Surface density of the charges induced on the bases of the cylinder

[rokiboxofficial Ref]

Homework Statement

The dielectric cylinder is in an external uniform electric field E, which is parallel to the axis of the cylinder. Find the surface density of the charges induced on the bases of the cylinder. Dielectric constant of the cylinder material is equal to ε. The height of a cylinder is much less than the radius of its bases.

Relevant Equations

$E_{in} = E_{out} - E_{ind}$
$E_{in} = \frac{E_{out}}{\varepsilon}$

The correct answer to this problem is: $\sigma = \varepsilon_0E\frac{\varepsilon-1}{\varepsilon}$
Here is my attempt to solve it, please tell me what is my mistake?
$E_{in} = E_{out} - E_{ind}$
$E_{ind} = E_{out} - E_{in}$
$E_{in} = \frac{E_{out}}{\varepsilon}$
$E_{ind} = E_{out} - \frac{E_{out}}{\varepsilon}$
$E = \frac{\sigma}{2\varepsilon_0\varepsilon}$
$E_{ind} = E_{ind+} + E_{ind-}$
$E_{ind} = \frac{\sigma}{2\varepsilon_0\varepsilon} + \frac{\sigma}{2\varepsilon_0\varepsilon} = \frac{\sigma}{\varepsilon_0\varepsilon}$
$\sigma = E_{ind}\varepsilon_0\varepsilon$
$\sigma = \left( E_{out} - \frac{E_{out}}{\varepsilon} \right)\varepsilon_0\varepsilon$
$\sigma = E_{out}\varepsilon_0\varepsilon- \frac{E_{out}\varepsilon_0\varepsilon}{\varepsilon}$
$\sigma = E_{out}\varepsilon_0\varepsilon- E_{out}\varepsilon_0$
$\sigma = E_{out}\varepsilon_0\left( \varepsilon- 1 \right)$

 

[kuruman]

I think $~E_{in} = \frac{E_{out}}{\varepsilon}~$ should be $~E_{in} = \frac{\varepsilon_0 E_{out}}{\varepsilon}~$ otherwise the dimensions are incorrect. See if that fixes the problem.

 

[rokiboxofficial Ref]

I think $~E_{in} = \frac{E_{out}}{\varepsilon}~$ should be $~E_{in} = \frac{\varepsilon_0 E_{out}}{\varepsilon}~$ otherwise the dimensions are incorrect. See if that fixes the problem.

No, it doesn't fix it:

$E_{in} = \frac{\varepsilon_0E_{out}}{\varepsilon}$
$E_{ind} = E_{out} - E_{in}$
$E_{ind} = E_{out} - \frac{\varepsilon_0E_{out}}{\varepsilon}$
$\sigma = \left( E_{out} - \frac{\varepsilon_0E_{out}}{\varepsilon} \right)\varepsilon_0\varepsilon$
$\sigma = E_{out}\varepsilon_0\varepsilon- \frac{\varepsilon_0E_{out}\varepsilon_0\varepsilon}{\varepsilon}$
$\sigma = E_{out}\varepsilon_0\varepsilon- E_{out}{\varepsilon_0}^2$
$\sigma = E_{out}\varepsilon_0\left( \varepsilon- \varepsilon_0 \right)$

And the formula $E_{in} = \frac{E_{out}}{\varepsilon}$ was given to me in the training material for the task, I think that it is most likely correct.

 

[TSny]

The correct answer to this problem is: $\sigma = \varepsilon_0E\frac{\varepsilon-1}{\varepsilon}$
Here is my attempt to solve it, please tell me what is my mistake?
$E_{in} = E_{out} - E_{ind}$
$E_{ind} = E_{out} - E_{in}$
$E_{in} = \frac{E_{out}}{\varepsilon}$
$E_{ind} = E_{out} - \frac{E_{out}}{\varepsilon}$
$E = \frac{\sigma}{2\varepsilon_0\varepsilon}$
$E_{ind} = E_{ind+} + E_{ind-}$
$E_{ind} = \frac{\sigma}{2\varepsilon_0\varepsilon} + \frac{\sigma}{2\varepsilon_0\varepsilon} = \frac{\sigma}{\varepsilon_0\varepsilon}$
$\sigma = E_{ind}\varepsilon_0\varepsilon$
$\sigma = \left( E_{out} - \frac{E_{out}}{\varepsilon} \right)\varepsilon_0\varepsilon$
$\sigma = E_{out}\varepsilon_0\varepsilon- \frac{E_{out}\varepsilon_0\varepsilon}{\varepsilon}$
$\sigma = E_{out}\varepsilon_0\varepsilon- E_{out}\varepsilon_0$
$\sigma = E_{out}\varepsilon_0\left( \varepsilon- 1 \right)$

