When can a charged sphere be treated as a point charge?

Click For Summary
SUMMARY

A charged sphere can be treated as a point charge when its charge distribution is spherically symmetric, meaning it is uniformly distributed over its volume or surface. According to Gauss's law, the electric field outside such a sphere is equivalent to that of a point charge located at its center. However, this approximation holds only if the sphere is isolated; the presence of external charges can distort the field, making the point charge model invalid. If the sphere's radius is significantly smaller than the characteristic lengths of the system, it can still be approximated as a point charge even in the presence of external influences.

PREREQUISITES
  • Understanding of Gauss's law in electromagnetism
  • Familiarity with electric field concepts and point charges
  • Knowledge of charge distribution types (uniform vs. non-uniform)
  • Basic principles of symmetry in physics
NEXT STEPS
  • Study the application of Gauss's law in various charge configurations
  • Explore the method of images for solving electrostatic problems
  • Learn about Green's functions in electrostatics
  • Investigate the effects of induced charges on non-isolated conductors
USEFUL FOR

Physics students, electrical engineers, and anyone studying electrostatics or working with charged objects in various applications.

annms
Messages
15
Reaction score
3
Not much more to add to the title. I was just wondering, under what conditions can a charged sphere be treated as a point charge?
 
Physics news on Phys.org
The electric field outside a (uniformly) charged sphere is equal to the electric field that would have been created by a point charge with the same total charge.
 
annms said:
I was just wondering, under what conditions can a charged sphere be treated as a point charge?
When the charge distribution is spherically symmetric, the field outside the sphere is exactly that of a point charge at the center of the sphere. (Where the charge of the point charge equals the total charge of the sphere.)
 
So in cases when the charge is uniformly distributed over its volume or its surface, the sphere can be treated as a point charge?
 
yep. Doc Al's response is slightly more accurate than mine - as long as the charge distribution is symmetric, the electric field outside will be same as that created by a point charge.

So any symmetric distribution you can think of will have this remarkable property.

Edit: For clarity, symmetric just means that the charge distribution will only depend on the radius. (Well, that's probably not the technical meaning of symmetric, but that's what it means in this example).
 
Apply the Gauss law in Maxwell's equations. Integral form, ∫E·dA=Q/ε0, where Q is the charge bounded inside the close surface. For any spherical Gaussian surface, E4πr2=Q/ε0, and is exactly the same as coulomb's law for point charge.

This means you can apply to whatever the charge distribution, it can be modeled by a equivalent point charge.
 
"When can a charged sphere be treated as a point charge? "

This is only true if the sphere is isolated from everything else. If another charge, or another object held at some potential, is brought near the sphere, it will induce surface charges in the sphere that are not spherically symmetric, and therefore you can no longer treat it as a point charge. But if the radius of the sphere is very small compared to characteristic lengths of the system, then you can treat it as a point charge even if it is not isolated because the induced charges will be spherically symmetric as an approximation if the sphere is very small.
 
chrisbaird said:
"When can a charged sphere be treated as a point charge? "

This is only true if the sphere is isolated from everything else. If another charge, or another object held at some potential, is brought near the sphere, it will induce surface charges in the sphere that are not spherically symmetric, and therefore you can no longer treat it as a point charge. But if the radius of the sphere is very small compared to characteristic lengths of the system, then you can treat it as a point charge even if it is not isolated because the induced charges will be spherically symmetric as an approximation if the sphere is very small.

But it can still be viewed as an point charge by taking a new gaussian surface where the centre shifts to the new equivalent point charge and take a larger sphere. Everywhere outside the surface the model is still valid.

But when there is an external charge the field is again distorted. Actually the net field should be modeled by another equivalent charge and a even larger sphere as gaussian surface.
 
ZealScience said:
But it can still be viewed as an point charge by taking a new gaussian surface where the centre shifts to the new equivalent point charge and take a larger sphere. Everywhere outside the surface the model is still valid.

But when there is an external charge the field is again distorted. Actually the net field should be modeled by another equivalent charge and a even larger sphere as gaussian surface.

I think you are confused. The Guassian-surface trick only works if you have a constant field over the Gaussian surface which will only happens if the charge density if there is symmetry (a spherical Gaussian surface requires a spherically symmetric charge density, a cylindrical Guassian surface requires...,etc). As soon as the symmetry is lost, the Guassian-surface trick no longer works and you have to go to more advanced methods (method of images, Green functions, etc).
 
  • #10
ZealScience said:
But it can still be viewed as an point charge by taking a new gaussian surface where the centre shifts to the new equivalent point charge and take a larger sphere. Everywhere outside the surface the model is still valid.

But when there is an external charge the field is again distorted. Actually the net field should be modeled by another equivalent charge and a even larger sphere as gaussian surface.

No, no, no. Even if you shift the centre and use a larger gaussian sphere, the electric field at each point on that gaussian sphere would not be equal. (Because the distribution is no longer symmetric).
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 15 ·
Replies
15
Views
2K