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Chapter 6: Problem 10

1Discuss whether Gauss's law can be applied to other forces, and if so, which ones.

Short Answer

Expert verified
Gauss's law can be applied to gravitational forces, as they follow an inverse-square law and have a conservative field. To apply Gauss's law to gravitational forces, the electric field is replaced with the gravitational field, and the enclosed charge is replaced with the enclosed mass. The law is derived from the Coulomb force, which also satisfies the necessary properties. However, magnetic forces do not satisfy the requirements for Gauss's law, but they have a similar law called Gauss's law for magnetism which states that the net magnetic flux through any closed surface is zero.

Step by step solution

01

Understanding Gauss's Law

Gauss's law is given by the equation: \(\Phi_E = \oint \vec{E} \cdot d\vec{A} = \frac{Q_{enclosed}}{\epsilon_0}\) where \(\Phi_E\) is the electric flux through the closed surface, \(\vec{E}\) is the electric field, \(d\vec{A}\) is the differential area vector, \(Q_{enclosed}\) is the charge enclosed by the Gaussian surface, and \(\epsilon_0\) is the electric constant. Gauss's law relies on the following properties: 1. The force must have a conservative field, meaning the work done by the force is path-independent. 2. The force must exhibit an inverse-square law dependence on distance, which is typical of electric and gravitational forces.
02

Identifying Other Forces

Some forces that could potentially satisfy Gauss's law are: 1. Gravitational force: The force between two objects due to their masses 2. Magnetic force: The force between magnetic poles 3. Coulomb force: The force between charged particles
03

Analyzing Applicability of Gauss's Law to Other Forces

We will now analyze each of these forces to determine if Gauss's law can be applied to them. 1. Gravitational Force: The gravitational force follows an inverse-square law and has a conservative field. Therefore, Gauss's law can be applied to the gravitational force. When using Gauss's law for gravitational forces, we replace the electric field with the gravitational field and the enclosed charge with the enclosed mass. 2. Magnetic Force: Magnetic forces do not follow an inverse-square law, and the magnetic field is not conservative. Thus, Gauss's law cannot be applied to magnetic forces. However, an analogous law can be derived for magnetic fields: Gauss's law for magnetism states that the net magnetic flux through any closed surface is zero. 3. Coulomb Force: The Coulomb force is essentially the force between point charges, and it follows an inverse-square law and has a conservative field. In fact, Gauss's law is derived from the Coulomb force for electric charges. In conclusion, Gauss's law can be applied to other forces such as gravitational forces, and it is derived from the Coulomb force. Magnetic forces do not strictly satisfy the requirements for Gauss's law but have a similar law, called Gauss's law for magnetism.

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Key Concepts

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

Electric Flux
Electric flux is a measure of how the electric field passes through a given surface. You can imagine electric flux as the amount of electric field lines penetrating a surface. It's much like how the breeze flows through an open window in a room.

Specific factors influence the electric flux:
  • The strength of the electric field (\(\vec{E}\)).
  • The area of the surface (\(d\vec{A}\)).
  • The angle between the field lines and the surface.
Gauss's Law uses the concept of electric flux to relate the enclosed charge within a surface to the electric field. It is an integral law which expresses this as: \[ \Phi_E = \oint \vec{E} \cdot d\vec{A} = \frac{Q_{enclosed}}{\epsilon_0} \] This equation helps simplify complex problems, particularly in cases where the symmetry of the charge distribution makes the integral easy to evaluate.

Understanding electric flux is foundational when working with electric fields and charges, as it helps predict how electric fields will interact with different materials and environments.
Inverse-Square Law
The inverse-square law is a pivotal principle in physics, particularly when discussing forces like gravity and electricity. This law states that the strength of these forces is inversely proportional to the square of the distance from the source. In simple terms, as you move away from the source, the strength of the force decreases rapidly. The mathematical representation of this is:\[ F \propto \frac{1}{r^2} \] where \( F \) denotes force and \( r \) is the distance from the source.

Some force fields that naturally follow this law include:
  • Gravitational field: Governing the attraction between masses.
  • Electric field: Dictating the force between electric charges, especially as described in Coulomb's law.
This principle is a cornerstone for applying Gauss's Law since Gauss's Law uses surfaces where such inverse-square dependencies make the calculations predictable and solvable. Only forces that follow this law simplify the mathematical complexity of estimating fields over large symmetrical surfaces.
Conservative Field
Conservative fields are a key concept in understanding forces like gravity and electricity. The main property of a conservative field is that the work done by the force in moving a particle between two points is independent of the path taken; it depends only on the position of the start and end points. This property implies energy conservation within the system.Mathematically, if a field is conservative, the following line integral over a closed path is zero:\[ \oint \vec{F} \cdot d\vec{r} = 0 \] where \( \vec{F} \) denotes force along the path \( d\vec{r} \).

Some important aspects of conservative fields include:
  • Potential energy can be defined within the field, which helps calculate the work done without needing to know the path.
  • This helps simplify many practical problems as it allows the use of scalar potentials, rather than having to integrate vector fields.
  • Electric and gravitational forces are perfect examples of conservative fields.
Gauss's Law applies to fields that are conservative, as it allows the enclosed charge or mass to be calculated easily using field lines that don't depend on the path. This makes problem-solving with Gauss's Law both intuitive and manageable, especially for symmetric cases.

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Most popular questions from this chapter

A circular area \(S\) is concentric with the origin, has radius \(a,\) and lies in the \(y\) z-plane. Calculate \(\int_{S} \overrightarrow{\mathbf{E}} \cdot \hat{\mathbf{n}} d A\) for \(\overrightarrow{\mathbf{E}}=3 z^{2} \hat{\mathbf{i}}\).

Two equal and opposite charges of magnitude \(Q\) are located on the \(x\) -axis at the points \(+a\) and \(-a,\) as shown below. What is the net flux due to these charges through a square surface of side \(2 a\) that lies in the \(y\) z-plane and is centered at the origin? (Hint: Determine the flux due to each charge separately, then use the principle of superposition. You may be able to make a symmetry argument.)

Two large rectangular aluminum plates of area \(150 \mathrm{cm}^{2}\) face each other with a separation of \(3 \mathrm{mm}\) between them. The plates are charged with equal amount of opposite charges, \(\pm 20 \mu \mathrm{C}\). The charges on the plates face each other. Find the flux through a circle of radius \(3 \mathrm{cm}\) between the plates when the normal to the circle makes an angle of \(5^{\circ}\) with a line perpendicular to the plates. Note that this angle can also be given as \(180^{\circ}+5^{\circ}\).

Is the term \(\overrightarrow{\mathbf{E}}\) in Gauss's law the electric field produced by just the charge inside the Gaussian surface?

A total charge \(Q\) is distributed uniformly throughout a spherical shell of inner and outer radii \(r_{1}\) and \(r_{2}\) respectively. Show that the electric field due to the charge is $$\begin{array}{ll} \overrightarrow{\mathbf{E}}=\overrightarrow{0} & \left(r \leq r_{1}\right) \\ \overrightarrow{\mathbf{E}} & =\frac{Q}{4 \pi \varepsilon_{0} r^{2}}\left(\frac{r^{3}-r_{1}^{3}}{r_{2}^{3}-r_{1}^{3}}\right) \hat{\mathbf{r}} \\ \overrightarrow{\mathbf{E}} & =\frac{Q}{4 \pi \varepsilon_{0} r^{2}} \hat{\mathbf{r}} \end{array}$$
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