Skip to main content
Physics

Return to Question

added 2 characters in body; edited tags
Source Link
Qmechanic ♦
Qmechanic ♦
  • 212.8k
  • 48
  • 589
  • 2.3k

I'm working with Griffiths Electrodynamics, and he introduces a uniqueness theorem:

First Uniqueness Theorem: The potential $V$ in a volume $\Omega$ is uniquely determined if (a) the charge density throughout the region, and (b) the value of $V$ on the boundary $\partial\Omega$, are specified.

First Uniqueness Theorem: The potential $V$ in a volume $\Omega$ is uniquely determined if (a) the charge density throughout the region, and (b) the value of $V$ on the boundary $\partial\Omega$, are specified.

I'm a little confused how Griffiths uses this theorem in examples: In the classic image problem (finding the potential due to a point charge $q$ a distance $d$ above an infinite, grounded conducting plane), the trick is to forge the original problem and the configuration of $q$ and its mirror image of the point charge $q$ through the plane (and this new charge is $-q$). The boundary conditions are given ($V=0$ on the plane, and $V\rightarrow 0$ far from the point charge), but how do we know that the charge distribution $\rho$ is the same in this second scenario as it is in the first?

I'm working with Griffiths Electrodynamics, and he introduces a uniqueness theorem:

First Uniqueness Theorem: The potential $V$ in a volume $\Omega$ is uniquely determined if (a) the charge density throughout the region, and (b) the value of $V$ on the boundary $\partial\Omega$, are specified.

I'm a little confused how Griffiths uses this theorem in examples: In the classic image problem (finding the potential due to a point charge $q$ a distance $d$ above an infinite, grounded conducting plane), the trick is to forge the original problem and the configuration of $q$ and its mirror image of the point charge $q$ through the plane (and this new charge is $-q$). The boundary conditions are given ($V=0$ on the plane, and $V\rightarrow 0$ far from the point charge), but how do we know that the charge distribution $\rho$ is the same in this second scenario as it is in the first?

I'm working with Griffiths Electrodynamics, and he introduces a uniqueness theorem:

First Uniqueness Theorem: The potential $V$ in a volume $\Omega$ is uniquely determined if (a) the charge density throughout the region, and (b) the value of $V$ on the boundary $\partial\Omega$, are specified.

I'm a little confused how Griffiths uses this theorem in examples: In the classic image problem (finding the potential due to a point charge $q$ a distance $d$ above an infinite, grounded conducting plane), the trick is to forge the original problem and the configuration of $q$ and its mirror image of the point charge $q$ through the plane (and this new charge is $-q$). The boundary conditions are given ($V=0$ on the plane, and $V\rightarrow 0$ far from the point charge), but how do we know that the charge distribution $\rho$ is the same in this second scenario as it is in the first?

Source Link
user153582
user153582
  • 729
  • 6
  • 14

Conductors and Uniqueness Theorem

I'm working with Griffiths Electrodynamics, and he introduces a uniqueness theorem:

First Uniqueness Theorem: The potential $V$ in a volume $\Omega$ is uniquely determined if (a) the charge density throughout the region, and (b) the value of $V$ on the boundary $\partial\Omega$, are specified.

I'm a little confused how Griffiths uses this theorem in examples: In the classic image problem (finding the potential due to a point charge $q$ a distance $d$ above an infinite, grounded conducting plane), the trick is to forge the original problem and the configuration of $q$ and its mirror image of the point charge $q$ through the plane (and this new charge is $-q$). The boundary conditions are given ($V=0$ on the plane, and $V\rightarrow 0$ far from the point charge), but how do we know that the charge distribution $\rho$ is the same in this second scenario as it is in the first?