How Does Gauss's Law Apply to Nonuniform Spherical Charge Distributions?

  • Thread starter Thread starter kbwelch17
  • Start date Start date
  • Tags Tags
    Calculus Charge
Click For Summary
The discussion focuses on applying Gauss's Law to a nonuniform spherical charge distribution defined by a specific charge density. Participants are tasked with calculating the total charge, electric fields in various regions, and identifying the maximum electric field. The initial confusion revolves around setting up the integral for the charge density and understanding the appropriate areas for integration. Clarifications suggest using the volume element for charge density integration rather than Gauss's Law for the total charge calculation. The conversation emphasizes the need for precise definitions of variables and integration limits to solve the problem effectively.
kbwelch17
Messages
7
Reaction score
0

Homework Statement



A nonuniform, but spherically symmetric, distribution of charge has a charge density ρ(r) given as follows:
ρ(r) = ρ0(1-r/R) for r≤R
ρ(r) = 0 for r≥R

where ρ0 = 3Q/∏R^3 is a positive constant.

(a.) Show that the total charge contained in the charge distribution is Q.
(b.) Show that the electric field in the region r≥R is identical to that produced by a point charge Q at r = 0.
(c.) Obtain an expression for the electric field in the region r≤R.
(d) Graph the electric field magnitude E as a function of r.
(e.) Find the value of r at which the electric field is maximum, and find the value of that maximum field.

Homework Equations



Gauss' Law : Flux = ∫E dot dA = Qencl/ε 0
Electric Field of a point charge: E = k*(q/r^2)

The Attempt at a Solution



I am really just having trouble setting up the integral. You would be integrating the electric field over a sphere, so you would need to choose dA (is this a slice of the area of the sphere?) and a formula to find electric field for a given r. In order to find the latter you would need to use Gauss' Law in which case a give r would give the electric field

ρ(r) = q /A1 (is this the area or the volume of the sphere in question?)
q = ρ(r)*A1

E*A2 (what area is this?)= q/ε0
E = ρ(r)*A1 / A2*ε0

There are a few things I am confused about in here so please feel free to ask me to clarify anything. I need to know if I am on the right track here.

Thanks!
 
Physics news on Phys.org
kbwelch17 said:

Homework Statement



A nonuniform, but spherically symmetric, distribution of charge has a charge density ρ(r) given as follows:
ρ(r) = ρ0(1-r/R) for r≤R
ρ(r) = 0 for r≥R

where ρ0 = 3Q/∏R^3 is a positive constant.

(a.) Show that the total charge contained in the charge distribution is Q.
(b.) Show that the electric field in the region r≥R is identical to that produced by a point charge Q at r = 0.
(c.) Obtain an expression for the electric field in the region r≤R.
(d) Graph the electric field magnitude E as a function of r.
(e.) Find the value of r at which the electric field is maximum, and find the value of that maximum field.

Homework Equations



Gauss' Law : Flux = ∫E dot dA = Qencl/ε 0
Electric Field of a point charge: E = k*(q/r^2)

The Attempt at a Solution



I am really just having trouble setting up the integral. You would be integrating the electric field over a sphere, so you would need to choose dA (is this a slice of the area of the sphere?) and a formula to find electric field for a given r. In order to find the latter you would need to use Gauss' Law in which case a give r would give the electric field

ρ(r) = q /A1 (is this the area or the volume of the sphere in question?)
q = ρ(r)*A1

E*A2 (what area is this?)= q/ε0
E = ρ(r)*A1 / A2*ε0

There are a few things I am confused about in here so please feel free to ask me to clarify anything. I need to know if I am on the right track here.

Thanks!
For part (a), you want to use ##dq = \rho\,dv## and integrate over the entire sphere to find the total charge. You don't use Gauss's law here.
 
  • Like
Likes 1 person
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

Replies
6
Views
1K
Replies
10
Views
3K
  • · Replies 11 ·
Replies
11
Views
1K
  • · Replies 4 ·
Replies
4
Views
968
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 7 ·
Replies
7
Views
3K
Replies
23
Views
4K
Replies
28
Views
5K
  • · Replies 12 ·
Replies
12
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K