Leakage resistance of capacitor theory help?

[notsosmart]

leakage resistance of capacitor theory help??

Homework Statement

i have worked out the answer to my question, yet it seems silly so have convinced myself its wrong. Please help, or tell me if I'm being stupid if you can??

a parallel plate capacitor Er=12, length=10mm, width=5mm, C=100pF, rho=10`14.

calculate the leakage resistance of the capacitor?

Homework Equations

R= rho( length/area) C= (ErEo x A)/ d

The Attempt at a Solution

no idea. got as far as area 5 x 10`-5
EoEr = 1.062 x 10`-10

i just can't figure out how to get back to the capacitors leakage resistance??

 

[berkeman]

Homework Statement

i have worked out the answer to my question, yet it seems silly so have convinced myself its wrong. Please help, or tell me if I'm being stupid if you can??

a parallel plate capacitor Er=12, length=10mm, width=5mm, C=100pF, rho=10`14.

calculate the leakage resistance of the capacitor?

Homework Equations

R= rho( length/area) C= (ErEo x A)/ d

The Attempt at a Solution

no idea. got as far as area 5 x 10`-5
EoEr = 1.062 x 10`-10

i just can't figure out how to get back to the capacitors leakage resistance??

Could you please write out your full R calculation, including units along the way? Your equation for R is correct, but it will help a lot to see the calculation with units (so we don't have to guess or do the full calc).

 

[notsosmart]

sorry, to clarify properly;
the question is three part.
a) i am given an area of SHEET material with dimensions 10mm x 20mm with rho=10`3ohm/m. i have to calculate the thickness required to give a resistance of 5kohms.
answer; l=A(R/rho)
l =0.01 x 0.02 x (5000/10`3)
l=1mm

b) i am now given a capacitor that is formed by two parallel metal plates separated by a dielectric of Er=12 with plates dimensions of 0.01m x 0.005m. I have to calculate the thickness of the dielectric required to create a capacitor of 100pF.
answer; E=EoEr
E=(8.85x10`-12) x 12
E=1.062 x 10`-10

(Eo is given as a general value in my notebook)

d= EA/C
d= (1.062x10`-12 x (0.01 x 0.005))/100x10-5
d= 5.31x10`-5m

c)I am now asked (this is where i am struggling),to find the leakage resistanceof A non ideal capacitor having having a 'leaky' dielectric. The values given are Er=12, rho=10`14 and C=100pF.

I have asked my tutor for help, and she said to look at the formula used in the 2 previous answers, to work out the leakage resistance. I am struggling to do this. I think she also wants me to incorporate the answers worked out.

Any help greatly appreciated. I am not looking for the answer, just guidance as how to get there.

Thanks for taking the time to read.

 

[berkeman]

sorry, to clarify properly;
the question is three part.
a) i am given an area of SHEET material with dimensions 10mm x 20mm with rho=10`3ohm/m. i have to calculate the thickness required to give a resistance of 5kohms.
answer; l=A(R/rho)
l =0.01 x 0.02 x (5000/10`3)
l=1mm

b) i am now given a capacitor that is formed by two parallel metal plates separated by a dielectric of Er=12 with plates dimensions of 0.01m x 0.005m. I have to calculate the thickness of the dielectric required to create a capacitor of 100pF.
answer; E=EoEr
E=(8.85x10`-12) x 12
E=1.062 x 10`-10

(Eo is given as a general value in my notebook)

d= EA/C
d= (1.062x10`-12 x (0.01 x 0.005))/100x10-5
d= 5.31x10`-5m

c)I am now asked (this is where i am struggling),to find the leakage resistanceof A non ideal capacitor having having a 'leaky' dielectric. The values given are Er=12, rho=10`14 and C=100pF.

I have asked my tutor for help, and she said to look at the formula used in the 2 previous answers, to work out the leakage resistance. I am struggling to do this. I think she also wants me to incorporate the answers worked out.

Any help greatly appreciated. I am not looking for the answer, just guidance as how to get there.

Thanks for taking the time to read.

The leakage resistance of the cap is just defined by the volume dimensions of the dielectric block in the cap, and the resistivity of the dielectric block. You have most of those numbers, and you just need to keep your units straight in the calcs (convert to mks throughout). To show the equation a little more clearly in latex:

$$R = \rho \frac{L}{A}$$

 

[notsosmart]

Is this simply;

R= 10'14 x ((5.31x10'-5)/ 0.01x0.005)

which equals 1.062x10'14 ohms??

This is the value I calculated first time but was not convinced?? Thanks for helping