Use Gauss' Law to find the magnitude of E in a sphere

In summary: You need to find an expression for the electric field inside a sphere of radius r, r \leq R_0. There are a few equations you can use, but the one that comes up most often is Gauss's law.
  • #1
Bluewrath
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0
Problem
Problem 27.56

A sphere of radius R_0 has total charge Q. The volume charge density (C\m^3) within the sphere is [tex]\rho[/tex](r) = C/(r^2), where C is a constant to be determined.

Part A
The charge within a small volume dV is dq = [tex]\rho[/tex]dV. The integral of [tex]\rho[/tex]dV over the entire volume of the sphere is the total charge Q. Use this fact to determine the constant C in terms of Q and R_0.
Express your answer in terms of pi, Q, and R_0.

Part B
Use Gauss's law to find an expression for the magnitude of the electric field E inside the sphere, r [tex]\leq[/tex] R_0.
Express your answer in terms of Q, R_0, r and appropriate constants.

Relevant equations and whatnot :-)
Gauss' Law for a sphere: 4pi(r^2)E = Q_in/[tex]\epsilon[/tex]_0
Q = [tex]\rho[/tex] * V

Solution attempt(s) (and the miserable failure(s))
I've already found Part A, through [tex]\rho[/tex] = Q/v; the answer is C = Q/(4*pi*R_0).

For Part B, I've tried the following:

Use Gauss' Law to set up:

E = Q_in/(A*[tex]\epsilon[/tex]_0)

Define Q_in as:

Q_in / Q = (4/3)*pi*r^3 / (4/3)*pi*((R_0)^3) ==> Q_in = Q*r^3/((R_0)^3)

Define A as:

A = 4*pi*r^2

and plug everything into get:

E = Qr / ( (R_0)^3) * 4 * pi * [tex]\epsilon[/tex]_0 )

Which is apparently wrong :-(

So, I figured that they probably were intending to make the students use the solution to Part A, in which we found the C constant of [tex]\rho[/tex](r) = C/(r^2).

So, using the same setup and the same A and the same [tex]\epsilon[/tex]_0 (like that's going to change :-), I defined Q_in as:

Q_in = [tex]\rho[/tex](r) * V(r) = (C/r^2) * ((4/3)*pi*r^3) = (Qr/3*(R_0))

Again plugging in all the variables, I got stuck with:

E = Q / ( 3 * 4 * pi * r * (R_0) * [tex]\epsilon[/tex]_0 )

And that too is also wrong, apparently (unless there's a huge conspiracy against me involving Dr. Randall Knight and quite possibly the US government whom are intent on sabotaging my efforts). And now, I'm lost and have no idea what to do. There's a question in the book that asks for the electric field inside a uniformly charged sphere, and the answer is E = Qr / ( (R_0)^3) * 4 * pi * [tex]\epsilon[/tex]_0 ), which is what I got at my first try. Am I missing something here? Helpy please! :-D
 
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  • #2
The problem is that this is not a uniformly charged sphere. To find the charge enclosed within a Gaussian sphere of radius r, set up the integral of [itex]\rho dV[/itex]. (The integral is easy to solve.)
 
  • #3
Whats the answer for the second part then?
 

FAQ: Use Gauss' Law to find the magnitude of E in a sphere

How does Gauss' Law help to find the magnitude of E in a sphere?

According to Gauss' Law, the electric flux through a closed surface is equal to the charge enclosed by that surface divided by the permittivity of free space. By using this law, we can calculate the electric field at any point on the surface of a sphere by considering the charge enclosed within the sphere.

What is the formula for calculating the magnitude of electric field using Gauss' Law?

The formula for calculating the magnitude of electric field using Gauss' Law is E = Q/ε0A, where E is the electric field, Q is the charge enclosed by the surface, and ε0 is the permittivity of free space. A represents the area of the surface on which the electric field is being calculated.

Can Gauss' Law be used to find the electric field at any point on a sphere?

Yes, Gauss' Law can be used to find the electric field at any point on the surface of a sphere. However, it is important to note that the charge enclosed by the surface must be symmetrically distributed.

What is the relationship between the charge enclosed and the electric field using Gauss' Law?

The electric field at any point on a closed surface is directly proportional to the charge enclosed by that surface. This means that as the charge enclosed increases, the electric field also increases.

Can Gauss' Law be used to find the electric field inside a sphere?

Yes, Gauss' Law can be used to find the electric field inside a sphere if the charge distribution is spherically symmetric. In this case, the charge enclosed by the surface will also be spherically symmetric, making the calculation simpler.

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