Chapter 18: Problem 20
By Coulomb's law, if a point charge of \(q\) coulombs is located at the origin, then the force exerted on a unit charge at \((x, y, z)\) is given by \(\mathrm{F}(x, y, z)=\left(c q\|\mathbf{r}\|^{3}\right) \mathrm{r}\) where \(\mathbf{r}=x \mathbf{i}+y \mathbf{j}+z \mathbf{k}\). If \(S\) is any sphere with center at the origin, show that the flux of \(\mathbf{F}\) through \(S\) is \(4 \pi c q .\)
Short Answer
Step by step solution
Understanding the Problem
Calculating the Divergence of \( \mathbf{F} \)
Applying Gauss's Divergence Theorem
Considering the Point at the Origin
Conclusion
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Coulomb's Law
\[F = k \frac{|q_1 q_2|}{r^2}\]where:
- \(F\) is the magnitude of the force between the charges,
- \(k\) is Coulomb's constant,
- \(q_1\) and \(q_2\) are the magnitudes of the charges,
- \(r\) is the distance between the charges.
Understanding this law is a stepping stone to more complex electromagnetic concepts.
Vector Fields
- Each vector in the field has a direction and magnitude.
- The direction at any point often reflects the field's influence at that point.
- The magnitude can signify the strength or intensity of the field.
Divergence of a Vector Field
This operation involves partial derivatives and is used to describe sources or sinks within a vector field. In simple words:
- A positive divergence suggests that vectors are "spreading out" from a point (source).
- A negative divergence indicates vectors are "converging towards" a point (sink).
- A zero divergence means that there is no net flow out of or into any region (incompressible flow).
Flux Integral
\[\Phi = \iint_{S} \mathbf{F} \cdot \mathbf{n} \, dS\]where:
- \( \mathbf{F} \) is the vector field.
- \( \mathbf{n} \) is the unit normal vector to the surface.
- \( dS \) is the differential surface area element.
In the exercise, the flux through a sphere due to a point charge at the origin was calculated using this concept, leading to a result consistent with Coulomb’s law.