)_{e q}=-K\left(x-x_{e q}\right) $$ where the derivatives must be evaluated in the neighborhood of \(x=x_{e q}\). The circular frequency is \(\omega=(K / m)^{1 / 2}\), where $$ K=\left(\frac{d^{2} W}{d x^{2}}\right)_{e q}=-\left(\frac{d F}{d x}\right)_{\because \|} $$ Show that $$ \omega=\left[\frac{k-\left(\epsilon_{0} A V^{2} / x_{e q}^{3}\right)}{m}\right]^{1 / 2} $$ This gives a frequency of \(6.16\) hertz.

Step by step solution

01

Expression for Potential Energy

First we need to express the potential energy of the system. The gravitational potential energy is given by the height above the reference point, hence, we have term from gravity: mg(x-x_0). The spring's potential energy is given by the formula for elastic potential energy: (1/2)k(x-x_0)^2. Electrical potential energy stored in the capacitor is given by: (1/2)(ε_0AV^2)(1/x - 1/x_0). The potential energy at x_0 is set to zero, which leads to the formula W = mg(x-x_0) + (1/2)k(x-x_0)^2 - (1/2)(ε_0AV^2)(1/x - 1/x_0).

02

Equilibrium Condition

To find the equilibrium condition, we set the derivative of the potential energy W with respect to x to zero, which represents the force on the plate being zero. The expression leads to mg + k(x-x_0) + (1/2)(ε_0AV^2)/x^2 = 0. This is the equilibrium condition of the system.

03

Expression for Simple Harmonic Motion

Small oscillations around the stable equilibrium point can be approximated by simple harmonic motion. For this condition, the potential energy curve near the equilibrium is approximated by a parabola. The force is given by the negative derivative of the potential energy with respect to x, leading to F = -dW/dx. For small displacements from stable equilibrium, the force is proportional to the displacement, F = -K(x-x_eq), where K is the effective spring constant for small oscillations.

04

Calculating the Effective Spring Constant

The effective spring constant K at the stable equilibrium is found by taking the second derivative of the potential energy S, expressing the second derivative of the force with respect to displacement. K = (d^2W/dx^2)_eq, which reduces to K = k - (ε_0AV^2/x_eq^3).

05

Finding the Frequency of Oscillations

Finally, to find the frequency of the oscillations, we use the formula for the circular frequency ω=(K/m)^(1/2). We substitute the expression for K to get ω = [(k-(ε_0AV^2/x_eq^3))/m]^(1/2), which determines the frequency of the small oscillations near the stable equilibrium point.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Potential Energy in Capacitor Systems

Understanding potential energy in capacitor systems is crucial as it allows us to analyze how energy is stored and behaves within an electrical system. In our given exercise, we explore a parallel-plate capacitor with a moving plate supported by a spring. This setup allows us to consider not only the electrical energy stored in the capacitor but also the gravitational potential energy of the upper plate and the elastic potential energy of the spring.

When the capacitor is charged by applying a voltage V, an electrostatic force is generated between the plates, which influences the equilibrium position of the movable plate. The electrical potential energy within the capacitor is represented by the term \text{-(1/2)(\epsilon_0AV^2)(1/x - 1/x_0)}\, where \(\epsilon_0\) is the permittivity of free space, A is the area of the plate, and x is the separation between the plates. As the plate moves away from the equilibrium position, the electrical energy changes because the capacitance varies with the plate separation.

In terms of optimizing the exercise for student comprehension, breaking down the equation for potential energy into its constituent parts and explaining each term individually can be highly beneficial. Emphasizing the three types of potential energy—gravitational, elastic, and electrical—and showing how each depends on the plate separation (x) assists in developing a deeper understanding. For visual learners, sketching a graph of potential energy against plate separation and highlighting the equilibrium points can help illustrate how the system behaves energetically.

Simple Harmonic Motion

Simple harmonic motion (SHM) is a type of oscillatory motion that is particularly important in physics and engineering because of its predictability and the fact it is one of the simplest forms of oscillating system behavior. In our exercise, we delve into SHM by looking at a system that oscillates when slight disturbances are introduced.

In the context of the capacitor-spring system, SHM can occur when the system is moved slightly from its equilibrium position and then released. The restoring force, which is directly proportional to the displacement from the equilibrium but in the opposite direction, causes the system to oscillate. The motion is 'simple' because the restoring force is linearly proportional to the displacement, and so, at small amplitudes, it can be described using the simple harmonic oscillator model.

The exercise encourages students to understand that the frequency of oscillation is directly related to the effective spring constant K and inversely related to the mass of the oscillating plate. By exploring the formula \(\omega=(K / m)^{1 / 2}\), where \(\omega\) represents the angular frequency, students learn to associate higher spring constants or lower masses with faster oscillations. Incorporating an explanation of how to derive the effective spring constant and how it relates to frequency can further enhance student learning and grasp of SHM principles.

Equilibrium in Electrical Systems

Achieving equilibrium in an electrical system, such as our capacitor arrangement, involves finding a state where forces are balanced and there is no net motion. In the scenario of our exercise, equilibrium occurs when the electrostatic force exerted by the charged capacitor plates is balanced by the mechanical forces from gravity and the spring supporting the movable plate.

The process of identifying the equilibrium condition mathematically requires setting the total force acting on the system to zero. This integral step leads to the equation \(m g + k(x-x_{0}) + (1/2) (\epsilon_{0} A V^{2}) / x^{2} = 0\), which is a condition that students must understand to solve the problem effectively.

It is also valuable for students to recognize that equilibrium points can be stable or unstable. In the exercise, we find two such points: the stable equilibrium at \(x_{eq}=3.46 \times 10^{-3}\) meter, where slight perturbations result in SHM, and the unstable equilibrium at a smaller separation that, if disturbed, would lead to destabilization rather than oscillation. When teaching these concepts, emphasizing differences between stable and unstable equilibria can deepen a student's comprehension. Illustrating with diagrams how potential energy varies with position and identifying the minima (stable) and maxima (unstable) of the curve can make the abstract concept of equilibrium more concrete for students.