I am trying to indicate the charge distribution of a rolled capacitor as following:

enter image description here

The essential feature is that (in the most layers) charge can be on the inner and outer side of each electrode (here blue minus, red plus and gray a dielectric) if compared to an unrolled situation. Such the capacity is (roughly) twice the value of $\epsilon_0 \epsilon_r \frac{A}{d}$.

However I am not sure of how to indicate the charges on the outermost and innermost layer. In the picture above I didn't indicate $+$ signs on the outer side of the outermost $+$ layer (but on both sides of other layers).

How (and why) can I get the charge distribution (maybe by also varying the density of the $+$ and $-$ signs) on the boundaries (outermost and innermost) more realistically?

Best would be results of a simulation for the charge density.

If it helps to answer with a simulation:

Here is how I parametrized the picture:

$$ (r + 4d\cdot \frac{t}{2\pi} - k\cdot d)\cdot (\sin(t),\cos(t)) $$ where $r$ is the radius, $d$ the thickness of a layer and $k=0$ for red, $k=1$ for gray1, $k=2$ for blue and $k=3$ for gray2. And domain for $t$ chosen as $[0;N\cdot 2\pi]$ where $N$ is the number of turns.


1 Answer 1


You indicated the charges on the metal layers of the rolled capacitor well. There is no charge on the outer side metal layer and the inside metal layer. If the thickness $d$ is much smaller than the smallest radius of the metal sheet, a very good approximation of the total capacitance should be $$C=\epsilon_0\epsilon_r (\frac {2WL_o}{d}+\frac {WL_{no}}{d})$$ where $L_o$ is the length of the inner overlapping part and $L_{no}$ is the length of the not overlapping outer and inner parts of the roll, and $W$ is the width of the capacitor roll. Any more sophisticated calculation of the capacitance will not give any practically meaningful deviation from this approximation.


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