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The capacitance per unit area of a parallel-plate capacitor is $\frac{\epsilon_0}{d}$.

But what if one of the plates is a mesh, but the distance $d$ is much much greater than the size of the holes in the mesh?.

My intuitive thinking would be that from the perspective of the solid plate the charge density on the mesh is perfectly uniform, and hence the capacitance will be the same as for a parallel-plate capacitor with two solid plates, but I don't know how to prove/disprove it.

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    $\begingroup$ Your intuition is correct. You could simply do the experiment to "prove it". Measure the capacitance of two plate capacitors, one with solid plates (aluminum foil will do), one with a fine mesh. If you want to do theory, then you will have to start thinking about what the electric field behind a mesh looks like. Since the charges on the plates have to balance out, the charge on the mesh has to be the same as on the solid plate, but it is concentrated on the metal wires. From that charge distribution you can calculate the field behind the mesh. $\endgroup$
    – FlatterMann
    Commented Apr 21, 2023 at 16:47
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    $\begingroup$ "The capacitance per unit area of a parallel-plate capacitor is $\frac{\epsilon_0 A}{d}$" Shouldn't you divide by A?. $\endgroup$
    – Bob D
    Commented Apr 21, 2023 at 16:48
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    $\begingroup$ Check out Chapter 7-5 of the FLP (he treats a grid, but the same method apples for a 2D mesh). Essentially, use Fourier series, the potential of the constant mode will dominate at long distance due to the increasingly fast exponential decay in distance of the higher modes. $\endgroup$
    – LPZ
    Commented Apr 21, 2023 at 16:50
  • $\begingroup$ $\frac{\epsilon_0 A}{d}$ is total capacitance, not capacitance per unit area. $\endgroup$
    – Ken Mellendorf
    Commented Apr 21, 2023 at 16:54
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    $\begingroup$ @LPZ thanks a lot, this is exactly what I was looking for $\endgroup$
    – user9413641
    Commented Apr 21, 2023 at 18:11

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As current in one plate equals current out the other, the total charge would be the same on both plates. Electrons do not "like" to be near each other, so charging the mesh plate would be more difficult that for a solid plate. As the mesh holes get smaller, this effect becomes less important. Also, the electric field paths near the mesh would initially be leaning toward the holes before straightening out toward the other plate. The charge density and field strength are larger near the metal mesh, but the effect weakens once a short distance from the mesh. So long as the distance from plate to plate is large compared to holes in the mesh, electric field due to charge would be essentially equal to two solid plates.

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