Electric field lines of a sphere in a uniform electric field

AI Thread Summary
The discussion revolves around drawing electric field lines of a sphere in a uniform electric field using LaTeX/TikZ, focusing on equations in polar coordinates. Participants clarify that the sphere is a dielectric, not a conductor, which leads to a surface charge distribution due to polarization. The electric field inside the sphere is uniform, represented by parallel lines, while the charge distribution on the surface is influenced by the surrounding medium's susceptibility. The equation for the field inside the sphere is provided, highlighting the relationship between the dielectric constants of the materials involved. The conversation emphasizes the importance of understanding the physical properties of the sphere to accurately depict the electric field lines.
sergiokapone
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I need to draw electric field lines of sphere in uniform electric field with LaTeX/TikZ. Can anyone show me equation of field lines in polar coordinates?
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sergiokapone said:
I need to draw electric field lines of sphere in uniform electric field with LaTeX/TikZ. Can anyone show me equation of field lines in polar coordinates?
View attachment 216702
Is this an uncharged conducting sphere? If so, why the field lines inside it, and what direction should the field lines be at the external surface?
 
Is the uncharged dielectric sphere. I found some equation
##r = \frac{k - \sin\phi}{E_0\sin\phi}## and plot the lines:
Code:
    \begin{tikzpicture}
        \clip(-4,-3) rectangle (4,3);
        \draw[red] (-4,0) -- (4,0);
        \foreach \i in {0,0.5,...,3} {
        \draw [thick, color=red, domain=0.05:3.1, samples=200, smooth]
        plot (xy polar cs:angle=\x r, radius={  (\i-sin(\x r))/(1*sin(\x r))  });
        \draw [thick, color=red, domain=0.05:3.1, samples=200, smooth]
        plot (xy polar cs:angle=\x r, radius={  (-\i+sin(\x r))/(1*sin(\x r))  });
        }
        \fill[white, draw=blue] (0,0) circle (1);
\end{tikzpicture}

I get

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Ok, so it is not a conductor. Looks reasonable, except that I am not sure what the figure of eight inside the sphere represents, or what the field lines are doing there.
 
haruspex said:
that I am not sure what the figure of eight inside the sphere represents, or what the field lines are doing there.

I am also not sure, but the nubber 8 inside is the sequence of eqution.
 
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sergiokapone said:
I am also not sure, but the nubber 8 inside is the sequence of eqution.
But... the same equation would not apply inside, would it?
 
haruspex said:
But... the same equation would not apply inside, would it?
Yes, the same equation would not apply inside. For lines inside I just draw series of parallel lines.
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sergiokapone said:
Yes, the same equation would not apply inside. For lines inside I just draw series of parallel lines.
View attachment 216839
What is the basis for making them parallel inside? Surely those would be distorted by what is going on outside. There is a symmetry here.
 
Parallel lines inside is because fileld inside sphere (as well as for every "ellipsoid-like" body) is uniform. It uniform because charge distribution around surface is ##\sigma \propto \cos\phi##. Such distribution governing by boundary conditions and supperppsition of external uniform field and dipole field of sphere.
 
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  • #10
sergiokapone said:
Parallel lines inside is because fileld inside sphere (as well as for every "ellipsoid-like" body) is uniform. It uniform because charge distribution around surface is ##\sigma \propto \cos\phi##. Such distribution governing by boundary conditions and supperppsition of external uniform field and dipole field of sphere.
You still have not clearly specified the problem. You have explained that the sphere contains dielectric material, but why should there be a surface charge as you describe? If it is a conductor there is no field inside; if an insulator why should there be a surface charge distribution?
Or maybe I am just out of my depth.
 
  • #11
Oh, sorry I forgetting to say, the dielectric sphere has the susceptibility ##\epsilon_2## differ from susceptibility of media ##\epsilon_1##, thus there is associated charges on the sphere (due to polarization)

##\sigma' = \frac{3}{4\pi} \frac{\epsilon_1 - \epsilon_2}{\epsilon_1 + 2\epsilon_2} \frac{\vec E_0\vec{r}}{R}.##

Of course, if ##\epsilon_2 =\epsilon_1## there is no charges around sphere.
The field inside sphere is ##E =\frac{3\epsilon_2}{\epsilon_2 + 2\epsilon_1}E_0##
 
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