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Chapter 3: Problem 70

Force and energy for two plates Calculate the electrical force that acts on one plate of a parallelplate capacitor. The potential difference between the plates is 10 volts, and the plates are squares \(20 \mathrm{~cm}\) on a side with a separation of \(3 \mathrm{~cm}\). If the plates are insulated so the charge cannot change, how much external work could be done by letting the plates come together? Does this equal the energy that was initially stored in the electric field?

Short Answer

Expert verified
The electrical force that acts on one plate of the capacitor is calculated as \( F \). The external work done to let the plates come together is \( W \) and the initial energy stored in the electric field is \( U \). If the external work done and the initial energy are equal, it verifies the energy conservation law.

Step by step solution

01

Calculation of charge in the plates

First, the surface charge density has to be calculated. This can be calculated using the formula \( \sigma = \varepsilon_0 * E \) where \( E = \frac{V}{d} \) is the electric field, \( V \) the voltage, \( d \) the separation, and \( \varepsilon_0 \) the permittivity of free space. Then, the total charge can be obtained by multiplying the surface charge density by the area of the plate, i.e., \( q = \sigma * A \) where \( A \) is the area of the plate.
02

Calculation of electrical force

The electrical force can be calculated using Coulomb's law, \( F = \frac{1}{4蟺\varepsilon_0} \frac{q^2}{d^2} \)
03

Calculation of external work

The external work that can be done by letting the plates come together can be calculated as the product of force and distance covered, i.e., \( W = F * d \)
04

Calculation of initial energy stored

The initial energy that was stored in the electric field can be calculated using the formula \( U = \frac{1}{2} CV^2 \) where \( C = \frac{q}{V} \) is the capacitance of the capacitor.
05

Compare the external work and initial energy

Finally, compare the calculated work done and the initial energy to verify if they are equal.

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Key Concepts

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

Parallel Plate Capacitor
A parallel plate capacitor is a simple yet fundamental component in the field of electronics. It consists of two conductive plates placed parallel to one another, separated by a small distance.
The plates are usually square or rectangular, and the material between the plates can be air or another type of insulator known as a dielectric.

The basic function of a parallel plate capacitor is to store electrical energy. By doing this, it temporarily holds electric charge and prevents rapid changes in voltage.
  • It stores energy in the form of an electrostatic field created by applying a voltage across the plates.
  • The area of the plates and the separation distance between them greatly influence the capacitance or storage capability of the capacitor.
Parallel plate capacitors are widely used in electronic devices for filtering, tuning circuits, and energy storage applications.
Electric Field
The electric field is a vector field surrounding an electric charge, exerting force on other charges in its vicinity. In a parallel plate capacitor, the electric field is uniform and is directed from the positive to the negative plate.

The electric field (E) in such a setup can be calculated using the formula:
\[ E = \frac{V}{d} \]
where \(V\) is the voltage across the plates and \(d\) is the separation distance.
This ensures the field is consistent between the plates.
  • The strength of the electric field determines the force exerted on charges situated within the field.
  • A stronger electric field means more force on the charges.
Understanding electric fields is crucial when dealing with capacitors, as they determine how the capacitor will store and discharge energy in a circuit.
Capacitance
Capacitance is a measure of a capacitor's ability to store charge. It is a crucial characteristic defining the capacity of the capacitor.

The capacitance \(C\) of a parallel plate capacitor is determined using the formula:
\[ C = \frac{\varepsilon_0 A}{d} \]
where \(\varepsilon_0\) is the permittivity of free space, \(A\) is the area of one of the plates, and \(d\) is the separation between the plates.
  • Larger plate areas increase capacitance, as there is more surface to store charge.
  • Reducing the distance between the plates also increases capacitance, enhancing charge storage ability.
Capacitance plays a fundamental role in the energy storage and management capabilities of capacitors, influencing how they are used in various applications.
Coulomb's Law
Coulomb's Law is a principle describing the electrostatic interaction between charged objects. It states that the force exerted by one charge on another is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

In the context of a parallel plate capacitor, the electrical force \( F \) between the plates is given by:
\[ F = \frac{1}{4\pi \varepsilon_0} \frac{q^2}{d^2} \]
where \( q \) is the charge on one of the plates and \( d \) is the separation distance.
  • This law helps in determining how much force the electric charge exerts, which is also critical in calculating work and understanding energy storage in capacitors.
  • The force calculated is used in the analysis of capacitors to explore their capabilities and limits under different conditions.
Coulomb's Law serves as a cornerstone in understanding forces in electrical circuits and the behavior of capacitors within them.

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Most popular questions from this chapter

Electron beam A beam of \(9.5\) megaelectron-volt (MeV) electrons \((\gamma \approx 20)\), amounting as current to \(0.05 \mu \mathrm{A}\), is traveling through vacuum. The transverse dimensions of the beam are less than \(1 \mathrm{~mm}\), and there are no positive charges in or near it. (a) In the lab frame, what is the average distance between an, electron and the next one ahead of it, measured parallel to the beam? What approximately is the average electric field strength 1 cm away from the beam? (b) Answer the same questions for the electron rest frame.

In the laboratory frame, a proton is at rest at the origin at \(t=0 .\) At that instant a negative pion (charge \(-e\) ) that has been traveling in along the \(x\) axis at a speed of \(0.6 c\) reaches the point \(x=0.01 \mathrm{~cm}\). There are no other charges around. What is the magnitude of the force on the pion? What is the magnitude of the force on the proton? What about Newton's third law? (We're getting a little ahead of ourselves with this last question, but see if you can answer it anyway.)

Van de Graaff current ? In a Van de Graaff electrostatic generator, a rubberized belt \(0.3 \mathrm{~m}\) wide travels at a velocity of \(20 \mathrm{~m} / \mathrm{s}\). The belt is given a surface charge at the lower roller, the surface charge density being high enough to cause a field of \(10^{6} \mathrm{~V} / \mathrm{m}\) on each side of the belt. What is the current in milliamps?

Capacitance of raindrops \(N\) charged raindrops with radius \(a\) all have the same potential. Assume that they are far enough apart so that the charge distribution on each isn't affected by the others (that is, it is spherically symmetric). What is the total capacitance of this system? How does this capacitance compare with the capacitance in the case where the drops are combined into one big drop?

Tapered rod Two graphite rods are of equal length. One is a cylinder of radius \(a\). The other is conical, tapering (or widening) linearly from radius \(a\) at one end to radius \(b\) at the other. Show that the end-to-end electrical resistance of the conical rod is \(a / b\) times that of the cylindrical rod. Hint: Consider the rod to be made up of thin, disk-like slices, all in series. (This result is actually only an approximate one, valid in the limit where the taper is slow. See Problem \(4.6\) for a discussion of this.)
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