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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.
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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