In the reported figure, a capacitor is formed by placing a compound dielectric between the plates of

Q: Why are the capacitors in series and not parallel?

Because the dielectrics are stacked one after another along the distance between the plates. The same charge would induce across each boundary, which is the characteristic of a series combination.

Q: What if the dielectrics were placed parallel to the plates, dividing the area?

If the dielectrics divided the area instead of the distance, they would act as capacitors in parallel, and you would use the formula $C_{eq} = C_1 + C_2 + C_3$.

Q: Why is the area $A$ the same for all three capacitors?

The dielectrics span the entire area of the plates. They only divide the distance $d$ between the plates, so the cross-sectional area $A$ remains constant for each section.

Q: Can I just add the dielectric constants to find an equivalent $K$?

No, you cannot simply add them. The equivalent dielectric constant depends on whether they are in series or parallel. For series, $\frac{d_{total}}{K_{eq}} = \frac{d_1}{K_1} + \frac{d_2}{K_2} + \dots$

Q: What happens if one of the slabs is a metal?

If one slab is a metal, its dielectric constant $K$ is effectively infinity. This means its capacitance becomes infinite, and it acts as a simple conducting wire, effectively reducing the total distance between the plates.

Question

In the reported figure, a capacitor is formed by placing a compound dielectric between the plates of parallel plate capacitor. The expression for the capacity of the said capacitor will be (Take, area of plate = $A$)

Options:

(A) $\frac{15}{34} \frac{K \varepsilon_0 A}{d}$

(B) $\frac{15}{6} \frac{K \varepsilon_0 A}{d}$

(C) $\frac{25}{6} \frac{K \varepsilon_0 A}{d}$

(D) $\frac{9}{6} \frac{K \varepsilon_0 A}{d}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are looking at a parallel plate capacitor, but instead of just air or a single material between the plates, there are three distinct blocks of dielectric materials stacked one after another.

Because these slabs are placed sequentially along the distance between the plates, the electric field lines must pass through each of them in turn. This physical arrangement means that the same charge would be induced across each boundary. Therefore, this setup behaves exactly like three separate capacitors connected in **series**.

### The Master Equation

To solve this, we need two fundamental tools. First, the capacitance of any parallel plate capacitor filled with a dielectric is given by:

$C = \frac{K \varepsilon_0 A}{t}$

where $K$ is the dielectric constant, $A$ is the area of the plates, and $t$ is the thickness of the dielectric slab.

Second, for capacitors connected in series, the equivalent capacitance $C_{eq}$ is found using the reciprocal sum formula:

$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$

### Calculating Individual Capacitances

Let's break down the compound capacitor into its three individual components. Notice that the cross-sectional area $A$ remains the same for all three sections because the slabs span the entire area of the plates.

For the first section, the thickness is $d$ and the dielectric constant is $K$:
$C_1 = \frac{K \varepsilon_0 A}{d}$

For the second section, the thickness is $2d$ and the dielectric constant is $3K$:
$C_2 = \frac{3K \varepsilon_0 A}{2d}$

For the third section, the thickness is $3d$ and the dielectric constant is $5K$:
$C_3 = \frac{5K \varepsilon_0 A}{3d}$

### Final Calculation

Now, we substitute these individual capacitances into our series combination formula:

$\frac{1}{C_{eq}} = \frac{d}{K \varepsilon_0 A} + \frac{2d}{3K \varepsilon_0 A} + \frac{3d}{5K \varepsilon_0 A}$

To make the math cleaner, let's factor out the common term $\frac{d}{K \varepsilon_0 A}$:

$\frac{1}{C_{eq}} = \frac{d}{K \varepsilon_0 A} \left( 1 + \frac{2}{3} + \frac{3}{5} \right)$

Next, we simply add the fractions inside the parenthesis. The least common multiple (LCM) of the denominators $3$ and $5$ is $15$:

$\frac{1}{C_{eq}} = \frac{d}{K \varepsilon_0 A} \left( \frac{15 + 10 + 9}{15} \right)$

$\frac{1}{C_{eq}} = \frac{34d}{15 K \varepsilon_0 A}$

Finally, to find the equivalent capacitance $C_{eq}$, we take the reciprocal of both sides:

**$C_{eq} = \frac{15 K \varepsilon_0 A}{34 d}$**

And there we have it! A seemingly complex compound dielectric problem beautifully unravels into a simple exercise in series combinations.

Animated Solution for Physics - Electrostatics: In the reported figure, a capacitor is formed by placing a compound dielectric between the plates of parallel plate capacitor. The expression for the capacity of the said capacitor will be (Take, area of plate = $A$ )

Select Answer:

Visualized Solution

Visualizing the Capacitor Arrangement

Formulas for Capacitance

Calculating Individual Capacitances

Applying the Series Formula

Simplifying the Expression

Adding the Fractions

Final Equivalent Capacitance

The Way Forward

The Sigma Insight: Capacitance and Capacitors

Analyzing the Setup

The Master Equation

Calculating Individual Capacitances

Final Calculation

Similar Questions

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If $q_{f}$ is the free charge on the capacitor plates and $q_{b}$ is the bound charge on the dielectric slab of dielectric constant $K$ placed between the capacitor plates, then bound charge $q_{b}$ can be expressed as

Select Answer:

Visualized Solution

Visualizing the Capacitor Arrangement

Formulas for Capacitance

Calculating Individual Capacitances

Applying the Series Formula

Simplifying the Expression

Adding the Fractions

Final Equivalent Capacitance

The Way Forward

The Sigma Insight: Capacitance and Capacitors

Analyzing the Setup

The Master Equation

Calculating Individual Capacitances

Final Calculation

Similar Questions

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A capacitor is made of two square plates each of side $a$ making a very small angle $α$ between them, as shown in figure. The capacitance will be close to