A parallel plate capacitor has plate area 100 m^2 and plate separation of 10 m. The space between th

Q: Why does the effective distance decrease when a dielectric is inserted?

The dielectric material polarizes in the presence of the external electric field, creating an opposing internal field. This reduces the net electric field inside the dielectric, which is mathematically equivalent to reducing the physical distance between the plates.

Q: What does the term $t/K$ represent in the formula?

The term $t/K$ represents the equivalent air thickness of the dielectric slab. Since the electric field inside the dielectric is reduced by a factor of $K$, a thickness $t$ of dielectric has the same potential drop as a thickness $t/K$ of air.

Q: Can we solve this by treating it as two capacitors in series?

Absolutely! You can model this system as an air-filled capacitor of thickness $d-t$ in series with a dielectric-filled capacitor of thickness $t$. Calculating their equivalent capacitance will yield the exact same formula.

Q: Why do we multiply by $10^{12}$ to get the answer in picoFarads?

The standard SI unit of capacitance is the Farad (F). One picoFarad (pF) is equal to $10^{-12}$ Farads. Therefore, to convert from Farads to picoFarads, we divide the value by $10^{-12}$, which is equivalent to multiplying by $10^{12}$.

Q: What would happen if the dielectric completely filled the space between the plates?

If the dielectric completely filled the space, then $t = d$. Substituting this into our formula gives $C = \frac{\varepsilon_0 A}{d - d + d/K} = \frac{K \varepsilon_0 A}{d}$, which is the standard formula for a fully filled capacitor.

Question

A parallel plate capacitor has plate area $100	ext{ m}^2$ and plate separation of 10 m. The space between the plates is filled upto a thickness 5 m with a material of dielectric constant of 10. The resultant capacitance of the system is $x	ext{ pF}$. The value of $\varepsilon_0 = 8.85 	imes 10^{-12}	ext{ F m}^{-1}$. The value of $x$ to the nearest integer is …… .

Sigma Percentile · Accepted Answer

### Visualizing the Setup

Imagine a giant parallel plate capacitor. The plates have a massive area of $100	ext{ m}^2$ and are separated by a distance of $10	ext{ m}$. Now, we slide a dielectric slab of thickness $5	ext{ m}$ and dielectric constant $10$ between them.

This physical insertion fundamentally alters the electric field inside the capacitor. The dielectric material polarizes, creating an internal electric field that opposes the external one, effectively reducing the overall potential difference for a given charge.

### The Master Equation

To find the new capacitance, we use the standard formula for a partially filled capacitor. The effective distance between the plates decreases because the dielectric reduces the electric field inside it.

The master formula is:

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

This elegant equation perfectly captures how the physical distance $d$ is modified by the presence of the dielectric slab of thickness $t$ and constant $K$.

### Substituting the Values

Let's carefully substitute the given values into our master equation. We know $\varepsilon_0 = 8.85 	imes 10^{-12}	ext{ F m}^{-1}$, $A = 100	ext{ m}^2$, $d = 10	ext{ m}$, $t = 5	ext{ m}$, and $K = 10$.

Substituting these gives us the raw structure:

$$C = \frac{8.85 	imes 10^{-12} 	imes 100}{10 - 5 + \frac{5}{10}}$$

Don't rush through this; let the raw structure sink in before we start crunching the numbers.

### Simplifying the Effective Distance

Now, let's simplify the denominator, which represents the **effective distance** between the plates.

We calculate $10 - 5 = 5$, and the dielectric term $\frac{5}{10} = 0.5$. Adding them together gives us an effective distance of $5.5	ext{ m}$.

Notice how the $10	ext{ m}$ physical gap behaves electrically like a $5.5	ext{ m}$ gap because of the dielectric!

### Final Calculation

Moving to the numerator, multiplying by $100$ changes $10^{-12}$ to $10^{-10}$. So we have:

$$C = \frac{8.85 	imes 10^{-10}}{5.5}	ext{ F}$$

Dividing $8.85$ by $5.5$ gives approximately $1.609$. So the capacitance is $1.609 	imes 10^{-10}	ext{ F}$.

To convert this to picoFarads ($	ext{pF}$), we shift the decimal two places to the right, giving $160.9	ext{ pF}$. Rounding to the nearest integer, we get our final answer: **$161	ext{ pF}$**.

### The Way Forward

We've found our answer, but let's think a bit deeper. What if the slab we inserted wasn't a dielectric, but a perfect conductor?

For a metal, $K 	o \infty$, making the $\frac{t}{K}$ term zero. The effective distance would simply become $d - t$. How would that change the final capacitance? This is a classic JEE concept, so keep it in your mental toolkit!

Enter Numerical Value:

Visualized Solution

Visualizing the Capacitor

Capacitance Formula

Substituting Values

Simplifying the Denominator

Calculating Capacitance

Final Answer in pF

The Way Forward

The Sigma Insight: Capacitance and Capacitors

Visualizing the Setup

The Master Equation

Substituting the Values

Simplifying the Effective Distance

Final Calculation

The Way Forward

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