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

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Hello.
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

## 1 Answer

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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 by (28,746 points)
edited Aug 9