How to Calculate Charge Content of a Sphere with Spherical Symmetry?

Click For Summary
SUMMARY

The discussion focuses on calculating the total charge content of a sphere with a charge density defined as ρ = ρ₀ r/R for 0 ≤ r ≤ R. The user initially attempted to derive the total charge Q using spherical coordinates and integrating the charge density over the volume. However, the integration approach was incorrect, leading to an erroneous result. The correct method involves considering the charge on infinitesimally thin spherical shells and integrating from 0 to R, ultimately yielding Q = π ρ₀ R³.

PREREQUISITES
  • Understanding of spherical coordinates in calculus
  • Familiarity with charge density concepts in electrostatics
  • Knowledge of integration techniques for volume calculations
  • Basic principles of electrostatics, particularly Gauss's law
NEXT STEPS
  • Review the derivation of charge density in spherical coordinates
  • Study the application of Gauss's law in spherical symmetry
  • Learn about integrating functions in three-dimensional space
  • Explore examples of charge distributions and their total charge calculations
USEFUL FOR

Students studying electromagnetism, physics educators, and anyone interested in mastering charge distribution calculations in spherical coordinates.

Hannisch
Messages
114
Reaction score
0

Homework Statement


A charge distribution with spherical symmetry has the density \rho = \rho _0 r/R for 0≤ r ≤ R. Determine the total charge content of the sphere.

Homework Equations



\rho = Q / V

The Attempt at a Solution



I started by thinking of the charge dQ of a small volume dV, since

\rho = dQ / dV

I used spherical coordinates to define dV, and said that dV would be

dV = (r d \varphi )(r d \theta) dr

Where \varphi goes from 0 to 2*pi, \theta goes from -pi/2 to pi/2, and r goes from 0 to R, thus covering the entire sphere.

So:

dQ = \rho dV = \rho r^2 d \varphi d \theta dr = \frac{\rho _0 r}{R} r^2 d \varphi d \theta dr = \frac{\rho _0 r^3}{R} d \varphi d \theta dr

I then integrated over this as:

Q = \int ^ {2 \pi} _ {0} \int ^ {\pi /2} _ {-\pi /2} \int ^ {R} _ {0} \frac{\rho _0 r^3}{R} d \varphi d \theta dr

Q = 2 \pi (\pi /2 + \pi /2) \frac{\rho _0 }{R} \int ^ {R} _ {0} r^3 dr = 2 \pi ^2 \frac{\rho _0 }{R} \frac{R^4}{4} = \frac{1}{2} \pi ^2 \rho _0 R^3

And this is not correct, and I can't figure out where I've gone wrong. (It's supposed to be only \pi \rho _0 R^3
 
Physics news on Phys.org
consider any spherical shell inside the sphere of radius x(<R) and thicknedd dx
find charge on it and then integrate it dx from 0 to R
 
Hannisch said:
I used spherical coordinates to define dV, and said that dV would be

dV = (r d \varphi )(r d \theta) dr

Where \varphi goes from 0 to 2*pi, \theta goes from -pi/2 to pi/2, and r goes from 0 to R, thus covering the entire sphere.

Check dV. It is wrong.

ehild
 

Similar threads

Replies
6
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 17 ·
Replies
17
Views
2K
Replies
11
Views
2K
Replies
6
Views
3K
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 63 ·
3
Replies
63
Views
5K
  • · Replies 43 ·
2
Replies
43
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K