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A parallel-plate capacitor has a charge \(Q\) and plates of area \(A\). What force acts on one plate to attract it toward the other plate? Because the electric field between the plates is \(E=Q / A \epsilon_{0},\) you might think that the force is \(F=Q E=Q^{2} / A \epsilon_{0} .\) This is wrong, because the field \(E\) includes contributions from both plates, and the field created by the positive plate cannot exert any force on the positive plate. Show that the force exerted on each plate is actually \(F=Q^{2} / 2 \epsilon_{0} A .\) (Suggestion: Let \(C=\epsilon_{0} A / x\) for an arbitrary plate separation \(x ;\) then require that the work done in separating the two charged plates be \(W=\int F \, d x\).) The force exerted by one charged plate on another is sometimes used in a machine shop to hold a workpiece stationary.

Step by step solution

01

Expression for the Electric Field

The electric field, \(E\), between two plates of a parallel-plate capacitor with a charge \(Q\) and plates of area \(A\) is given by \(E=Q / A \epsilon_{0}\), where \(\epsilon_{0}\) is the permittivity of free space. Since the electric field line travels from positive to negative, the field due to the positive plate will not exert any force on the positive plate itself.

02

Expression for the Electric Force

The force, \(F\), exerted by the negative plate on the positive plate is given by \(F=E Q / 2\). This is one-half of \(Q^2 / A \epsilon_{0}\) because the field \(E\) includes contributions from both plates and the field created by the positive plate does not exert any force on the positive plate.

03

Expression for Work Done in Separating the Plates

The work done, \(W\), in separating the two charged plates is given by the integral of the force, \(F\), over the separation, \(x\), such that \(W=\int F dx\). Let \(C=\epsilon_{0} A / x\). We have \(dW = Cdx\), and therefore \(W = \int Cdx\).

04

Derive the Force Exerted on Each Plate

Computing the integral results in \(W = Q^2/2C\). Equating this expression for work done with the previous one, yields \(Q^2/2C = Q^2/2 \epsilon_{0} A\), which simplifies to \(F=Q^{2} / 2 \epsilon_{0} A\). Therefore, that is the force exerted on each plate.

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Key Concepts

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

Electric Field

Understanding the electric field is crucial when studying parallel-plate capacitors. An electric field represents the force that would be exerted on a positive test charge placed within the field. For a parallel-plate capacitor with plates of charge Q and area A, the electric field (E) can be calculated using the formula \( \frac{Q}{A \epsilon_{0}} \), where \( \epsilon_{0}\) represents the permittivity of free space.

In simpler terms, think of the electric field as an invisible pressure pushing charged particles around. This field is evenly distributed between the plates, assuming a vacuum or air between them, and its strength depends on the amount of charge and the size of the plates. The electric field concept is essential for understanding how capacitors store and release energy.

Electric Force

The electric force is what actually moves charges in an electric field. It's the tug or push experienced by a charge placed in the field. Referring to our capacitor, the force (F) on one plate due to the other can be misleading if we consider the total field, because each plate contributes to the field and a plate can't influence itself. That's why we halve the influence, calculating the force as \( \frac{E Q}{2} \).

Here's a simpler picture: imagine two people throwing a ball to one another. The ball is the charge (Q), and the effort in the throw is akin to the electric force. Now, if one person suddenly disappeared, the ball wouldn't get thrown back—that's similar to why we only count the force from one plate.

Capacitance

Capacitance is a measure of how much electric charge a capacitor can store per unit of electric potential difference between its plates. The formula for the capacitance (C) of a parallel-plate capacitor is \( \frac{ \epsilon_{0} A}{x} \), with x being the separation between the plates. Think of capacitance as a container's ability to hold water: a bigger container (capacitor with larger plates or smaller separation) can hold more water (charge) before spilling (reaching its voltage limit).

Exercise Improvement

When solving for capacitance in exercises, visualize it as a capacity indicator—it's a fundamental property that dictates how much charge can be held at a certain voltage. The relationship between capacitance, plate area, and separation is directly related to how capacitors are used in circuits to store and release energy.

Permittivity of Free Space

Permittivity of free space (\( \epsilon_{0} \)) is a constant value that denotes how much resistance is encountered when forming an electric field in a vacuum. It's a fundamental concept in electromagnetism, setting the scale for electric charge and force calculations in classical physics.

Think of it as the quality of the 'fabric' of space. Just like certain materials might be more difficult to push a hand through, permittivity represents how 'thick' or 'thin' free space is to the formation of electric fields. When students encounter \( \epsilon_{0} \), it serves as a reminder that even in empty space, there's a fundamental property affecting electric phenomena.

Work Done in Physics

In physics, work done is the energy transferred when a force causes an object to move. It's calculated as the integral of force over the distance through which it acts. For our capacitor, separating the charged plates requires work, which is done against the electric force, and can be expressed as \( \int F dx \).

In daily life, work can be seen when lifting objects against gravity—lifting a book onto a shelf requires work. Similarly, separating the plates of a capacitor, against the pull of the electric force, requires work. Understanding this concept helps explain energy conservation and transfer within electric systems, such as the energy stored in a capacitor being used to power a device.