How Does Smearing Charge dQ Uniformly on a Sphere Maintain Constant Energy?

Click For Summary
SUMMARY

The discussion centers on the uniform distribution of charge dQ on a sphere of radius r, which maintains constant energy as it is deposited from infinity. The potential energy of the charge dQ is expressed as Q dQ/4πε₀r. The key points include that the electric field (E) due to the total charge Q behaves as if it were a point charge, and the tangential smearing of dQ does not alter the energy due to the symmetry of the electric field around the sphere.

PREREQUISITES
  • Understanding of electrostatics and electric fields
  • Familiarity with potential energy concepts in physics
  • Knowledge of spherical charge distributions
  • Basic calculus for understanding potential energy equations
NEXT STEPS
  • Study the derivation of electric fields from point charges
  • Explore the concept of potential energy in electrostatics
  • Learn about charge distribution on conductors and insulators
  • Investigate the implications of symmetry in electric fields
USEFUL FOR

This discussion is beneficial for first-year physics undergraduates, educators in electromagnetism, and anyone interested in the principles of electrostatics and energy conservation in electric fields.

hasan_researc
Messages
166
Reaction score
0
A sphere of radius r contains a charge Q distributed uniformly throughout its volume. Charge dQ is brought in from infinity and deposited uniformly over the surface. The potential energy of the charge dQ is Q dQ/4\pi\epsilon<sub>0</sub>.

This follows naturally from the result for the potential of 2 point charges after noting
(a) E due to Q is the same as a point charge and
(b) smearing dQ around the sphere (i.e., tangential to the radial direction) doesn’t change the energy.

Why/how does "smearing dQ around the sphere (i.e., tangential to the radial direction) not change the energy"?

Thanks in advance for any help!

[N B : I am a first year physics undergraduate.]
 
Physics news on Phys.org
I am sorry for the poor latex code. The potential energy expression should read Q dQ/4\pi\epsilon_{0}
 
Mistaken for the second time now. The potential energy expression should read Q dQ/4\pi\epsilon_{0}r
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 11 ·
Replies
11
Views
5K