In a parallel plate capacitor set up, the plate area of capacitor is 2 m^2 and the plates are separa

Q: Can I treat this as two capacitors in series?

Absolutely! You can consider it as one capacitor with dielectric $K$ and thickness $t$, and another with air (dielectric constant 1) and thickness $d-t$, connected in series. The formula $C = \frac{\varepsilon_0 A}{d - t + t/K}$ is actually derived from this exact series combination!

Q: Why do we subtract $t$ and add $t/K$ in the denominator?

The total distance is $d$. The air gap is $d-t$, which contributes to the resistance against electric field lines. The dielectric slab of thickness $t$ effectively acts like an air gap of thickness $t/K$ because the dielectric constant $K$ reduces the electric field inside it. So the effective "air equivalent" distance is $(d-t) + t/K$.

Q: What if the dielectric slab was a metal slab?

For a metal, the dielectric constant $K$ is infinity. So the term $t/K$ becomes zero. The capacitance would simply be $C = \frac{\varepsilon_0 A}{d - t}$, which means the effective distance between the plates is reduced by the thickness of the metal slab.

Q: Does the position of the dielectric slab matter?

No, as long as the slab is parallel to the plates, its exact position (left, right, or middle) doesn't change the equivalent capacitance. The series combination of the air gaps and the dielectric remains the same.

Q: I got stuck in the decimal calculation. Is there an easier way?

Yes! Converting decimals to fractions often helps. $0.5$ is $1/2$, and $3.2$ is $32/10$ or $16/5$. Using fractions can make the cancellation much smoother and reduce calculation errors.

Question

In a parallel plate capacitor set up, the plate area of capacitor is $2	ext{ m}^2$ and the plates are separated by $1	ext{ m}$. If the space between the plates are filled with a dielectric material of thickness $0.5	ext{ m}$ and area $2	ext{ m}^2$ (see figure) the capacitance of the set-up will be ...... $\varepsilon_0$.
(Dielectric constant of the material $= 3.2$)
(Round off to the nearest integer)

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a standard parallel plate capacitor. Normally, the space between the plates is just filled with air (or vacuum). But in this problem, we've spiced things up! We have introduced a dielectric slab into the gap.

However, the slab doesn't fill the entire space. The total distance between the plates is $d = 1	ext{ m}$, but our dielectric slab only has a thickness of $t = 0.5	ext{ m}$. This means half of the capacitor is filled with the dielectric (with a dielectric constant $K = 3.2$), and the other half is just air.

### The Master Equation

When a dielectric slab of thickness $t$ is placed between the plates of a capacitor (where $t < d$), the new equivalent capacitance is given by a very elegant formula:

$$C = \frac{\varepsilon_0 A}{d - t + \frac{t}{K}}$$

Why does this formula look the way it does? You can actually think of this setup as **two capacitors connected in series**:
1. One capacitor filled with the dielectric of thickness $t$.
2. Another capacitor filled with air of thickness $d - t$.

If you calculate their individual capacitances ($C_1 = \frac{K\varepsilon_0 A}{t}$ and $C_2 = \frac{\varepsilon_0 A}{d-t}$) and use the series combination formula $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}$, you will arrive at this exact same master equation!

### Executing the Calculation

Let's substitute the given values into our master equation. We know:
- Area, $A = 2	ext{ m}^2$
- Total separation, $d = 1	ext{ m}$
- Dielectric thickness, $t = 0.5	ext{ m}$
- Dielectric constant, $K = 3.2$

Plugging these in, we get:

$$C = \frac{\varepsilon_0 (2)}{1 - 0.5 + \frac{0.5}{3.2}}$$

Now, let's simplify the denominator. First, $1 - 0.5 = 0.5$.

$$C = \frac{2\varepsilon_0}{0.5 + \frac{0.5}{3.2}}$$

To make the math smoother, let's factor out the $0.5$ from the denominator:

$$C = \frac{2\varepsilon_0}{0.5 \left(1 + \frac{1}{3.2}ight)}$$

Since $0.5$ is just $\frac{1}{2}$, dividing by $0.5$ is the same as multiplying the numerator by $2$. This gives us $4\varepsilon_0$ on top. Inside the bracket, we can simplify the fraction:

$$C = \frac{4\varepsilon_0}{1 + \frac{1}{3.2}} = \frac{4\varepsilon_0}{\frac{3.2 + 1}{3.2}} = \frac{4\varepsilon_0}{\frac{4.2}{3.2}}$$

Flipping the fraction in the denominator, we get:

$$C = \frac{4 	imes 3.2}{4.2} \varepsilon_0$$

### Final Conclusion

Now it's just basic arithmetic.

$$C = \frac{12.8}{4.2} \varepsilon_0 \approx 3.047 \varepsilon_0$$

The question asks us to find the coefficient of $\varepsilon_0$ and round it off to the **nearest integer**.

Looking at $3.047$, the nearest integer is clearly $3$.

**Final Answer: 3**

Enter Numerical Value:

Visualized Solution

\text{Visualizing the Setup}

\text{Formula for Partially Filled Capacitor}

\text{Substituting Given Values}

\text{Simplifying the Denominator}

\text{Further Simplification}

\text{Final Calculation}

\text{Rounding Off}

\text{Alternative Approach: Series Combination}

The Sigma Insight: Capacitance and Capacitors

Analyzing the Setup

The Master Equation

Executing the Calculation

Final Conclusion

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