# Determine the capacitance between two surfaces

• Engineering
Homework Statement:
Determine capacitance between the two surfaces (see picture). The electric field is assumed to be radial.
Relevant Equations:
1) C = Q/V. 2) Gauss law (cylinder); Qenc = integral of E*e0 dA , where dA is small element of the surface. 3) V = -gradient of E
For my solution I'm skipping writing out all the vectors, I just want to see if I'm in the right way or totally off.

Attempt at solution:
Qenc = ∫ E(r)*e0 ds = ∫ E(r)*e0 *h* r*dtheta, we integrate from 0 to phi0. This will give us Q = E(r)*e0*h*r*phi0.

Now we find V by integrating E from a to b with respect to r.
V = ∫ E(r) dr = Q / (e0*h*phi0) * ∫ 1/r dr = Q * ln(b/a) * 1/(e0*h*phi0).

And lastly we have C = Q/V = ln(b/a) * 1(e0*h*phi0).

Does this look reasonable? Unfortunately I don't have the right answer to this task.

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Just a small correction on
And lastly we have C = Q/V = ln(b/a) * 1(e0*h*phi0).
$$C=\epsilon_0 \frac{h \phi_0}{ln\frac{b}{a}}$$

berkeman
I could be wrong but, why are you using gauss law for this charge distribution, I don't see any symmetry in this problem for a gaussian surface to be applied

Last edited:
If the surfaces are not ##\phi_0## parts but 2##\pi## full cylinders, are you all right on symmetry ?

Yeah, actually in this case it doesn't seem to be symmetrical so my attempt was probably wrong.

If the cylinders are whole (2 pi angle) then it is symmetrical.

If the cylinders are whole (2 pi angle) then it is symmetrical.

By symmetry you get conductance of whole cylinders that is #2 answer with ##\phi_0=2\pi##
Then cut the cake or pizza of thus charged cylinders with ##\phi_0## and ##2\pi-\phi_0## parts.
Conductance is divided to the parts with the ratio of ##\phi_0## and ##2\pi-\phi_0##.