Potential of sphere in electric field

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SUMMARY

The discussion focuses on calculating the electric potential and electric fields inside and outside a sphere in an electric field, specifically addressing the conditions for a conducting sphere. The potential equations derived are: for inside the sphere, \(\phi_{in} = - \frac{\sigma_0 + 3\epsilon_0\epsilon_{r2}E_0}{\epsilon_0 (\epsilon_{r1} + 2\epsilon_{r2})} r \cos \theta\), and for outside, \(\phi_{out} = \left[ -E_0 r + \frac{\left( E_0 \epsilon_0 (\epsilon_{r1} - \epsilon_{r2}) - \sigma_0\right)}{\epsilon_0 (\epsilon_{r1} + 2\epsilon_{r2})} \frac{a^3}{r^2} \right] \cos \theta\). The electric fields are expressed as \(E_{in} = \frac{3E_0 + 3\epsilon_{r2} E_0}{\epsilon_{r1} + 2\epsilon_{r2}} \cos \theta\) for inside and \(E_{out} = \left[ 1 + \frac{2(\epsilon_{r1} - \epsilon_{r2} - 3)}{\epsilon_{r1} + 2\epsilon_{r2}} \left( \frac{a}{r}\right)^3 \right] E_0 \cos \theta\) for outside. The discussion also clarifies the behavior of charges in a conducting sphere under an electric field.

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  • Understanding of electric potential and electric fields
  • Familiarity with boundary conditions in electrostatics
  • Knowledge of conducting and non-conducting materials
  • Basic proficiency in vector calculus, specifically gradient operations
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Homework Statement



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Part (a): Find potential inside the sphere and outside of the sphere.
Part (b): Find the electric fields in these two cases. Show for the first case it is identical to a conducting sphere in an electric field.

Homework Equations


The Attempt at a Solution



I have found potential inside and outside of sphere by solving the boundary conditions: ##\phi_in = \phi_out## and ##\epsilon_{r1} \epsilon_0 \frac{\partial \phi_{in}}{\partial r} - \epsilon_{r2} \epsilon_0 \frac{\partial \phi_{out}}{\partial r} = \sigma##:

\phi_{in} = - \frac{\sigma_0 + 3\epsilon_0\epsilon_{r2}E_0}{\epsilon_0 (\epsilon_{r1} + 2\epsilon_{r2})} r cos \theta

\phi_{out} = \left[ -E_0 r + \frac{\left( E_0 \epsilon_0 (\epsilon_{r1} - \epsilon_{r2} \right) - \sigma_0}{\epsilon_0 (\epsilon_{r1} + 2\epsilon_{r2})} \frac{a^3}{r^2} \right] cos \thetaFor ##\sigma_0 = 3\epsilon_0 E_0##:

E_{in} = -\nabla \phi_{in} = \frac{\sigma_0 + 3\epsilon_0\epsilon_{r2}E_0}{\epsilon_0 (\epsilon_{r1} + 2\epsilon_{r2})} cos \theta
= \frac{3E_0 + 3\epsilon_{r2} E_0}{\epsilon_{r1} + 2\epsilon_{r2}} cos \theta

And for outside:
E_{out} = \left[ 1 + \frac{2(\epsilon_{r1} - \epsilon_{r2} - 3)}{\epsilon_{r1} + 2\epsilon_{r2}} \left( \frac{a}{r}\right)^3 \right] E_0 cos \theta

I'm not sure what I'm supposed to look out for?

For ##\sigma_0 = -(\epsilon_r -1)\epsilon_0 E_0##:

I'm assuming outside is free space with ##\epsilon_{r2} = 1##:

E_{in} = \frac{\sigma_0 + 3\epsilon_r E_0 \epsilon_0}{\epsilon_0 (\epsilon_r + 2} cos \theta
= \frac{1 - \epsilon_r + 3}{2 + \epsilon_r} E_0 cos\theta
= \frac{4 + \epsilon_r}{2 + \epsilon_r} E_0 cos \theta

For outside:
E_{out} = \left[ E_0 + \frac{2\left( E_0 \epsilon_0 (\epsilon_r - 1) - \sigma_0 \right)}{\epsilon_0 (\epsilon_r + 2)} \left( \frac{a}{r}\right)^3 \right] cos \theta
= \left[ 1 + \frac{2(\epsilon_r - 1) + (\epsilon_r - 1)}{\epsilon_r + 2}\left( \frac{a}{r}\right)^3 \right] cos \theta
= \left[ 1 + \frac{3(\epsilon_r - 1)}{\epsilon_r + 2} \left( \frac{a}{r}\right)^3 \right] cos \theta

What's the deal with the conductor in an electric field?
I know that for a conducting sphere, the positive charges migrate to +z while the negative charges migrate to -z and they are all spread on the surface of the sphere.
 
Last edited:
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I have searched and found online that when ##\sigma_0 = 3\epsilon_0 E_0##, the field inside would simply me ##-E_0 r cos \theta##.

This clearly isn't the case here, I'm not sure why!
 

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