Capacitance and Resistance of a fluid

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Homework Help Overview

The problem involves a parallel plate capacitor with a specified capacitance, immersed in a fluid with a known resistivity. The objective is to calculate the resistance between the capacitor plates based on these parameters.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to relate the capacitance and resistance using the equations for capacitance and resistance, questioning the correctness of their calculations.
  • Some participants question the interpretation of the capacitance unit, prompting a correction regarding the value of pico Farads.
  • There is a progression in the calculations as the original poster revises their resistance value based on the corrected capacitance.

Discussion Status

The discussion has evolved with participants providing feedback on the calculations. The original poster has adjusted their values based on input, leading to a revised resistance calculation that appears to be accepted by others.

Contextual Notes

There was a misunderstanding regarding the unit conversion for capacitance, which was clarified during the discussion. The original poster's calculations were based on initial assumptions that were later corrected.

TFM
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Homework Statement



A parallel plate air-spaced capacitor has a capacitance of 100pF. It is immersed in a fluid of resistivity 10 m. Calculate the resistance between the plates.

Homework Equations



[tex]C = \frac{\epsilon_0A}{d}[/tex]

[tex]R = \frac{\rho l}{A}[/tex]

The Attempt at a Solution



I seemed to get this question too easily...

First of all, I said that for the fluid, the area is the area of the pates, and l is the distance d between them, then I rearranged the resistivity equation into terms of a/d:

[tex]\frac{A}{d} = \frac{\rho}{R}[/tex]

then I substituted this equation into the capacitance equation:

[tex]C = \epsilon_0\frac{\rho}{R}[/tex]

rearrange for R:

[tex]R = \epsilon_0 \frac{\rho}{C}[/tex]

Insert values:

[tex]R = \epsilon_0 \frac{10}{100*10^{-6}}[/tex]

this gives a value of [tex]8.85 * 10^{-7}[/tex] Ohms

Does this look correct?

TFM
 
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You're almost there. What does the "p" in pF mean? Hint: it's not 10-6.
 
pF is a pico Farad, I believe

Iyt is 10^-12, not -6, that was illy of me - I should have remebered that 10^-6 is micro:redface:

TFM
 
Putting in my new value for C, I get the resistance to be:

0.885 Ohms

Does this look right now?

TFM
 
Yes, that looks good.
 
Excellent

Thanks for your Assistance, Redbelly98, Most Appreciated :smile:

TFM
 

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