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The net electric flux crossing an open surface is never zero. True or false?

Short Answer

Expert verified
False. The net electric flux crossing an open surface can be zero under certain conditions, such as when the electric field lines are perpendicular to the surface or if there are equal amounts of electric field lines entering and leaving the open surface.

Step by step solution

01

Understanding Electric Flux

Electric flux, denoted by Φ, is a measure of the electric field passing through a given surface. It is defined as the dot product of the electric field (E) and the area vector (A): Φ = E • A. The area vector A is given by the product of the area of the surface and the unit vector perpendicular to the surface.
02

Statement of Gauss's law

Gauss's law states that the net electric flux through a closed surface is proportional to the net charge enclosed within that surface. Mathematically, it is represented as: \[ \oint_\textnormal{closed surface} E \cdot dA = \frac{Q_\textnormal{enclosed}}{\epsilon_0} \] where Q_enclosed is the total charge enclosed within the closed surface, dA is the infinitesimal area vector, ε₀ is the permittivity of free space, and the integral is taken over the entire closed surface.
03

Flux through an open surface

Now, consider an open surface instead of a closed one. In this case, we cannot apply Gauss's law directly, because it is only applicable to closed surfaces. However, we can still discuss the possibility of having zero or non-zero electric flux through the surface. A net zero electric flux through an open surface is possible when the electric field lines passing into or out of the open surface cancel each other out. For example, when an open surface is perpendicular to the electric field lines, or if there are equal amounts of electric field lines entering and leaving the open surface, the net electric flux will be zero.
04

Answer to the exercise

The statement "The net electric flux crossing an open surface is never zero" is False. It is possible that the net electric flux through an open surface can be zero under certain conditions, as explained in the previous step.

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Key Concepts

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

Gauss's Law
Gauss's Law is a fundamental principle of electromagnetism that relates the electric flux through a closed surface to the charge enclosed by that surface. This law highlights the symmetrical nature of electric fields and simplifies the calculation of electric fields when high symmetry is present. Gauss’s Law is mathematically expressed as: \[\oint_{\text{closed surface}} E \cdot dA = \frac{Q_{\text{enclosed}}}{\epsilon_0}\]
  • Here, \(E\) is the electric field vector.
  • \(dA\) is an infinitesimally small area on the closed surface.
  • \(Q_{\text{enclosed}}\) is the charge enclosed by the surface.
  • \(\epsilon_0\) is the permittivity of free space, a constant that characterizes the ability of a vacuum to permit electric field lines.
The integral sign with a circle, \(\oint\), indicates that the integration is over a closed surface.
One key aspect to remember is Gauss's Law applies only to closed surfaces. The symmetry of the surface and the distribution of charge are crucial for effectively using Gauss's Law.
Open Surface
An open surface, unlike a closed surface, does not completely enclose a volume. Think of an open surface like a piece of paper that's not sealed into a box. This concept is important to understand when discussing electric flux because the behavior of electric fields across these surfaces can be different. Electric flux through an open surface is computed similarly to a closed surface but does not have the same simplifications provided by Gauss's Law, as it doesn't enclose any volume completely.
For open surfaces, the flux can be zero, non-zero, or even change over time, depending on how the field lines interact with the surface.

When is Flux Zero?

The net electric flux through an open surface can become zero under certain conditions:
  • If the electric field lines entering and leaving the open surface are equal and opposite, they cancel each other out.
  • When the surface is perpendicular to the electric field lines, resulting in no net flow through it.
It’s essential to consider the exact orientation and position of the surface relative to the electric field to determine the actual flux.
Electric Field
An electric field is a vector field around charged particles and is responsible for the force experienced by other charges in the vicinity. It indicates the direction and magnitude of the force a positive test charge would experience at any given point in space. Understanding electric fields is crucial as they are the primary mediators of electric flux.The electric field \(E\) at a point is given by:\[E = \frac{F}{q}\]
  • \(F\) is the force experienced by the charge.
  • \(q\) is the magnitude of the charge experiencing the force.
The direction of the electric field is the direction of the force that a positive charge would feel. This is important when considering open surfaces and determining the resulting electric flux.
Electric field lines represent the field visually:
  • They start on positive charges and end on negative charges.
  • The density of these lines corresponds to the magnitude of the electric field.
This visualization helps in predicting how electric fields interact with both open and closed surfaces.

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Most popular questions from this chapter

The infinite slab between the planes defined by \(z=-a / 2\) and \(z=a / 2\) contains a uniform volume charge density \(\rho\) (see below). What is the electric field produced by this charge distribution, both inside and outside the distribution?

A charge \(q\) is placed in the cavity of a conductor as shown below. Will a charge outside the conductor experience an electric field due to the presence of \(q\) ?

The electric flux through a spherical surface is \(4.0 \times 10^{4} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C} .\) What is the net charge enclosed by the surface?

A spherical rubber balloon carries a total charge \(Q\) distributed uniformly over its surface. At \(t=0,\) the radius of the balloon is \(R\). The balloon is then slowly inflated until its radius reaches \(2 R\) at the time \(t_{0} .\) Determine the electric field due to this charge as a function of time (a) at the surface of the balloon, (b) at the surface of radius \(R,\) and \((\mathrm{c})\) at the surface of radius \(2 R\). Ignore any effect on the electric field due to the material of the balloon and assume that the radius increases uniformly with time.

Two non-conducting spheres of radii \(R_{1}\) and \(R_{2}\) are uniformly charged with charge densities \(\rho_{1}\) and \(\rho_{2}\) respectively. They are separated at center-to-center distance \(a\) (see below). Find the electric field at point \(P\) located at a distance \(r\) from the center of sphere 1 and is in the direction \(\theta\) from the line joining the two spheres assuming their charge densities are not affected by the presence of the other sphere. (Hint: Work one sphere at a time and use the superposition principle.)

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