# Homework Help: Capacitor "resistance"?

1. Jun 18, 2014

### Razvan

1. The problem statement, all variables and given/known data
I was told that I can measure the "resistance" of a capacitor using 3 methods:
1. "The direct method"
$$R = \int_1^2 \frac{\rho}{A} ds$$

2. "Ohm's Law"

$$R=\frac{U}{I}$$
$$U=\int_1^2 E ds$$
$$E=\rho J = \frac{\rho i}{A}$$

3. "The analogy between electrostatics and electrokinetics"
$\epsilon$ is replaced by $\sigma$
and
$C$ is replaced by $G$

2. Relevant equations

What does this really mean, the resistance of the capacitor?
Why is $E=\rho J$?
What is this analogy between electrostatics and electrokinetics?

3. The attempt at a solution

Last edited: Jun 18, 2014
2. Jun 18, 2014

### Simon Bridge

It means exactly what it says... probably the easiest way to view it is to look at the Ohm's law approach.

An ideal capacitor has infinite DC resistance ... when an ideal DC voltage source is connected via ideal conducting wires, the charge flows from one plate to the other in zero time giving rise to an infinite current.

This does not happen in real life though.
IRL it takes a finite amount of time for the charges to transfer ...

You should be able to derive that equation ... do you know what the letters stand for?

The analogy in question is just a handy way of getting electrokinetic equations from the static ones.
Most people can remember the electrostatics.

3. Jun 27, 2014

### rude man

There is series resistance Rs and there is parallel reistance Rp. The former is usually << 1 ohm, the latter >> 1 megohm for say a C = 0.1 uF ceramic capacitor. So we can say that if we apply a voltage V across C then I = V/Rp after a very short time of say 10 Rs*C time constants, maybe 0.1 μs.

The electric field inside the capcitive part of C depends on how it's fabricated. In all cases, ∫Edl = V = Q/C with Q the charge and C the capacitance. Inside Rp we have E = i/σ where i is current density and σ = conductivity of the resistive material for Rp.