A parallel plate capacitor of capacitance C has spacing d between two plates having area A. The regi

Q: Why are the layers considered to be in series and not parallel?

Because the layers are stacked one after another between the plates. The electric field lines pass through each layer sequentially, and the total potential difference is the sum of the potential drops across each individual layer. This is the exact definition of a series combination.

Q: How did the discrete ratio $m/N$ magically become $x/d$?

It's not magic, it's geometry! The total distance is $d$, divided into $N$ equal layers, so each layer is $d/N$ thick. The $m^{\text{th}}$ layer is located at a distance $x = m \times (d/N)$ from the start. Rearranging this simple equation gives $m/N = x/d$.

Q: Why do we integrate $1/dC$ instead of just integrating $dC$?

Remember the formula for capacitors in series: $1/C_{eq} = 1/C_1 + 1/C_2 + \dots$ In the continuous limit, this discrete summation of reciprocals turns into an integral of the reciprocal: $\int (1/dC)$. If they were in parallel, we would integrate $dC$ directly.

Q: I messed up the integration. Where did the extra '$d$' in the numerator come from?

When you integrate $\int \frac{1}{1 + ax} dx$, the result is $\frac{1}{a} \ln(1 + ax)$. Here, your '$a$' is $1/d$. So, dividing by $1/d$ brings a $d$ up into the numerator. It's a classic chain rule reversal!

Q: What if $N$ was a small number, like $N=2$?

If $N$ is small, you cannot use integration because the layers are discrete blocks, not infinitesimally thin slices. You would simply calculate $C_1$ and $C_2$ using the given formula and then use the standard series formula $1/C_{eq} = 1/C_1 + 1/C_2$.

Question

A parallel plate capacitor of capacitance C has spacing d between two plates having area A. The region between the plates is filled with N dielectric layers, parallel to its plates, each with thickness $\delta = \frac{d}{N}$. The dielectric constant of the $m^{	ext{th}}$ layer is $K_m = K\left(1 + \frac{m}{N}ight)$. For a very large N ($> 10^3$), the capacitance C is $\alpha \left(\frac{K \epsilon_0 A}{d \ln 2}ight)$. The value of $\alpha$ will be _____.

Sigma Percentile · Accepted Answer

This problem is a beautiful intersection of electrostatics and calculus. It takes the familiar concept of capacitors in series and pushes it to the continuous limit, transforming a discrete summation into an elegant integral.

### The Transition from Discrete to Continuous

Imagine a capacitor filled with thousands of incredibly thin dielectric layers. The problem states that we have $N$ layers, and $N$ is very large ($> 10^3$). As the number of layers approaches infinity, we can no longer treat them as discrete blocks. Instead, the dielectric constant becomes a continuous function of the position $x$ between the plates.

Let's find the position $x$ of the $m^{	ext{th}}$ layer. Since each layer has a thickness of $\delta = \frac{d}{N}$, the distance $x$ from the left plate is simply $m$ times $\frac{d}{N}$.

$$x = m \left(\frac{d}{N}ight)$$

Rearranging this, we get the ratio $\frac{m}{N} = \frac{x}{d}$. This is our crucial substitution that bridges the discrete world with the continuous one.

### Slicing the Capacitor

Now, look at the given formula for the dielectric constant of the $m^{	ext{th}}$ layer: $K_m = K\left(1 + \frac{m}{N}ight)$. By substituting $\frac{m}{N}$ with $\frac{x}{d}$, we transform the discrete $K_m$ into a continuous function $K(x)$.

$$K(x) = K\left(1 + \frac{x}{d}ight)$$

This tells us exactly how the dielectric constant increases linearly as we move across the plates. Let's isolate one of these infinitesimally thin layers of thickness $dx$. It acts as a tiny capacitor on its own. Using the standard formula for a parallel plate capacitor, its capacitance $dC$ is:

$$dC = \frac{K(x) \epsilon_0 A}{dx}$$

### The Series Combination Integral

Since these layers are stacked one after another, they are in a **series combination**. For capacitors in series, we add their reciprocals. So, the reciprocal of the equivalent capacitance is the integral of $\frac{1}{dC}$, evaluated from $x = 0$ to $x = d$.

$$\frac{1}{C_{eq}} = \int_0^d \frac{1}{dC} = \int_0^d \frac{dx}{K\left(1 + \frac{x}{d}ight) \epsilon_0 A}$$

### Conquering the Logarithmic Integral

Now for the calculus. We can pull the constants $K$, $\epsilon_0$, and $A$ outside the integral. We are left with integrating $dx$ over $1 + \frac{x}{d}$. This is a standard logarithmic integral, but we must watch out for the coefficient of $x$, which is $\frac{1}{d}$.

$$\frac{1}{C_{eq}} = \frac{1}{K \epsilon_0 A} \int_0^d \frac{dx}{1 + \frac{x}{d}}$$

Integrating gives us the natural log of $1 + \frac{x}{d}$, but we must multiply by $d$ to account for the chain rule in reverse.

$$\frac{1}{C_{eq}} = \frac{d}{K \epsilon_0 A} \left[ \ln\left(1 + \frac{x}{d}ight) ight]_0^d$$

Now, we apply the upper limit $d$ and the lower limit $0$. Substituting $d$ gives $\ln(2)$, and substituting $0$ gives $\ln(1)$, which is just $0$.

$$\frac{1}{C_{eq}} = \frac{d}{K \epsilon_0 A} [\ln(2) - 0] = \frac{d \ln 2}{K \epsilon_0 A}$$

Flipping both sides, we get the final equivalent capacitance:

$$C_{eq} = \frac{K \epsilon_0 A}{d \ln 2}$$

Comparing this with the expression given in the question, $\alpha \left(\frac{K \epsilon_0 A}{d \ln 2}ight)$, it is crystal clear that **$\alpha = 1$**.

Enter Numerical Value:

Visualized Solution

\text{Continuous Limit of Dielectric Layers}

\text{Variable Transformation}

\text{Dielectric Constant Function}

\text{Capacitance of Elemental Layer}

\text{Series Combination Setup}

\text{Integration}

\text{Evaluating the Integral}

\text{Final Equivalent Capacitance}

\text{The Way Forward}

The Sigma Insight: Capacitance and Capacitors

The Transition from Discrete to Continuous

Slicing the Capacitor

The Series Combination Integral

Conquering the Logarithmic Integral

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