Is this all you have to do for this problem

  • Thread starter Thread starter charlies1902
  • Start date Start date
Click For Summary

Homework Help Overview

The problem involves calculating the total charge of a nonconducting solid sphere with a volume charge density that varies with distance from the center, specifically described by the equation ρ = Ar, where A is a constant and r is the distance from the center of the sphere. The context is rooted in electrostatics and involves integration to find the total charge.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the need to integrate the charge density over the volume of the sphere, with some suggesting the use of a triple integral. There are questions about the correct form of the differential volume element and the limits of integration.

Discussion Status

There is an ongoing exploration of different methods to approach the problem, including the use of a triple integral and the integration of the charge density. Participants are clarifying the relationship between the charge density and the volume element, with some guidance provided on the integration process.

Contextual Notes

Participants are navigating the implications of the varying charge density and the appropriate mathematical techniques to apply, including the need for proper limits of integration and the definition of the volume element in spherical coordinates.

charlies1902
Messages
162
Reaction score
0

Homework Statement


A nonconducting solid sphere of radius R has a volume charge density that is proportional to the distance from the center. That is rho=Ar for r is less than/equal to , A is a constant. Find the total charge on the sphere.


Homework Equations




The Attempt at a Solution


rho=Ar=Q/V
Q=Ar(4/3)*pi*R^3

Is that it?
 
Physics news on Phys.org
No.

Integrate the charge density over the volume of the sphere.
 
would've that just be the integral of (Ar dV) which equals Ar*V?
 
Use dV = 4πr2dr .

Integrate (A)r , with respect to r, from r = 0 to r = R .
 
I went in for help on this one and my TA told me that I could do a triple integral for this one or with the integral you provided, but I chose to do the triple integral.

What I did was take the triple integral of Ar*r^2 * sin(theta) in this order: dr,dtheta,dphi
with limits of integration: 0 to R, 0 to pi, 0 to 2pi, respectively
For the answer I got AR^4 * pi

Can you explain how you get the dV = 4πr2dr
 
The surface area of a sphere of radius r, is 4πr2.

So the volume of a spherical shell of thickness dr is given by multiplying the thickness by the surface area.

dV = 4πr2 dr.

What's the integral of 4πr2 dr, from 0 to R ?
 

Similar threads

Replies
11
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
6
Views
2K
Replies
7
Views
2K
  • · Replies 36 ·
2
Replies
36
Views
2K
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
4
Views
2K
Replies
6
Views
2K