Two Conducting Spheres Connected

AI Thread Summary
When two conducting spheres of different radii are connected by an ideal wire, they reach the same electric potential, leading to unequal charge densities on each sphere. This occurs because the smaller sphere, having a higher curvature, will have a greater surface charge density despite the intuitive assumption that charges would distribute evenly. The application of Gauss's Law confirms this behavior, although it may seem counterintuitive at first. The discussion highlights the importance of understanding how charge distribution works in three-dimensional space, particularly in relation to curvature. Ultimately, the concept of charge density in this context becomes clearer through deeper exploration of electrostatic principles.
LikwidN2
Messages
6
Reaction score
0
When a charge is applied to a conducting sphere and connected to another conducting sphere by an ideal thin wire (infinite distance apart), the potential at the surface of each sphere is the same.

However, that means (for spheres of different radii) that the charge density on each sphere would be unequal. This seems counterintuitive. Why wouldn't the charges maintain maximum (and therefor equal) distances from each other, giving equal charge density.

It's been proved to me using Gauss's Law, and the numbers work right, but to me that's the equivalent of saying the sky is blue because a camera detects it as being blue (apologies for the awful analogy, but I think you'll get what I mean - I want something more concrete and intuitive).

Is there any such explanation? I'm not doubting Gauss or my physics textbook, just trying to understand this seemingly fundamental concept.

Thanks in advance,
Likwid

P.S - Lucky you guys, there's actually no problem attatched to this question.
 
Physics news on Phys.org
Naively, if the surface charge density on each sphere was the same, then the charges would be much more crowded on the smaller sphere (since it curves more into 3D).
 
cesiumfrog said:
Naively, if the surface charge density on each sphere was the same, then the charges would be much more crowded on the smaller sphere (since it curves more into 3D).

The surface charge density is actually more on the smaller sphere...
 
Thanks for both the replies all the replies. I think I finally get it.
 
Shooting star said:
The surface charge density is actually more on the smaller sphere...

ah. More crowded in 3D, but importantly less crowded tangentially. :smile: This naivity thing isn't working out so good anymore..
 
Thread 'Gauss' law seems to imply instantaneous electric field propagation'
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
Thread 'A scenario of non-uniform circular motion'
(All the needed diagrams are posted below) My friend came up with the following scenario. Imagine a fixed point and a perfectly rigid rod of a certain length extending radially outwards from this fixed point(it is attached to the fixed point). To the free end of the fixed rod, an object is present and it is capable of changing it's speed(by thruster say or any convenient method. And ignore any resistance). It starts with a certain speed but say it's speed continuously increases as it goes...
Maxwell’s equations imply the following wave equation for the electric field $$\nabla^2\mathbf{E}-\frac{1}{c^2}\frac{\partial^2\mathbf{E}}{\partial t^2} = \frac{1}{\varepsilon_0}\nabla\rho+\mu_0\frac{\partial\mathbf J}{\partial t}.\tag{1}$$ I wonder if eqn.##(1)## can be split into the following transverse part $$\nabla^2\mathbf{E}_T-\frac{1}{c^2}\frac{\partial^2\mathbf{E}_T}{\partial t^2} = \mu_0\frac{\partial\mathbf{J}_T}{\partial t}\tag{2}$$ and longitudinal part...
Back
Top