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Laplace Equation boundary condition - Why are the current densities equal 0?

1 vote

In my homework I was given the following problem:
A wire that has a current I is connected to a perfect electric conductor sphere with radius a. The bottom half of the sphere sunk to the ground. A layer with a width of (b-a) and a conductivity of sigma - was created around the bottom half. The earth's conductivity is sigma-E.

I was asked what all the boundary conditions for the electrostatic potential in the ground are. One of them was:

I am not quite sure why this is true. I do think there's a radial current in this angle of theta, why could there not be an angular current as well?
Any explanation would be greatly appreciated!

asked Jul 24, 2019 in Physics Problems by Guy (110 points)

1 Answer

0 votes

There ought to be a diagram showing how the angle $\theta$ is defined. Nevertheless, the boundary condition which you have been given is confusing.

The diagram above shows a vertical plane through the centre of the spherical electrode. There is cylindrical symmetry here, so a cylindrical co-ordinate system is the obvious choice. If the plane is rotated through any azimuthal angle $\theta$ about the vertical axis $z$ through the centre of the sphere, then all measurements (potentials, current densities, etc) would be the same at all points with the same cylindrical co-ordinates $\rho, z$. So there could be no difference in potential or current density around any circle of radius $\rho$ which is centred on and perpendicular to the $z$ axis.

Inside the conducting electrode there is no resistance (infinite conductivity) so there could be a current of constant density around such a circle inside the electrode, without there being any change in potential around any azimuthal circle : $$J_{\theta}=\text{constant}$$

Inside the regions of finite conductivity 1 & 2, there could not be any current around an azimuthal circle, because there would have to be a change in potential between the start and the end points, which are the same, so this is impossible :$$J_{1\theta}=J_{2\theta}=0$$

In all three regions the boundary condition applies for all values of $\theta$ and particular values of $\rho, z$. There is nothing special about $\theta=\frac12 \pi$. The boundary condition given in the book is confusing. It suggests that there is something special about the azimuthal angle $\theta=\frac12\pi$ but there is nothing in the problem which supports this.

Perhaps a spherical polar system is being used. If so, and if $\phi$ is the polar angle, then $\phi=\frac12 \pi$ defines the horizontal plane through the centre of the sphere. The above boundary condition is still true for this value of $\phi$ and all values of $r$ : $$J_{1\theta}(\phi=\frac12 \pi)=J_{2\theta}(\phi=\frac12 \pi)=0$$ But there is nothing special about this plane. The boundary condition applies for all other planes perpendicular to the axis, and all values of $\phi$ and $r$ provided that $\rho=r\sin\phi$ and $z=r\cos\phi$ are constant.

answered Jul 30, 2019 by sammy gerbil (28,896 points)
edited Aug 9, 2019 by sammy gerbil