Gauss' Law, Spherical Charge Distribution

In summary: Good luck! In summary, the problem involves a spherically symmetric charge distribution surrounded by a conducting shell. The electric field is found using Gauss's Law and depends on the distance from the center of the distribution. The correct answers are obtained by correctly applying Gauss's Law and taking into account the charge enclosed within a certain radius.
  • #1
Fronzbot
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Before I get into the question I'd just like to state that this is not homework, but questions in my book that I'm going through to prepare myself for the midterm in one week. I got stuck at a few questions, here's the first one. I won't ask the next until I'm done with this and so forth. Normally I'd go to my professor for help but I can literally not understand him when he speaks. It's awful. Luckily he's a lenient grader (at least compared to other EM Fields profs), but yeah. Ok, onto the question then...

Homework Statement


A spherical distribution of charge [tex]\rho = \rho_{0}[1-(R^{2}/b^{2}][/tex] exists in the region [tex]0\leqR\leqb[/tex]. This charge distribution is concentrically surrounded by a conducting shell with inner radius [tex]R_{i} (>b)[/tex] and outer radius [tex]R_{o}[/tex]. Determine E everywhere.

Homework Equations


[tex]
\int E \cdot ds = Q_{enc} / \epsilon_{0}
[/tex]


The Attempt at a Solution


I have the right answers (from the back of the book) but cannot figure out how to get there. Before I write my solution down, here are the correct answers:
[tex]
For 0 \leq R \leq b:[/tex]
[tex]
E_{R1} = \frac{\rho_{0}R}{\epsilon_{0}}(\frac{1}{3}-\frac{R^{2}}{5b^{2}})
[/tex]

[tex]
For b \leq R < R_{i}:
[/tex]
[tex]
E_{R2} = \frac{2 \rho_{0} b^{3}}{15 \epsilon_{0} R^{2}}
[/tex]

[tex]
For R_{i}<R<R_{o}
[/tex]
[tex]
E_{R3} = 0 [/tex]

[tex]
For R > R_{o}:
[/tex]
[tex]
E_{R4} = \frac{2 \rho_{0} b^{3}}{15 \epsilon_{0}R^{2}}
[/tex]

Now I didn't get very far, but here's what I have:
[tex]
Q_{enc} = \int \rho dV = \rho_{0}[1 - \frac{R^2}{b^2}]\frac{4\pi R^{3}}{3}
[/tex]
[tex]
\int E \cdot ds = E_{R1} 4 \pi R^{2}[/tex]
[tex]
E_{R1} = \frac{\rho [1-\frac{R^2}{b^2}]R}{3\epsilon_{0}}[/tex]
[tex]
E_{R1} = \frac{\rho R}{\epsilon}(\frac{1}{3}-\frac{R^2}{3b^2})

[/tex]

I have some work for the next two, but I'd rather go one step at a time here to make sure I completely what's going on. At least this answer is close, but I'm not sure where the book's 5b^2 came from.
 
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  • #2


Dear student,

I understand that you are preparing for your midterm and have encountered some difficulties with a question from your textbook. I appreciate your effort in trying to solve the problem on your own before seeking help.

Firstly, it is important to understand the concept of spherical symmetry in this problem. Since the charge distribution is spherically symmetric, the electric field will also have spherical symmetry. This means that the electric field will only depend on the distance from the center of the distribution, and not on the direction.

Next, let's take a closer look at the charge distribution. The charge density, \rho, is given by \rho_{0}[1-(R^{2}/b^{2}] and is only dependent on the distance from the center, R. This means that the charge enclosed within a radius R is given by Q_{enc} = \rho_{0}[1-(R^{2}/b^{2})] \frac{4\pi R^{3}}{3}. This is the expression you have used in your attempt to find the electric field.

Now, let's move on to the correct answers. For 0 \leq R \leq b, the electric field is given by E_{R1} = \frac{\rho_{0}R}{\epsilon_{0}}(\frac{1}{3}-\frac{R^{2}}{5b^{2}}). To get this answer, we need to use Gauss's Law, which states that \int E \cdot ds = Q_{enc}/\epsilon_{0}. In this case, the charge enclosed is given by Q_{enc} = \rho_{0}[1-(R^{2}/b^{2})] \frac{4\pi R^{3}}{3}. Plugging this into Gauss's Law and solving for E_{R1}, we get the correct answer.

Similarly, for b \leq R < R_{i}, we use Gauss's Law again, but this time the charge enclosed is given by Q_{enc} = \rho_{0}b^{3} \frac{4\pi}{3}. Plugging this into Gauss's Law and solving for E_{R2}, we get the correct answer of E_{R2} = \frac{2\rho_{0}b^{3}}{15\epsilon_{0}R^{2}}.

I hope this explanation helps you understand the problem better and prepares
 

FAQ: Gauss' Law, Spherical Charge Distribution

1. What is Gauss' Law and how does it relate to spherical charge distributions?

Gauss' Law is a fundamental law of electromagnetism that relates the electric flux through a closed surface to the charge enclosed by that surface. In the case of a spherical charge distribution, Gauss' Law states that the electric flux through a closed surface surrounding the distribution is equal to the charge enclosed divided by the permittivity of free space. This relationship holds true regardless of the distribution's shape or size, as long as it is spherically symmetric.

2. How do you calculate the electric field at a point outside of a spherical charge distribution?

To calculate the electric field at a point outside of a spherical charge distribution, you can use Gauss' Law in integral form. First, choose a Gaussian surface that encloses the charge distribution, such as a spherical surface. Then, use the formula E = Q/(4πε0r²) to calculate the electric field at the surface. Finally, use the inverse square law to determine the electric field at the desired point outside the surface.

3. What is the difference between a point charge and a spherical charge distribution?

A point charge is a theoretical concept that represents a single, infinitesimally small charge. On the other hand, a spherical charge distribution is a real-world object that consists of multiple point charges distributed on the surface or within a sphere. While both can be described using Gauss' Law, the calculations for a point charge are simpler since it is a single charge, while a spherical charge distribution requires integration over the entire distribution.

4. How does the electric field vary within a spherical charge distribution?

Within a spherical charge distribution, the electric field varies depending on the distance from the center of the distribution. At the center, the electric field is zero since all the charges are equidistant from the point. As you move towards the edge of the distribution, the electric field increases, reaching its maximum at the surface. Beyond the surface, the electric field follows the inverse square law and decreases with distance.

5. Can Gauss' Law be applied to non-spherical charge distributions?

Yes, Gauss' Law can be applied to non-spherical charge distributions as long as they have some form of symmetry. For example, if a charge distribution has cylindrical symmetry, a cylindrical Gaussian surface can be used to apply Gauss' Law. However, the calculations become more complex as the shape and symmetry of the distribution deviate from a spherical shape.

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