Calculating Charge Density and Total Charge Using Gauss's Law

In summary, the conversation discusses finding the charge density and total charge contained in a sphere of radius R with an electric field given in spherical coordinates. The process involves using the formula for the divergence of a vector in spherical basis, and integrating over the volume of the sphere using the integral form of Gauss's law. Both methods yield the same result, 4\pi \epsilon_{0} R^{5}, for the total charge.
  • #1
spaghetti3451
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Homework Statement



Suppose the electric field in some region is found to be [itex]\vec{E} = kr^{3} \hat{r}[/itex], in spherical coordinates (k is some constant).

(a) Find the charge density ρ.

(b) Find the total charge contained in a sphere of radius R, centered at the origin. (Do it two different ways.)

Homework Equations



The Attempt at a Solution



(a) Use the formula for the divergence of a vector in spherical basis to get [itex]\nabla . \vec{E} = 5kr^{2}[/itex] so that [itex]\rho = 5k\epsilon_{0}r^{2}[/itex].

(b) Q is the volume integral of [itex]\rho[/itex] over the volume of the sphere. So, we integrate over d[itex]\phi[/itex], integrate sinθ over dθ, integrate the [itex]\rho[/itex] times [itex]r^{2}[/itex] over dr and multiply the three results. The process gives us 4[itex]\pi \epsilon_{0} R^{5}[/itex].

Q can also be found using the integral form of Gauss's law, where the surface integral of the electric field is taken with the infinitesimal area, which is R2 sinθ dθ d∅ r, where r is the unit vector in the radial direction. We take the constants out of the integral and integrate 1 over phi and sinθ over theta to obtain the same Q as above.

Please could you check if the process and the answers are correct?
 
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  • #2
a) Looks good to me. I get the same.
b) Is it a hollow spherical shell. Then you can use that the electric field is constant over the surface(fixed r) and only multiply by the area of the sphere(shell) because:
[itex] \oint \vec{E}\cdot\vec{da} = \oint E\,\text{d}a = E\oint 1\text{d}a = E 4\pi R^2 = Q/\epsilon_0 \Leftrightarrow Q = 4\pi \epsilon_0 kR^5[/itex]
The dot product [itex]\vec{E}\cdot\vec{da}[/itex] is just da times E cause they point in the same direction.
 

Related to Calculating Charge Density and Total Charge Using Gauss's Law

What is Gauss's law in action?

Gauss's law in action is a fundamental law in electromagnetism that relates the electric flux through a closed surface to the charge enclosed by that surface. It is represented by the equation ∫SE · dA = Qenc0, where E is the electric field, S is the closed surface, Qenc is the enclosed charge, and ε0 is the permittivity of free space.

How is Gauss's law applied in real-life situations?

Gauss's law can be applied in various real-life situations, such as calculating the electric field inside a capacitor, determining the electric field produced by a point charge, or finding the electric field near a charged conducting surface. It is also used in the design of electrical circuits and devices.

What is the significance of Gauss's law in physics?

Gauss's law is one of the four Maxwell's equations that describe the behavior of electric and magnetic fields, and it is an essential tool for understanding the principles of electromagnetism. It has a wide range of applications in physics, engineering, and other fields of science.

How does Gauss's law relate to Coulomb's law?

Coulomb's law and Gauss's law are closely related, as they both describe the behavior of electric charges. Coulomb's law gives the magnitude and direction of the electric force between two point charges, while Gauss's law relates the electric flux through a closed surface to the charge enclosed by that surface. In some cases, Gauss's law can be used to derive Coulomb's law.

Can Gauss's law be applied to non-uniform electric fields?

Yes, Gauss's law can be applied to non-uniform electric fields as long as the enclosed charge is known. In such cases, the electric field at a point is calculated by considering an imaginary Gaussian surface that encloses the point and integrating the electric field over that surface. This allows for the determination of the electric field at any point in a non-uniform field.

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