What is the capacitance formula between two (2) evenly curved plates?

Imagine these plate halves curved around on opposite ends of a cylindrical dielectric material. Assume the curvature angle as 'theta' where 0 <= theta <=180 degrees. @theta=180 degrees, they become standard parallel plate capacitors whose formula is well known.

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  • \$\begingroup\$ You mean like a Leyden Jar? Seem like a homework question to me. \$\endgroup\$
    – Oldfart
    Jan 18, 2018 at 19:16
  • \$\begingroup\$ Curvature angle?!? I think you need to add a diagram that explains the geometry you're talking about. \$\endgroup\$
    – Dave Tweed
    Jan 18, 2018 at 19:46
  • \$\begingroup\$ What about the capacitance between concentric spheres. Would that help because it is solved and might provide insight. \$\endgroup\$
    – Andy aka
    Jan 18, 2018 at 19:49
  • \$\begingroup\$ @Oldfart ...this is NOT a homework issue but a design problem I'm working on for a customer. \$\endgroup\$
    – David M
    Jan 18, 2018 at 20:33
  • 1
    \$\begingroup\$ If the diagram is at all to scale, the capacitance will be dominated by the flat areas on each side. \$\endgroup\$
    – MikeP
    Jan 18, 2018 at 21:12

1 Answer 1


Oh, the formula is easy: it's $$C = Q/V$$ and you find V by solving for the E field with a test charge +Q on one plate, -Q on the other, and integrating (line integral) $$V = \int \overline E \cdot d\overline L $$ from any convenient point on one electrode, to the other.

The charge isn't equally distributed over the area of the plates, its location IS on the outer surface, with E inside the outer surface set to zero.

It gets more complicated because of the dielectric and/or nearby metal structures, and the geometry is not self-shielding, so technically you need to know all the surroundings out to infinity to complete the E-field calculation.

The electrode shapes are boundaries, with boundary conditions that amount to the E-field component parallel to the surface vanishing at zero distance from the surface. As boundaries go, these are not simple shapes. The field solution, therefore, is not going to be simple.

If the problem can be made two-dimensional (i.e. very long device, with constant cross-section), there are conformal mapping solutions that can create symmetry and make the on-paper problem tractable. Otherwise, the use of finite-element modeling would give good results, but not a general formula. Practical capacitors have uniform field in the dielectric, and negligible field outside the dielectric, so approximate solutions are very simple. This, however, is not a practical capacitor in that sense.


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