Leakage Resistance of 100pF Capacitor with Mica Dielectric

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The discussion focuses on calculating the leakage resistance of a 100pF capacitor with a mica dielectric. The relevant formulas for capacitance and resistance are provided, with capacitance depending on the relative permittivity and area, and resistance based on resistivity and dimensions. An initial calculation yields a resistance of 1.2 x 10^25 Ohm, but a correction highlights the omission of vacuum permittivity in the capacitance formula. The importance of unit consistency is emphasized, leading to a more accurate understanding of the capacitor's leakage resistance. Accurate calculations are crucial for evaluating the performance of capacitors with leaky dielectrics.
Kev1n
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Homework Statement


A non ideal capacitor having a 'leaky' dielectric where RL is the leakage resistance measured across the dielectric. A 100pF parallel plate capacitor has a mica dielectric of relative permittivelty Er = 12 and resistivity of p = 1014 ohm m.
Calculate the leakage resistance of this capacitor





2. C = Er A/d & R = p d/A



3. Ans
C = 100 pF, resistivity = 1014 ohm m
Er = 12
C = Er A/d ---- (1)
where A is the area of cross section of the plate and d is the distance between the plates ( or the length of the dielectric)
R = p d/A ----(2)
From (1), d/A = Er/C = 12/(100 x 10-12) = 12 x 1010
Substituting in (2)
R = 1014 x 12 x 1010
Or R = 12 x 1024 Ohm
Or R = 1.2 x 1025 Ohm

The Equation seems right however answer looks wrong
 
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You have the right idea. However, in the formula for the capacitance you are missing the vacuum permittivity E_0 = 8.85419 * 10^-12 F/m

C = E_0 E_r A/d

Always check your units!
 
Roger, of course - thanks for pointing that out, appreciated
 
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