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Q.The space between the plates of a parallel plate capacitor is completely filled in two ways. In the first case, it is filled with a slab of dielectric constant KK. In the second case, it is filled with two slabs of equal thickness and dielectric constants K1K_1 and K2K_2 respectively as shown in the figure. The capacitance of the capacitor is same in the two cases. Obtain the relationship between KK, K1K_1 and K2K_2.

Figure — CBSE 2020 55/1/1 Q22
Figure
CBSECBSE Class XII Board 2020Subjective· 2mImportance★★★★★
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The key idea is that inserting two dielectrics of equal thickness is equivalent to placing two capacitors in series. Equating the capacitance of the single-slab case to the two-slab case gives the harmonic mean relation: K=2K1K2K1+K2K = \frac{2K_1K_2}{K_1+K_2}.

Why this approach works

When you fill a parallel-plate capacitor completely with a dielectric, the capacitance simply multiplies by the dielectric constant KK: C=K⋅ε0AdC = K \cdot \frac{\varepsilon_0 A}{d}. That’s straightforward.

But when you fill the gap with two different dielectrics of equal thickness, something interesting happens. Each dielectric slab occupies half the gap, so each slab acts like its own tiny capacitor — with half the plate separation. And because these two "sub-capacitors" share the same conducting plates (the original plates are at the top and bottom), the charge on the top plate must flow through both dielectrics to reach the bottom plate. That’s the hallmark of a series combination.

So the problem reduces to: find the equivalent capacitance of two capacitors in series, each with plate separation d/2d/2 and dielectric constants K1K_1 and K2K_2, then set that equal to the single-slab case.

Step-by-step solution

1. Write the capacitance for Case 1 (single slab)

For a parallel-plate capacitor of plate area AA and plate separation dd, completely filled with a dielectric of constant KK:

C1=K⋅ε0AdC_1 = K \cdot \frac{\varepsilon_0 A}{d}

This is our reference.

2. Model Case 2 as two capacitors in series

In the second case, the gap dd is split into two equal layers, each of thickness d/2d/2, stacked one on top of the other as shown in the figure — the top layer has dielectric constant K1K_1, the bottom layer K2K_2.

Figure — CBSE 2020 55/1/1 Q22
Figure — CBSE 2020 55/1/1 Q22

Each layer forms a capacitor with the same plate area AA but half the separation. So:

  • Capacitance of the top slab:

Ctop=K1⋅ε0Ad/2=2K1ε0AdC_{\text{top}} = K_1 \cdot \frac{\varepsilon_0 A}{d/2} = \frac{2K_1 \varepsilon_0 A}{d}

  • Capacitance of the bottom slab:

Cbottom=K2⋅ε0Ad/2=2K2ε0AdC_{\text{bottom}} = K_2 \cdot \frac{\varepsilon_0 A}{d/2} = \frac{2K_2 \varepsilon_0 A}{d}

Since these two capacitors share the same conducting plates (the top plate is one electrode, the bottom plate is the other, and the interface between dielectrics is an equipotential surface), they are in series.

Watch out

A common mistake is to treat the two dielectrics as being in parallel. That would happen if the dielectrics were placed side-by-side (filling different horizontal strips), not stacked vertically. Here they are stacked, so the charge must pass through both — that’s series.

3. Combine the series capacitors

For two capacitors in series, the equivalent capacitance is: …

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