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The capacitor of the 2 cylinders exists.

As shown in the diagram, the righter space has been filled with the dielectric of \$ \epsilon_{1} \$ and the lefter space has been filled with the dielectric of \$ \epsilon_{2} \$

What I want to do is to deduce the equation of surface charge density of the inner surface of the outer cylinder.

$$ a :=\text{ radius of the inner cylinder } $$

$$ b :=\text{ radius of outer cylinder } $$

We'll observe what happens after the grounding of the outer cylinder and giving the charge \$ \lambda \$ per unit length against the inner cylinder is done.

The electric field is uniform by distance(which means that the dielectric where point belongs is whatever).

$$ \epsilon_{1}E =D_{1} :=\small\text{ electric flux density of the right dielectric at the inner surface of the outer cylinder} $$

$$ \epsilon_{2}E =D_{2} :=\small\text{ electric flux density of the left dielectric at the inner surface of the outer cylinder} $$

$$ \sigma_{1} :=\small\text{ surface charge density of the right side of the inner surface of the outer cylinder } $$

$$ \sigma_{2} :=\small\text{ surface charge density of the left side of the inner surface of the outer cylinder } $$

$$ D_{1} =\sigma_{1}= \epsilon_{1} E $$

$$ D_{2} =\sigma_{2} =\epsilon_{1} E$$

To find out E, I've used Gauss law.

$$ \lambda = \epsilon_{1}E \left( \pi b \cdot 1 \right) + \epsilon_{2}E \left( \pi b \cdot 1 \right) $$

$$ \left( \pi b \cdot 1 \right) ~~ \leftarrow~~ \text{half area of flank of the cylinder} $$

Here I used \$b\$ to calculate the area of flank of the cylinder, actually I took the approximation. Strictly speaking, I have to take the radius of \$b-\delta_{x} \$ to avoid touching the outer cylinder(conductor).

$$ \lambda =E\left( \pi b \cdot 1\right) \left( \epsilon_{1}+ \epsilon_{2} \right) $$

$$ \therefore ~~ E= \frac{ \lambda }{ \pi b \left( \epsilon_{1}+ \epsilon_{2} \right) } $$

$$ \therefore ~~ \sigma_{1} = \frac{ \lambda \epsilon_{1} }{ \pi b \left( \epsilon_{1}+ \epsilon_{2} \right) } $$

However the official description says that the above equation takes the opposite sign.

$$ \text{official equation} \rightarrow ~~ \sigma_{1} = \frac{ - \lambda \epsilon_{1} }{ \pi b \left( \epsilon_{1}+ \epsilon_{2} \right) } $$

Where I made mistakes?

I assumed that the outside the outer cylinder can't take non-zero value of electric field to hold the outer cylinder be zero potential.

And \$-\lambda\$ is induced to the outer cylinder per unit length to make the outer cylinder be \$0 \left[ \text{V} \right]\$

electrical apprentice
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1 Answers1

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You have to be explicit in defining what your Gaussian Surface is. If \$\lambda\$ is the total free charge per length on the inner cylinder, then for this application of Gauss's Law, your Gaussian surface must enclose the charge density on that inner cylinder. Your equation relating \$E\$ and \$\lambda\$\$(λ=ϵ_1E(πb⋅1)+ϵ_2E(πb⋅1))\$is correct if your Gaussian surface is a simple cylinder of radius b. \$E\$, and so from this equation you can correctly calculate the electric field \$E\$.

Once you have \$E\$, and you want to formally calculate the charge density on the inner surface of the outer conductor, you need to define another Gaussian surface. A sensible choice would be a cylindrical annulus (cylindrical shell), with the inner surface just inside the inside surface of the outer conductor, and the outer surface (of the Gaussian cylindrical annulus) inside the conductor material of the outer conductor. In that case, when you take the dot product of \$E\$ with the surface normal element \$d\mathbf{A}\$, you pick up the negative sign you are missing, because the surface normal points toward the axis of the cylinder. The other curved surface of the cylinder does not contribute to the Gaussian integral because the field is zero, of course, inside the conductor.

Let me know if that is not clear and I will try to clean it up.

rpm2718
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