# Leakage Resistance

1. Jan 19, 2010

### Kev1n

1. The problem statement, all variables and given/known data
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

2. Jan 19, 2010

### rogerbacon

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

3. Jan 20, 2010

### Kev1n

Roger, of course - thanks for pointing that out, appreciated