Name: Gauss's Law: Problem-Solving
Uploaded: 2023-09-18T17:11:16
Description: Learn by watching this video about Gauss's Law: Problem-Solving at JoVE.com

Consider a Gaussian surface enclosing 30 electrons inside it. What would be the value of electric flux through the surface and the electric field at a distance of 0.6 meters from the center to its surface?
Recall the expression for Gauss&#39;s law: by multiplying the known values of the charge of an electron with the total electrons present inside it, the total charge can be obtained.
By substituting the known quantities&#8212;the total charge and the permittivity of free space&#8212;into the Gauss&#39;s law expression, the electric flux can be calculated.
To find the electric field on the surface, remember that flux is the product of the electric field times the area of the surface. Rearranging the expression and writing the area in terms of the radius, the electric field can be calculated.
Suppose there are additional charges in and around the Gaussian surface; then, the total flux through the surface can be obtained by summing up only those charges that are enclosed inside the surface divided by the permittivity of free space.

<ol>
	<li>OpenStax. (2019). University Physics Vol. 2. [Web version], pp 251 - 253. Retrieved from <a href="https://openstax.org/books/university-physics-volume-2@8ede2ba/pages/6-3-applying-gausss-law">https://openstax.org/books/university-physics-volume-2@8ede2ba/pages/6-3-applying-gausss-law</a></li>
</ol>

gauss's law: problem-solving

Gauss&#39;s law helps determine electric fields even though the law is not directly about electric fields but electric flux. In situations with certain symmetries (spherical, cylindrical, or planar) in the charge distribution, the electric field can be deduced based on the knowledge of the electric flux. In these systems, we can find a Gaussian surface S over which the electric field has a constant magnitude. Furthermore, suppose the electric field is parallel (or antiparallel) to the area vector everywhere on the surface. In that case, the flux integral transforms into the product of the electric field magnitude and an appropriate area. Thus, the equation representing Gauss&#39;s law simplifies to the following:
<img alt="Equation1" src="/files/ftp_upload/13717/13717_Equation_1.png" style="text-align: center; height: 40px;" />
When this flux is used in the expression for Gauss&#39;s law, an algebraic equation is obtained, which can be solved to find the magnitude of the electric field.
To summarize, when applying Gauss&#39;s law to solve a problem, the following steps are followed:
<ol>
	<li>Identify the spatial symmetry of the charge distribution. This is an important first step that allows the choice of the appropriate Gaussian surface. For example, an isolated point charge has spherical symmetry, whereas an infinite line of charge has cylindrical symmetry.</li>
	<li>Choose a Gaussian surface with the same symmetry as the charge distribution, and identify its consequences. With this choice, the electric flux can be easily determined over the Gaussian surface.</li>
	<li>Evaluate the flux through the surface. The symmetry of the Gaussian surface allows for factoring the electric field outside the integral.</li>
	<li>Determine the amount of charge enclosed by the Gaussian surface. This is an evaluation of the right-hand side of the equation representing Gauss&#39;s law. It is often necessary to perform an integration to obtain the net enclosed charge.</li>
	<li>Evaluate the electric field of the charge distribution.</li>
</ol>

Learn by watching this video about Gauss's Law: Problem-Solving at JoVE.com

Gauss's Law: Problem-Solving

Gauss&#39;s law helps determine electric fields even though the law is not directly about electric fields but electric flux. In situations with certain ...