A leaky parallel plate capacitor is filled completely with a material having dielectric constant K =

Q: Why do we model a leaky capacitor as a parallel RC circuit?

A real dielectric is not a perfect insulator. It allows some charge to flow through it, which is the definition of resistance. Since the same material block stores charge (capacitance) and conducts charge (resistance) across the same potential difference, they are effectively in parallel.

Q: Why did the area and distance cancel out in the time constant?

The capacitance is directly proportional to the area and inversely proportional to the distance, while the resistance is inversely proportional to the area and directly proportional to the distance. When multiplied, these geometric factors perfectly cancel out, leaving the time constant dependent only on the intrinsic properties of the material ($K$ and $\sigma$).

Q: Why is the leakage current the derivative of charge?

Current is defined as the rate of flow of charge. In a discharging capacitor, the charge $q$ on the plates is decreasing, so the rate of change $dq/dt$ is negative. The leakage current $i$ flowing through the dielectric is the magnitude of this rate of decrease, hence $i = -dq/dt$.

Q: What does the time constant physically represent?

The time constant $\tau$ represents the time it takes for the charge (or voltage) to drop to $1/e$ (about 37%) of its initial value. A larger time constant means the capacitor leaks charge more slowly.

Q: Can we use this model for any real-world capacitor?

Yes! All real-world capacitors have some leakage current, though for high-quality capacitors, the resistance is so high that the time constant is measured in days or even years. This parallel RC model is the standard way to represent real capacitors in circuit analysis.

Question

A leaky parallel plate capacitor is filled completely with a material having dielectric constant $K = 5$ and electrical conductivity $\sigma = 7.4 	imes 10^{-12} \Omega^{-1} 	ext{ m}^{-1}$. If the charge on the capacitor at instant $t = 0$ is $q = 8.85 \mu	ext{C}$, then calculate the leakage current at the instant $t = 12 	ext{ s}$.

Sigma Percentile · Accepted Answer

### The Leaky Capacitor Model

When we think of a capacitor, we usually imagine two conducting plates separated by a perfect insulator. But in the real world, no insulator is perfect.

The dielectric material between the plates has some small electrical conductivity, $\sigma$. This means it allows a tiny amount of charge to leak from one plate to the other.

To analyze this mathematically, we can model this "leaky" capacitor as an ideal capacitor $C$ connected in parallel with an ideal resistor $R$. The capacitor stores the charge, and the resistor provides the path for the leakage current.

### Finding the Time Constant

For any $RC$ circuit, the rate at which it charges or discharges is governed by its time constant, $	au_c = CR$.

Let's write down the expressions for both components. The capacitance is given by $C = \frac{K \varepsilon_0 A}{d}$, and the resistance of the dielectric block is $R = \frac{d}{\sigma A}$.

When we multiply them to find the time constant, something beautiful happens:
$$ 	au_c = \left(\frac{K \varepsilon_0 A}{d}ight) \left(\frac{d}{\sigma A}ight) = \frac{K \varepsilon_0}{\sigma} $$

Notice how the geometric factors—area $A$ and distance $d$—perfectly cancel out! The time constant depends **only** on the intrinsic properties of the material.

### Calculating the Leakage Current

Now, we can substitute the given values into our time constant formula:
$$ 	au_c = \frac{5 	imes 8.85 	imes 10^{-12}}{7.4 	imes 10^{-12}} \approx 5.98 	ext{ s} $$

Because the capacitor is discharging through its own internal resistance, the charge on the plates decays exponentially according to the equation $q(t) = q_0 e^{-t/	au_c}$.

The leakage current $i(t)$ is simply the rate at which this charge is decreasing. By differentiating the charge equation with respect to time, we get:
$$ i(t) = -\frac{dq}{dt} = \frac{q_0}{	au_c} e^{-t/	au_c} $$

Finally, we substitute $t = 12 	ext{ s}$, $q_0 = 8.85 \mu	ext{C}$, and $	au_c = 5.98 	ext{ s}$ to find the current at that specific instant:
$$ i(12) = \frac{8.85 	imes 10^{-6}}{5.98} e^{-12/5.98} $$

Evaluating this gives us $i(12) \approx 0.198 	imes 10^{-6} 	ext{ A}$, which is exactly **$0.198 \mu	ext{A}$**.

This is a classic problem that beautifully bridges the concepts of electrostatics and current electricity!

Enter Numerical Value:

Visualized Solution

Leaky Capacitor Model

Capacitance and Resistance

Time Constant $τ_{c}$

Substituting Values

Calculating $τ_{c}$

Discharging Current

Substituting for Current

Final Leakage Current

The Sigma Insight: RC Circuit

The Leaky Capacitor Model

Finding the Time Constant

Calculating the Leakage Current

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