Two charges, +q and -q, occupy two corners of an equilateral...

Two charges, $+q$ and $-q,$ occupy two corners of an equilateral triangle, as shown in FIGURE $19-51 .$ (a) If $q=1.8 \mu \mathrm{C}$ and $r=0.50 \mathrm{m}$ find the magnitude and direction of the electric field at point $A$ the other vertex of the triangle. Let the direction angle be measured counterclockwise from the positive $x$ axis. (b) What is the total electric flux through the surface indicated in the figure? (c) Explain why Gauss's law cannot easily be used to find the magnitude of the electric field at point A.

Two charges, $+q$ and $-q,$ occupy two corners of an equilateral
triangle, as shown in FIGURE $19-51 .$ (a) If $q=1.8 \mu \mathrm{C}$ and $r=0.50 \mathrm{m}$
find the magnitude and direction of the electric field at point $A$
the other vertex of the triangle. Let the direction angle be measured counterclockwise from the positive $x$ axis. (b) What is the
total electric flux through the surface indicated in the figure?
(c) Explain why Gauss's law cannot easily be used to find the magnitude of the electric field at point A.

Key Concepts

Coulomb's Law

Coulomb's law describes the interaction between two point charges, stating that the magnitude of the force (and hence the electric field produced by a charge) is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. This fundamental law underpins the calculation of the electric field due to a single charge and forms the basis for analyzing more complex systems through its quantitative relationship.

Superposition Principle for Electric Fields

The superposition principle states that the total electric field created by multiple charges is the vector sum of the fields produced by each individual charge. This principle is critical when dealing with systems involving more than one charge because it allows the separate contributions from each charge to be calculated independently and then combined, taking into account both magnitude and direction, to determine the overall field at any point.

Electric Flux and Gauss's Law

Electric flux represents the number of electric field lines passing through a given surface, and Gauss's law relates this flux to the charge enclosed by that surface. Specifically, Gauss's law states that the net flux through any closed surface is equal to the total charge enclosed divided by the permittivity of free space. This concept is useful for determining the net charge within a region and for solving problems with high symmetry where the field distribution can be easily integrated over the surface.

Limitations of Gauss's Law in Non-Symmetric Configurations

Gauss's law, while always true, is most useful for calculating electric fields when there is a high degree of symmetry (such as spherical, cylindrical, or planar symmetry) that simplifies the flux integration. In configurations lacking such symmetry, like those with irregular geometric arrangements of multiple charges, it becomes difficult to exploit Gauss's law to directly calculate the electric field at a specific point because the electric field may vary in both magnitude and direction over the surface in a non-trivial way.

Transcript

00:01 Question 93, states that two charges, positive q and negative q, occupy two corners of an equilateral triangle as shown in the figure here.

00:10 Note that the big circle is a gaussian surface.

00:13 So i'm going to label it here.

00:15 This circle, which assume it's a perfect circle because i can not draw very well.

00:21 Part a, the question asks if q equals 1 .8 microculems and r equals 0 .5 meters, find the magnitude and the direction of the electric field at point a.

00:30 Let the direction be measured counterclockwise from the positive x -axis.

00:37 So the scenario here, so the electric field we're looking for at point a, point a should say, well, at point a in part a.

00:47 We say electric field at a is due both to the electric field contributions from positive q and the electric field contributions from negative q.

01:01 So we can look at for each of these values.

01:04 So we have e times, sorry, the illustration field is k times q over r squared.

01:09 And we note that for the positive branch, the positive charges say the electric field line points towards it.

01:16 Whereas the electric field line from b points away from it.

01:20 Sorry, the negative charge points away from it.

01:26 So if you know it because both the magnitude to the charges are equivalent, the magnitude of the charges are equivalent, so the magnitude of the charges are equivalent.

01:34 So the electric field contributions, at least in terms of magnitude, will be equivalent...

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