I believe the mistake occurs where you wrote $$E = \frac{\sigma}{2\varepsilon_0\varepsilon}$$ for the electric field produced by $\sigma$ on one of the surfaces. You can treat the surface as an infinite plane. The electric field produced by an infinite plane of surface charge $\sigma$ can be found using Gauss' law.

 

[haruspex]

I think $~E_{in} = \frac{E_{out}}{\varepsilon}~$ should be $~E_{in} = \frac{\varepsilon_0 E_{out}}{\varepsilon}~$ otherwise the dimensions are incorrect. See if that fixes the problem.

Looks like $\varepsilon$ is being used for the relative permittivity, so is dimensionless.

 

[rokiboxofficial Ref]

I believe the mistake occurs where you wrote $$E = \frac{\sigma}{2\varepsilon_0\varepsilon}$$ for the electric field produced by $\sigma$ on one of the surfaces. You can treat the surface as an infinite plane. The electric field produced by an infinite plane of surface charge $\sigma$ can be found using Gauss' law.

Yes, you are right! Thank you a lot!
Is this solution a correct?
By Gauss' law:
$\frac{q}{\varepsilon_0} = ES_{full}$
$\frac{q}{\varepsilon_0} = 2ES$
$E = \frac{q}{2\varepsilon_0S}$
$E = \frac{\sigma}{2\varepsilon_0}$
$E_{ind} = E_{ind+} + E_{ind-}$
$E_{ind} = \frac{\sigma}{2\varepsilon_0} + \frac{\sigma}{2\varepsilon_0} = \frac{\sigma} {\varepsilon_0}$
$\sigma = E_{ind}\varepsilon_0$
$E_{ind} = E_{out} - \frac{E_{out}}{\varepsilon}$
$\sigma = \left( E_{out} - \frac{E_{out}}{\varepsilon} \right)\varepsilon_0$
$\sigma = E_{out}\varepsilon_0- \frac{E_{out}\varepsilon_0}{\varepsilon}$
$\sigma = E_{out}\varepsilon_0\left( 1 - \frac{1}{\varepsilon}\right)$
$\sigma = E_{out}\varepsilon_0\left( \frac{\varepsilon}{\varepsilon} - \frac{1}{\varepsilon}\right)$
$\sigma = E_{out}\varepsilon_0\left(\frac{\varepsilon - 1}{\varepsilon}\right)$

 

[TSny]

Yes, you are right! Thank you a lot!
Is this solution a correct?
By Gauss' law:
$\frac{q}{\varepsilon_0} = ES_{full}$
$\frac{q}{\varepsilon_0} = 2ES$
$E = \frac{q}{2\varepsilon_0S}$
$E = \frac{\sigma}{2\varepsilon_0}$
$E_{ind} = E_{ind+} + E_{ind-}$
$E_{ind} = \frac{\sigma}{2\varepsilon_0} + \frac{\sigma}{2\varepsilon_0} = \frac{\sigma} {\varepsilon_0}$
$\sigma = E_{ind}\varepsilon_0$
$E_{ind} = E_{out} - \frac{E_{out}}{\varepsilon}$
$\sigma = \left( E_{out} - \frac{E_{out}}{\varepsilon} \right)\varepsilon_0$
$\sigma = E_{out}\varepsilon_0- \frac{E_{out}\varepsilon_0}{\varepsilon}$
$\sigma = E_{out}\varepsilon_0\left( 1 - \frac{1}{\varepsilon}\right)$
$\sigma = E_{out}\varepsilon_0\left( \frac{\varepsilon}{\varepsilon} - \frac{1}{\varepsilon}\right)$
$\sigma = E_{out}\varepsilon_0\left(\frac{\varepsilon - 1}{\varepsilon}\right)$

Looks good.

 

[TSny]

Looks like $\varepsilon$ is being used for the relative permittivity, so is dimensionless.

Yes. $\varepsilon$ is given to be the "dielectric constant", which is the same as the relative permittivity.