Determining Electric field of Spherical charge distribution

[JasonZ]

Hey, this is from "Foundations of Electromagnetic Theory" by Reitz, et al. Problem 2-15.

I have had a really hard time trying to learn from this book as there are no examples to apply the equations they prove throughout the chapters. Anyhow, I don't really have anything down for this problem, which is as follows:

A spherical charge distribution has a volume charge density that is a function only of r, the distance from the center of the distribution. In other words, \rho = \rho (r). If \rho (r) is as given below, determine the electric field as a function of r. Integrate the result to obtain an expression for the electrostatic potential \phi (r), subject to the restriction that \phi (\infty) = 0.

(a) \rho = \frac {A}{r} with A a constant for 0 \leq r \leq R; \rho = 0 for r > R.

I assume this is a Guass law problem, I just don't understand how to solve the right hand integral, supposing it is indeed: \int \rho dv

Can anyone help me, I am quite stuck.

-Jason

 

[member 553083]

Hi Jason, The electric field produced by the charge distribution can be found by using Gauss's Law. The electric field at a distance r from the center of the spherical charge distribution is given by: E = \frac{1}{4\pi\epsilon_0}\int \rho(r)dV = \frac{A}{4\pi\epsilon_0r^2} where A is the constant in the problem statement. To find the electrostatic potential, you will need to integrate this expression. Using the boundary condition that \phi(\infty)=0, you can find that the electrostatic potential is given by: \phi(r) = -\frac{A}{4\pi\epsilon_0}\frac{1}{r} Hope this helps!

 

[M98Ranger]

Hi Jason,

I understand your frustration with the lack of examples in the book. It can be difficult to apply equations without seeing them in action. Let me try to break down the problem for you and provide some guidance.

First, let's review some important equations for determining the electric field and potential of a spherical charge distribution.

1. Electric field due to a point charge: E = kQ/r^2, where k is the Coulomb constant, Q is the charge, and r is the distance from the charge.

2. Electric field due to a continuous charge distribution: E = k \int \frac{dQ}{r^2}, where dQ is a small element of charge and the integral is taken over the entire charge distribution.

3. Electric potential due to a point charge: \phi = kQ/r, where k is the Coulomb constant, Q is the charge, and r is the distance from the charge.

4. Electric potential due to a continuous charge distribution: \phi = k \int \frac{dQ}{r}, where dQ is a small element of charge and the integral is taken over the entire charge distribution.

Now, let's apply these equations to the problem at hand.

We are given that the volume charge density, \rho, is a function of r only. This means that the charge is evenly distributed throughout the spherical volume, with the density decreasing as the distance from the center increases. We are also given that the charge density is given by \rho = A/r, where A is a constant for 0 \leq r \leq R and 0 for r > R.

To determine the electric field as a function of r, we can use the equation for a continuous charge distribution, E = k \int \frac{dQ}{r^2}. We can rewrite this equation as E = k \int \rho dv, where dv is a small volume element. Since the charge density is a function of r only, we can rewrite this as E = k \int_0^r \frac{A}{r^2} (4\pi r^2 dr), where the integral is taken from 0 to r to account for the varying charge density.

Simplifying this equation, we get E = \frac{4\pi kA}{r}. This is the electric field as a function of r for 0 \leq r \leq R. For r