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sammy gerbil
sammy gerbil
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Is What is the resistance of an ohmic sphere infinitewith surface potential $\phi_0\cos\theta$?

I'm trying to find the volumetric density of current in the interior of a sphere (made of ohmic material) with radius $a$, conductivity $g$ and surface potential $\phi=\phi_0cos\theta$.
This

This is the way I thought I could solve it:

First, it seems to me that the electric field should have only a $z$ component since the potential is symmetric with respect to the azimutal angle and increases when we move along the $z$ axis, having values $\phi=-\phi_0$ when $z=-a$ and $\phi=\phi_0$ when $z=a$. (Is this assumption correct?)
Secondly, I would compute the value of the resistance $R$ of the sphere using the formula $R=\frac{\Delta L}{gS}$ for a cylindric conductor of section $S$ and length $\Delta L$. In order to do that I have to sum up the resistance of all the cylinders of infinitesimal lenght $dz$, section $S(z)=\pi(a^2-z^2)$ and resistance $dR(z)=\frac{dz}{\pi g(a^2-z^2)}$. So, I would have: $$R=\int_{-a}^a\frac{dz}{\pi g(a^2-z^2)}$$ Then I would use Omh's law fo find $I$ (using that $\Delta\phi=2\phi_0$), and finally I would obtain the volumetric density of current $J$ as a function of $z$ using the formula $J=\frac{I}{S(z)}$.

Now, the problem is that the integral for the resiastance diverges, which would imply $J=0$ and it doesn't seem to be correct. What is wrong with my approach?

Is the resistance of an ohmic sphere infinite?

I'm trying to find the volumetric density of current in the interior of a sphere (made of ohmic material) with radius $a$, conductivity $g$ and surface potential $\phi=\phi_0cos\theta$.
This is the way I thought I could solve it:

First, it seems to me that the electric field should have only a $z$ component since the potential is symmetric with respect to the azimutal angle and increases when we move along the $z$ axis, having values $\phi=-\phi_0$ when $z=-a$ and $\phi=\phi_0$ when $z=a$. (Is this assumption correct?)
Secondly, I would compute the value of the resistance $R$ of the sphere using the formula $R=\frac{\Delta L}{gS}$ for a cylindric conductor of section $S$ and length $\Delta L$. In order to do that I have to sum up the resistance of all the cylinders of infinitesimal lenght $dz$, section $S(z)=\pi(a^2-z^2)$ and resistance $dR(z)=\frac{dz}{\pi g(a^2-z^2)}$. So, I would have: $$R=\int_{-a}^a\frac{dz}{\pi g(a^2-z^2)}$$ Then I would use Omh's law fo find $I$ (using that $\Delta\phi=2\phi_0$), and finally I would obtain the volumetric density of current $J$ as a function of $z$ using the formula $J=\frac{I}{S(z)}$.

Now, the problem is that the integral for the resiastance diverges, which would imply $J=0$ and it doesn't seem to be correct. What is wrong with my approach?

What is the resistance of an ohmic sphere with surface potential $\phi_0\cos\theta$?

I'm trying to find the volumetric density of current in the interior of a sphere (made of ohmic material) with radius $a$, conductivity $g$ and surface potential $\phi=\phi_0cos\theta$.

This is the way I thought I could solve it:

First, it seems to me that the electric field should have only a $z$ component since the potential is symmetric with respect to the azimutal angle and increases when we move along the $z$ axis, having values $\phi=-\phi_0$ when $z=-a$ and $\phi=\phi_0$ when $z=a$. (Is this assumption correct?)
Secondly, I would compute the value of the resistance $R$ of the sphere using the formula $R=\frac{\Delta L}{gS}$ for a cylindric conductor of section $S$ and length $\Delta L$. In order to do that I have to sum up the resistance of all the cylinders of infinitesimal lenght $dz$, section $S(z)=\pi(a^2-z^2)$ and resistance $dR(z)=\frac{dz}{\pi g(a^2-z^2)}$. So, I would have: $$R=\int_{-a}^a\frac{dz}{\pi g(a^2-z^2)}$$ Then I would use Omh's law fo find $I$ (using that $\Delta\phi=2\phi_0$), and finally I would obtain the volumetric density of current $J$ as a function of $z$ using the formula $J=\frac{I}{S(z)}$.

Now, the problem is that the integral for the resiastance diverges, which would imply $J=0$ and it doesn't seem to be correct. What is wrong with my approach?

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user.3710634
edit approved Oct 9, 2016 at 23:38
user.3710634
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Resistance Is the resistance of aan ohmic sphere is infinityinfinite?

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la flaca
la flaca
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Resistance of a ohmic sphere is infinity?

I'm trying to find the volumetric density of current in the interior of a sphere (made of ohmic material) with radius $a$, conductivity $g$ and surface potential $\phi=\phi_0cos\theta$.
This is the way I thought I could solve it:

First, it seems to me that the electric field should have only a $z$ component since the potential is symmetric with respect to the azimutal angle and increases when we move along the $z$ axis, having values $\phi=-\phi_0$ when $z=-a$ and $\phi=\phi_0$ when $z=a$. (Is this assumption correct?)
Secondly, I would compute the value of the resistance $R$ of the sphere using the formula $R=\frac{\Delta L}{gS}$ for a cylindric conductor of section $S$ and length $\Delta L$. In order to do that I have to sum up the resistance of all the cylinders of infinitesimal lenght $dz$, section $S(z)=\pi(a^2-z^2)$ and resistance $dR(z)=\frac{dz}{\pi g(a^2-z^2)}$. So, I would have: $$R=\int_{-a}^a\frac{dz}{\pi g(a^2-z^2)}$$ Then I would use Omh's law fo find $I$ (using that $\Delta\phi=2\phi_0$), and finally I would obtain the volumetric density of current $J$ as a function of $z$ using the formula $J=\frac{I}{S(z)}$.

Now, the problem is that the integral for the resiastance diverges, which would imply $J=0$ and it doesn't seem to be correct. What is wrong with my approach?