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Ahmed1029
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For the second uniqueness theorem of electrostatics to apply, does the outer boundary enclosing all the conductors have to be at a constant potential?
Because in the proof he assumes that V3 (the difference between the two supposed potentials) is constant at all surfaces and at the boundary, but that would mean that every supposed potential is constant at the outer boundary. What am I getting wrong?vanhees71 said:No, why?
oh so that's it! Thanks!Delta2 said:No the difference being constant doesn't necessarily imply that each potential is constant.
Take for example the functions $$\phi_1(x,y,z)=\frac {e^{-x-y-z}}{x^2+y^2+z^2}$$, $$\phi_2(x,y,z)=\frac {e^{-x-y-z}}{x^2+y^2+z^2}+1$$their difference is constant and equal to 1 (or -1) but the functions are not constant.
It just implies that their gradient is the same (under some other conditions too).
Sorry I have another question. Suppose now I have two supposed potential functions whose difference is not constant at the outer boundary, how does the theorem play out?Delta2 said:No the difference being constant doesn't necessarily imply that each potential is constant.
Take for example the functions $$\phi_1(x,y,z)=\frac {e^{-x-y-z}}{x^2+y^2+z^2}$$, $$\phi_2(x,y,z)=\frac {e^{-x-y-z}}{x^2+y^2+z^2}+1$$their difference is constant and equal to 1 (or -1) but the functions are not constant.
It just implies that their gradient is the same (under some other conditions too).
Not sure here but I think we need the condition that their gradients should be equal at the boundary (but again not necessarily constant gradients). Otherwise there is no uniqueness.Ahmed1029 said:Sorry I have another question. Suppose now I have two supposed potential functions whose difference is not constant at the outer boundary, how does the theorem play out?
If the boundary is my domain and the gradient of each function is the same, the functions only differ by a constant and there is no issue here. But why do the gradients have to be the same?Delta2 said:Not sure here but I think we need the condition that their gradients should be equal at the boundary (but again not necessarily constant gradients). Otherwise there is no uniqueness.
in Electrodynamics the scalar potentials can differ by any function ##\chi(x,y,z,t)##, but in electrostatics they can differ only by a constant at most right?vanhees71 said:Of course the solutions for the potentials need not be the same, because they are only defined up to a gauge transformation.
The second uniqueness theorem applies to electrostatic fields in regions where the charge density is zero.
The second uniqueness theorem is a principle in electrostatics that states that the electric field in a region with zero charge density is uniquely determined by the boundary conditions on the surface of the region.
The first uniqueness theorem applies to regions with non-zero charge density, while the second uniqueness theorem applies to regions with zero charge density. Additionally, the first uniqueness theorem only applies to regions with a single point charge, while the second uniqueness theorem applies to regions with any number of charges.
Yes, the second uniqueness theorem only applies to regions where the charge density is zero. It also does not apply to regions with time-varying fields or magnetic fields present.
The second uniqueness theorem is often used in solving boundary value problems in electrostatics, such as determining the electric field inside a capacitor or finding the potential on the surface of a conductor. It is also used in analyzing and designing electrical circuits and devices.